Binomial Distribution Word Problems

Solution of exercise 1, solution of exercise 2, solution of exercise 3, solution of exercise 4, solution of exercise 5, solution of exercise 6, solution of exercise 7.

Paolo

A coin is tossed four times. Calculate the probability of obtaining more heads than tails.

An agent sells life insurance policies to five equally aged, healthy people. According to recent data, the probability of a person living in these conditions for 30 years or more is 2/3. Calculate the probability that after 30 years:

1. All five people are still living.

2. At least three people are still living.

3. Exactly two people are still living.

If from six to seven in the evening one telephone line in every five is engaged in a conversation: what is the probability that when 10 telephone numbers are chosen at random, only two are in use?

The probability of a man hitting the target at a shooting range is 1/4. If he shoots 10 times, what is the probability that he hits the target exactly three times? What is the probability that he hits the target at least once?

There are 10 red and 20 blue balls in a box. A ball is chosen at random and it is noted whether it is red. The process repeats, returning the ball 10 times. Calculate the expected value and the standard deviation of this game.

It has been determined that 5% of drivers checked at a road stop show traces of alcohol and 10% of drivers checked do not wear seat belts. In addition, it has been observed that the two infractions are independent from one another. If an officer stops five drivers at random:

1. Calculate the probability that exactly three of the drivers have committed any one of the two offenses.

2. Calculate the probability that at least one of the drivers checked has committed at least one of the two offenses.

A pharmaceutical lab states that a drug causes negative side effects in 3 of every 100 patients. To confirm this affirmation, another laboratory chooses 5 people at random who have consumed the drug. What is the probability of the following events?

1. None of the five patients experience side effects.

2. At least two experience side effects.

3. What is the average number of patients that the laboratory should expect to experience side effects if they choose 100 patients at random?

how to solve binomial probability word problems

B(10, 1/5) p = 1/5 1 − p = 4/5

how to solve binomial probability word problems

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In question 3 : i think you have missed 9 . it should be (10*9)/2 (1/5)^2(4/5)^8

Can you give me 5 real-life problems involving random variables? i need it now please

A) 0.2668 B) 0.33965 C) 0,04575 D) 0.97175

Hello please Can you please help me with question 6? Thank you

p (AUB) = 0.05 + ( (1 – 0.05 ) x 0.1) = 0.145

in 6th question a part

Pls explain question six 2

For ex.2 part.3 there’s a rounding error – the answer is 0.165

Please help me work these cumulative binomial probability. In a particular strain of staphylococcus product abdominal cramps in 30% of person infected. At a clinic 10 persons ate contaminated food and were infected with the organisms, find: A) exactly three people will develop the symptoms. B) between 3 and 7 inclusively will develop the symptoms. C) more than 5 people will develop the symptoms. D) at least one person will develop the symptoms.

Binomial Probabilities Examples and Questions

In a binomial experiment, you have a number \( n \) of independent trials and each trial has two possible outcomes or several outcomes that may be reduced to two outcomes. The properties of a binomial experiment are: 1) The number of trials \( n \) is constant. 2) Each trial has 2 outcomes (or that can be reduced to 2 outcomes) only: "success" or "failure" , "true" or "false", "head" or "tail", ... 3) The probability \( p \) of a success in each trial must be constant. 4) The outcomes of the trials must be independent of each other. Examples of binomial experiments 1) Toss a coin \( n = 10 \) times and get \( k = 6 \) heads (success) and \( n - k \) tails (failure). 2) Roll a die \( n = 5\) times and get \( 3 \) "6" (success) and \( n - k \) "no 6" (failure). 3) Out of \( n = 10 \) tools, where each tool has a probability \( p \) of being "in good working order" (success), select 6 at random and get 4 "in good working order" and 2 "not in working order" (failure). 4) A newly developed drug has probability \( p \) of being effective. Select \( n \) people who took the drug and get \( k \) "successful treatment" (success) and \( n - k \) "not successful treatment" (failure).

Binomial Formula Explanations

The best way to explain the formula for the binomial distribution is to solve the following example.

Mean and Standard Deviation of a Binomial Distribution

Examples on the use of the binomial formula.

Example 2 A fair coin is tossed 5 times. What is the probability that exactly 3 heads are obtained? Solution to Example 2 The coin is tossed 5 times, hence the number of trials is \( n = 5\). The coin being a fair one, the outcome of a head in one toss has a probability \( p = 0.5 \) and an outcome of a tail in one toss has a probability \( 1 - p = 0.5 \) The probability of having 3 heads in 5 trials is given by the formula for binomial probabilities above with \( n = 5 \), \( k = 3 \) and \( p = 0.5\) \( \displaystyle P(3 \; \text{heads in 5 trials}) = {5\choose 3} (0.5)^3 (1-0.5)^{5-3} \\ = \displaystyle {5\choose 3} (0.5)^3 (0.5)^{2} \) Use formula for combinations to calculate \( \displaystyle {5\choose 3} = \dfrac{5!}{3!(5-3)!} = \dfrac{1 \times 2 \times 3 \times 4 \times 5}{(1 \times 2 \times 3)(1 \times 2)} = 10 \) Substitute \( P(3 \; \text{heads in 5 trials}) = 10 (0.5)^3 (0.5)^{2} = 0.3125 \)

Example 3 A fair die is rolled 7 times, find the probability of getting "\( 6 \) dots" exactly 5 times. Solution to Example 3 This is an example where although the outcomes are more than 2, we interested in only 2: "6" or "no 6". The die is rolled 7 times, hence the number of trials is \( n = 7\). In a single trial, the outcome of a "6" has probability \( p = 1/6 \) and an outcome of "no 6" has a probability \( 1 - p = 1 - 1/6 = 5/6 \) The probability of having 5 "6" in 7 trials is given by the formula for binomial probabilities above with \( n = 7 \), \( k = 5 \) and \( p = 1/6\) \( \displaystyle P(5 \; \text{5 "6" in 7 trials}) = \displaystyle {7\choose 5} (1/6)^5 (1-5/6)^{7-5} \\ = \displaystyle {7\choose 5} (1/6)^5 (5/6)^{2} \) Use formula for combinations to calculate \( \displaystyle {7\choose 5} = \dfrac{7!}{5!(7-5)!} = 21 \) Substitute \( P(5 \; \text{5 "6" in 7 trials}) = 21 (1/6)^5 (5/6)^{2} = 0.00187 \)

Example 4 A factory produces tools of which 98% are in good working order. Samples of 1000 tools are selected at random and tested. a) Find the mean and give it a practical interpretation. b) Find the standard deviation of the number of tools in good working order in these samples. Solution to Example 4 When a tool is selected, it is either in good working order with a probability of 0.98 or not in working order with a probability of 1 - 0.98 = 0.02. When selecting a sample of 1000 tools at random, 1000 may be considered as the number of trials in a binomial experiment and therefore we are dealing with a binomial probability problem. a) mean: \( \mu = n p = 1000 \times 0.98 = 980 \) In a sample of 1000 tools, we would expect that 980 tools are in good working order . b) standard deviation: \( \sigma = \sqrt{ n \times p \times (1-p)} = \sqrt{ 1000 \times 0.98 \times (1-0.98)} = 4.43\)

Example 5 Find the probability that at least 5 heads show up when a fair coin is tossed 7 times. Solution to Example 5 The number of trials is \( n = 7\). The coin being a fair one, the outcome of a head in one toss has a probability \( p = 0.5 \). Obtaining at least 5 heads; is equivalent to showing : 5, 6 or 7 heads and therefore the probability of showing at least 5 heads is given by \( P( \text{at least 5}) = P(\text{5 or 6 or 7}) \) Using the addition rule with outcomes mutually exclusive , we have \( P( \text{at least 5 heads}) = P(5) + P(6) + P(7) \) where \( P(5) \) , \( P(6) \) and \( P(7) \) are given by the formula for binomial probabilities with same number of trial \( n \), same probability \( p \) but different values of \( k \). \( \displaystyle P( \text{at least 5 heads} ) = {7\choose 5} (0.5)^5 (1-0.5)^{7-5} + {7\choose 6} (0.5)^6 (1-0.5)^{7-6} + {7\choose 7} (0.5)^7 (1-0.5)^{7-7} \\ = 0.16406 + 0.05469 + 0.00781 = 0.22656 \)

Example 6 A multiple choice test has 20 questions. Each question has four possible answers with one correct answer per question. What is the probability that a student will answer 10 or more questions correct (to pass) by guessing randomly? NOTE: this questions is very similar to question 5 above, but here we use binomial probabilities in a real life situation that most students are familiar with. Solution to Example 6 Each questions has 4 possible answers with only one correct. If a question is answered by guessing randomly, the probability of answering it correctly is \( p = 1/4 = 0.25 \). When an answer is selected randomly, it is either answered correctly with a probability of 0.25 or incorrectly with a probability of \( 1 - p = 0.75 \). This can be classified as a binomial probability experiment. The probability that a student will answer 10 questions or more (out of 20) correct by guessing randomly is given by \( P(\text{answer at least 10 questions correct}) = P(\text{10 or 11 or 12 or 13 or 14 or 15 or 16 or 17 or 18 or 19 or 20}) \) Using the addition rule, we write \( P(\text{answer at least 10 questions correct}) = P(10) + P(11) + .... + P(20) \) \( = \displaystyle {20\choose 10} \cdot 0.25^10 \cdot 0.75^{20-10} + {20\choose 11} \cdot 0.25^11 \cdot 0.75^{20-11} +.... + {20\choose 20} \cdot 0.25^20 \cdot 0.75^{20-20} \) \( = 0.00992 + 0.00301 + 0.00075 + 0.00015 + 0.00003 + 0 + 0 + 0 + 0 + 0 + 0 = 0.01386 \) Note 1) The last five probabilities are not exactly equal to 0 but negligible compared to the first 5 values. 2) According to the concept of probabilities, passing a test by guessing answers randomly does not work.

Example 7 A box contains 3 red balls, 4 white balls and 3 black balls. 6 times, a ball is selected at random, the color noted and then replaced in the box. What is the probability that the red color shows at least twice? Solution to Example 7 The event "the red color shows at least twice" is the complement of the event "the red color shows once or does not show"; hence using the complement probability formula, we write P("the red color shows at least twice") = 1 - P("the red color shows at most 1") = 1 - P("the red color shows once" or "the red color does not show") Using the addition rule P("the red color shows at least twice") = 1 - P("the red color shows once") + P("the red color does not show") Although there are more than two outcomes (3 different colors) we are interested in the red color only. The total number of balls is 10 and there are 3 red, hence each time a ball is selected, the probability of getting a red ball is \( p = 3/10 = 0.3\) and hence we can use the formula for binomial probabilities to find P("the red color shows once") = \( \displaystyle{6\choose 1} \cdot 0.3^1 \cdot (1-0.3)^{6-1} = 0.30253 \) P("the red color does not show") = \( \displaystyle{6\choose 0} \cdot 0.3^0 \cdot (1-0.3)^{6-0} = 0.11765 \) P("the red color shows at least twice") = 1 - 0.11765 - 0.30253 = 0.57982

Example 8 80% of the people in a city have a home insurance with "MyInsurance" company. a) If 10 people are selected at random from this city, what is the probability that at least 8 of them have a home insurance with "MyInsurance"? b) If 500 people are selected at random, how many are expected to have a home insurance with "MyInsurance"? Solution to Example 8 a) If we assume that we select these people, at random one, at the time, the probability that a selected person to have home insurance with "MyInsurance" is 0.8. This is a binomial experiment with \( n = 10 \) and p = 0.8. "at least 8 of them have a home insurance with "MyInsurance" means 8 or 9 or 10 have a home insurance with "MyInsurance" The probability that at least 8 out of 10 have have home insurance with the "MyInsurance" is given by \( P( \text{at least 8}) = P( \text{8 or 9 or 10}) \) Use the addition rule \( = P(8)+ P(9) + P(10) \) Use binomial probability formula calling "have a home insurance with "MyInsurance" as a "success". \( = P(8 \; \text{successes in 10 trials}) + P(9 \; \text{successes in 10 trials}) + P(10 \; \text{successes in 10 trials}) \) \( = \displaystyle{10\choose 8} \cdot 0.8^8 \cdot (1-0.8)^{10-8} + \displaystyle{10\choose 9} \cdot 0.8^9 \cdot (1-0.8)^{10-9} + \displaystyle{10\choose 10} \cdot 0.8^10 \cdot (1-0.8)^{10-10} \) \( = 0.30199 + 0.26843 + 0.10737 = 0.67779 \) b) It is a binomial distribution problem with the number of trials is \( n = 500 \). The number of people out of the 500 expected to have a home insurance with "MyInsurance" is given by the mean of the binomial distribution with \( n = 500 \) and \( p = 0.8 \). \( \mu = n p = 500 \cdot 0.8 = 400 \) 400 people out of the 500 selected at random from that city are expected to have a home insurance with "MyInsurance".

Questions and their Solutions

Solutions to the above questions, solution to question 1.

a) There are 3 even numbers out of 6 in a die. Hence if you roll a die once, the probability of getting an even number is \( p = 3/6 = 1/2 \) It is a binomial experiment with \( n = 5 \) , \( k = 3 \) and \( p = 0.5 \) \( P( \text{3 even numbers in 5 trials} ) = \displaystyle{5\choose 3} 0.5^3 (1-0.5)^{5-3} = 0.3125 \) b) \( P (\text{at least 3}) = P (3) + P(4) + P(5) = \displaystyle{5\choose 3} 0.5^3 (1-0.5)^{5-3} + {5\choose 4} 0.5^4 (1-0.5)^{5-4} + {5\choose 5} 0.5^5 (1-0.5)^{5-5} \) \( = 0.3125 + 0.15625 + 0.03125 = 0.5 \) c) \( P (\text{at most 3}) = P (0) + P(1) + P(2) = \displaystyle {5\choose 0} 0.5^0 (1-0.5)^{5-0} + {5\choose 1} 0.5^1 (1-0.5)^{5-1} + {5\choose 2} 0.5^2 (1-0.5)^{5-2} \) \( = 0.03125 + 0.15625 + 0.3125 = 0.5 \) Note The events "at least 3 even numbers are obtained" in part b) and "at most 2 even numbers are obtained " in part c) are complementary and the sum of their probabilities is be equal to 1.

Solution to Question 2

Because the card is replaced back, it is a binomial experiment with the number of trials \( n = 10 \) There are 26 red card in a deck of 52. Hence the probability of getting a red card in one trial is \( p = 26/52 = 1/2 \) The event A = "getting at least 3 red cards" is complementary to the event B = "getting at most 2 red cards"; hence \( P(A) = 1 - P(B) \) \( P(A) = P(3)+P(4) + P(5)+P(6) + P(7)+P(8) + P(9) + P(10) \) \( P(B) = P(0) + P(1) + P(2) \) The computation of \( P(A)\) needs much more operations compared to the calculations of \( P(B) \), therefore it is more efficient to calculate \( P(B) \) and use the formula for complement events: \( P(A) = 1 - P(B) \). \( P(B) = \displaystyle {10\choose 0} 0.5^0 (1-0.5)^{10-0} + {10\choose 1} 0.5^1 (1-0.5)^{10-1} + {10\choose 2} 0.5^2 (1-0.5)^{10-2} \\ = 0.00098 + 0.00977 + 0.04395 = 0.0547 \) \( P(\text{getting at least 3 red cards}) = P(A) = 1 - P(B) = 0.9453 \)

Solution to Question 3

Each question has 5 possible answers with one correct. Therefore the probability of getting a correct answer in one trial is \( p = 1/5 = 0.2 \) It is a binomial experiment with \( n = 20 \) and \( p = 0.2 \). \( P(\text{student answers 15 or more}) = P( \text{student answers 15 or 16 or 17 or 18 or 19 or 20}) \\ = P(15) + P(16) + P(17) + P(18) + P(19) + P(20) \) Using the binomial probability formula \( P(\text{student answers 15 or more}) = \displaystyle{20\choose 15} 0.2^{15} (1-0.2)^{20-15} + {20\choose 16} 0.2^{16} (1-0.2)^{20-16} \\ \quad\quad\quad\quad\quad + \displaystyle {20\choose 17} 0.2^{17} (1-0.2)^{20-17} + {20\choose 18} 0.2^{18} (1-0.2)^{20-18} \\ \quad\quad\quad\quad\quad + \displaystyle {20\choose 19} 0.2^{19} (1-0.2)^{20-19} + {20\choose 20} 0.2^{20} (1-0.2)^{20-20} \) \( \quad\quad\quad\quad\quad \approx 0 \) Conclusion: Answering questions randomly by guessing gives no chance at all in passing a test.

Solution to Question 4

In both cases, it is a binomial experiment with Canada: \( p = 0.618 \) and \( n = 200,000 \) mean : \( \mu = n p = 200,000 \cdot 0.618 = 123600 \) 123600 out of 200,000 are expected to have tertiary education in Canada. United Kingdom: \( p = 0.508 \) and \( n = 200,000 \) mean : \( \mu = n p = 200,000 \cdot 0.508 = 101600 \) 101600 out of 200,000 are expected to have tertiary education in the UK.

More References and links

Binomial Probability Calculations

Tips for when to use binomial probability formula, this video covers some really important tips for how to solve binomial probability distribution word problems, starting with how to know which problems you should be doing this to in the first place also covered are tricky situations where you have to use the formula a few times for "or more" type problems, as in "if you flip a coin 5 times, what is the probability of getting 3 or more heads".

how to solve binomial probability word problems

Binomial Probability Distribution Vocab

This video covers what all those letters in the binomial probability theorem mean, from the p's and q's, to the x's and n's, to that giant backwards-e-looking thing they call sigma. also covered are what these problems mean by "success" and "failure" and "trials"., binomial probability calculations using formula, this video is pretty long, but that's a good thing because it's so full of awesome examples plus it takes a long time to cancel all those crazy factorials (exclamation points). whether you're in stats, algebra or discrete math, if you're in a section that has anything to do with "binomial probability" -- or if you just honestly happen to wonder what the probability is of getting between 5 and 7 heads if you flip a coin exactly 13 times -- this video is for you, binomial probability using your calculator, this video covers another batch of binomial probability word problems, but this time we use a calculator to do the fancy combinatorics stuff because the numbers are way too big to handle doing factorials by hand., mean & expected value of binomial distributions, this video covers the basic formula that quickly spits out the mean, a.k.a. "expected value", of a binomial distribution. flipping coins or rolling dice, this formula quickly tells you how many heads or 5's you'd expect to get on average from n trials., variance & standard deviation of binomial distributions, this video covers the formulas for variance and standard deviation for a binomial distribution -- using nothing more than n, p, and q -- so that you can plug it into those "unusual value" problems., usual vs unusual events for binomial distribution, this is a common type of question which every book covers, every professor will ask at one point on a quiz or test, so definitely worth reviewing., privacy | cancellation policy | contact, free videos reviews your playlists tips & tricks help faq study tips common core.

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how to solve binomial probability word problems

Examples of Binomial Distribution Problems and Solutions

On this page you will learn:

  • Binomial distribution definition and formula.
  • Conditions for using the formula.
  • 3 examples of the binomial distribution problems and solutions.

Many real life and business situations are a pass-fail type. For example, if you flip a coin, you either get heads or tails. You either will win or lose a backgammon game. There are only two possible outcomes – success and failure , win and lose.

And the key element here also is that likelihood of the two outcomes may or may not be the same.

So, what is binomial distribution?

Let’s define it:

In simple words, a binomial distribution is the probability of a success or failure results in an experiment that is repeated a few or many times.

The prefix “bi” means two. We have only 2 possible incomes.

Binomial probability distributions are very useful in a wide range of problems, experiments, and surveys. However, how to know when to use them?

Let’s see the necessary conditions and criteria to use binomial distributions:

  • Rule 1: Situation where there are only two possible mutually exclusive outcomes (for example, yes/no survey questions).
  • Rule2: A fixed number of repeated experiments and trials are conducted (the process must have a clearly defined number of trials).
  • Rule 3: All trials are identical and independent (identical means every trial must be performed the same way as the others; independent means that the result of one trial does not affect the results of the other subsequent trials).
  • Rule: 4: The probability of success is the same in every one of the trials.

Notations for Binomial Distribution and the Mass Formula:

  • P is the probability of success on any trail.
  • q = 1- P – the probability of failure
  • n  – the number of trails/experiments
  • x  – the number of successes, it can take the values 0, 1, 2, 3, . . . n.
  • n C x  = n!/x!(n-x)  and denotes the number of combinations of n elements taken x at a time.

Assuming what the  n C x  means, we can write the above formula in this way:

Just to remind that the ! symbol after a number means it’s a factorial. The factorial of a non-negative integer x is denoted by x!. And x! is the product of all positive integers less than or equal to x. For example, 4! = 4 x 3 x 2 x 1 = 24.

Examples of binomial distribution problems:

  • The number of defective/non-defective products in a production run.
  • Yes/No Survey (such as asking 150 people if they watch ABC news).
  • Vote counts for a candidate in an election.
  • The number of successful sales calls.
  • The number of male/female workers in a company

So, as we have the basis let’s see some binominal distribution examples, problems, and solutions from real life.

Let’s say that 80% of all business startups in the IT industry report that they generate a profit in their first year. If a sample of 10 new IT business startups is selected, find the probability that exactly seven will generate a profit in their first year.

First, do we satisfy the conditions of the binomial distribution model?

  • There are only two possible mutually exclusive outcomes – to generate a profit in the first year or not (yes or no).
  • There are a fixed number of trails (startups) – 10.
  • The IT startups are independent and it is reasonable to assume that this is true.
  • The probability of success for each startup is 0.8.

We know that:

n = 10, p=0.80, q=0.20, x=7

The probability of 7 IT startups to generate a profit in their first year is:

This is equivalent to:

Interpretation/solution: There is a 20.13% probability that exactly 7 of 10 IT startups will generate a profit in their first year when the probability of profit in the first year for each startup is 80%.

The important points here are to know when to use the binomial formula and to know what are the values of p, q, n, and x.

Also, binomial probabilities can be computed in an Excel spreadsheet using the = BINOMDIST function .

Your basketball team is playing a series of 5 games against your opponent. The winner is those who wins more games (out of 5).

Let assume that your team is much more skilled and has 75% chances of winning. It means there is a 25% chance of losing.

What is the probability of your team get 3 wins?

We need to find out.

In this example:

n = 5, p=0.75, q=0.25, x=3

Let’s replace in the formula to get the answer:

Interpretation: the probability that you win 3 games is 0.264.

A box of candies has many different colors in it. There is a 15% chance of getting a pink candy. What is the probability that exactly 4 candies in a box are pink out of 10?

We have that:

n = 10, p=0.15, q=0.85, x=4

When we replace in the formula:

Interpretation: The probability that exactly 4 candies in a box are pink is 0.04.

The above binomial distribution examples aim to help you understand better the whole idea of binomial probability.

If you need more examples in statistics and data science area, our posts descriptive statistics examples and categorical data examples might be useful for you.

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how to solve binomial probability word problems

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Silvia Valcheva is a digital marketer with over a decade of experience creating content for the tech industry. She has a strong passion for writing about emerging software and technologies such as big data, AI (Artificial Intelligence), IoT (Internet of Things), process automation, etc.

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  • Math Article

Binomial Distribution

In probability theory and statistics, the binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either Success or Failure . For example, if we toss a coin, there could be only two possible outcomes: heads or tails, and if any test is taken, then there could be only two results: pass or fail. This distribution is also called a binomial probability distribution.

There are two parameters n and p used here in a binomial distribution. The variable ‘n’ states the number of times the experiment runs and the variable ‘p’ tells the probability of any one outcome. Suppose a die is thrown randomly 10 times, then the probability of getting 2 for anyone throw is ⅙. When you throw the dice 10 times, you have a binomial distribution of n = 10 and p = ⅙. Learn the formula to calculate the two outcome distribution among multiple experiments along with solved examples here in this article.

Table of Contents:

Negative Binomial Distribution

  • Mean and Variance

Binomial Distribution Vs Normal Distribution

  • Solved Problems

Practice Problems

Binomial probability distribution.

In binomial probability distribution, the number of ‘Success’ in a sequence of n experiments, where each time a question is asked for yes-no, then the boolean-valued outcome is represented either with success/yes/true/one (probability p) or failure/no/false/zero (probability q = 1 − p). A single success/failure test is also called a Bernoulli trial or Bernoulli experiment, and a series of outcomes is called a Bernoulli process . For n = 1, i.e. a single experiment, the binomial distribution is a Bernoulli distribution . The binomial distribution is the base for the famous binomial test of statistical importance.

In probability theory and statistics, the number of successes in a series of independent and identically distributed Bernoulli trials before a particularised number of failures happens. It is termed as the negative binomial distribution. Here the number of failures is denoted by ‘r’. For instance, if we throw a dice and determine the occurrence of 1 as a failure and all non-1’s as successes. Now, if we throw a dice frequently until 1 appears the third time, i.e., r = three failures, then the probability distribution of the number of non-1s that arrived would be the negative binomial distribution.

Binomial Distribution Examples

As we already know, binomial distribution gives the possibility of a different set of outcomes. In real life, the concept is used for:

  • Finding the quantity of raw and used materials while making a product.
  • Taking a survey of positive and negative reviews from the public for any specific product or place.
  • By using the YES/ NO survey, we can check whether the number of persons views the particular channel.
  • To find the number of male and female employees in an organisation.
  • The number of votes collected by a candidate in an election is counted based on 0 or 1 probability.

Also, read:

Binomial Distribution Formula

The binomial distribution formula is for any random variable X, given by;

n = the number of experiments

x = 0, 1, 2, 3, 4, …

p = Probability of Success in a single experiment

q = Probability of Failure in a single experiment = 1 – p

The binomial distribution formula can also be written in the form of n-Bernoulli trials, where n C x = n!/x!(n-x)!. Hence,

P(x:n,p) = n!/[x!(n-x)!].p x .(q) n-x

Binomial Distribution Mean and Variance

For a binomial distribution, the mean, variance and standard deviation for the given number of success are represented using the formulas

Mean, μ = np

Variance, σ 2 = npq

Standard Deviation σ= √(npq)

Where p is the probability of success

q is the probability of failure, where q = 1-p

The main difference between the binomial distribution and the normal distribution is that binomial distribution is discrete, whereas the normal distribution is continuous. It means that the binomial distribution has a finite amount of events, whereas the normal distribution has an infinite number of events. In case, if the sample size for the binomial distribution is very large, then the distribution curve for the binomial distribution is similar to the normal distribution curve.

Properties of Binomial Distribution

The properties of the binomial distribution are:

  • There are two possible outcomes: true or false, success or failure, yes or no.
  • There is ‘n’ number of independent trials or a fixed number of n times repeated trials.
  • The probability of success or failure remains the same for each trial.
  • Only the number of success is calculated out of n independent trials.
  • Every trial is an independent trial, which means the outcome of one trial does not affect the outcome of another trial.

Binomial Distribution Examples And Solutions

Example 1: If a coin is tossed 5 times, find the probability of:

(a) Exactly 2 heads

(b) At least 4 heads.

(a) The repeated tossing of the coin is an example of a Bernoulli trial. According to the problem:

Number of trials: n=5

Probability of head: p= 1/2 and hence the probability of tail, q =1/2

For exactly two heads:

P(x=2) = 5 C2 p 2 q 5-2 = 5! / 2! 3! × (½) 2 × (½) 3

P(x=2) = 5/16

(b) For at least four heads,

x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5)

P(x = 4) = 5 C4 p 4 q 5-4 = 5!/4! 1! × (½) 4 × (½) 1 = 5/32

P(x = 5) = 5 C5 p 5 q 5-5 = (½) 5 = 1/32

P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16

Example 2: For the same question given above, find the probability of:

a) Getting at most 2 heads

Solution: P (at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1) + + P (X = 2)

P(X = 0) = (½) 5 = 1/32

P(X=1) = 5 C 1 (½) 5. = 5/32

P(x=2) = 5 C2 p 2 q 5-2 = 5! / 2! 3! × (½) 2 × (½) 3 = 5/16

P(X ≤ 2) = 1/32 + 5/32 + 5/16 = 1/2

A fair coin is tossed 10 times, what are the probability of getting exactly 6 heads and at least six heads.

Let x denote the number of heads in an experiment.

Here, the number of times the coin tossed is 10. Hence, n=10.

The probability of getting head, p ½

The probability of getting a tail, q = 1-p = 1-(½) = ½.

The binomial distribution is given by the formula:

P(X= x) = n C x p x q n-x , where = 0, 1, 2, 3, …

Therefore, P(X = x) = 10 C x (½) x (½) 10-x

(i) The probability of getting exactly 6 heads is:

P(X=6) = 10 C 6 (½) 6 (½) 10-6

P(X= 6) = 10 C 6 (½) 10

P(X = 6) = 105/512.

Hence, the probability of getting exactly 6 heads is 105/512.

(ii) The probability of getting at least 6 heads is P(X ≥ 6)

P(X ≥ 6) = P(X=6) + P(X=7) + P(X= 8) + P(X = 9) + P(X=10)

P(X ≥ 6) = 10 C 6 (½) 10 + 10 C 7 (½) 10  + 10 C 8 (½) 10  + 10 C 9 (½) 10  + 10 C 10 (½) 10

P(X ≥ 6) = 193/512.

Solve the following problems based on binomial distribution:

  • The mean and variance of the binomial variate X are 8 and 4 respectively. Find P(X<3).
  • The binomial variate X lies within the range {0, 1, 2, 3, 4, 5, 6}, provided that P(X=2) = 4P(x=4). Find the parameter “p” of the binomial variate X.
  • In binomial distribution, X is a binomial variate with n= 100, p= ⅓, and P(x=r) is maximum. Find the value of r.

Frequently Asked Questions on Binomial Distribution

What is meant by binomial distribution.

The binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either success or failure.

Mention the formula for the binomial distribution.

The formula for binomial distribution is: P(x: n,p) = n C x p x (q) n-x Where p is the probability of success, q is the probability of failure, n= number of trials

What is the formula for the mean and variance of the binomial distribution?

The mean and variance of the binomial distribution are: Mean = np Variance = npq

What are the criteria for the binomial distribution?

The number of trials should be fixed. Each trial should be independent. The probability of success is exactly the same from one trial to the other trial.

What is the difference between a binomial distribution and normal distribution?

The binomial distribution is discrete, whereas the normal distribution is continuous.

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  1. Binomial Probability Distribution Word Problems

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COMMENTS

  1. Using the Binomial Formula to Solve a Basic Word Problem

    Step 1 : Identify what makes up one trial, what a success is, and what a failure is. Step 2: Identify n, the number of trials; p, the probability of success; and x, the number of successes. Step...

  2. How to Solve Probability Word Problems

    How to Solve Probability Word Problems | P (A and B) | P (A or B) | Binomial Probability - YouTube © 2024 Google LLC...

  3. Using the Binomial Formula to Solve an Advanced Word Problem

    Solving Advanced Word Problems with the Binomial Formula Step 1: Identify the probability of success on a single trial, p, the number of trials, the lower bound of the number of...

  4. Binomial Distribution Word Problems

    A coin is tossed four times. Calculate the probability of obtaining more heads than tails. Exercise 2. An agent sells life insurance policies to five equally aged, healthy people. According to recent data, the probability of a person living in these conditions for 30 years or more is 2/3. Calculate the probability that after 30 years: 1.

  5. Binomial Probabilities Examples and Questions

    The best way to explain the formula for the binomial distribution is to solve the following example. Example 1 A fair coin is tossed 3 times. Find the probability of getting 2 heads and 1 tail. Solution to Example 1 When we toss a coin we can either get a head H or a tail T.

  6. Binomial probability (basic) (article)

    problem A If she makes 2 of the free-throws, how many free-throws does that mean she needs to miss? Choose 1 answer: 0 misses A 0 misses 1 miss B 1 miss 2 misses C 2 misses 3 misses D 3 misses problem b Find the probability that she makes her first 2 free-throws and misses her third free-throw.

  7. Binomial Distribution Word Problem 1

    Binomial Distribution Word Problem 1 Mathbyfives 142K subscribers Subscribe Subscribed 401 Share 65K views 10 years ago algebra This is the introductory example for solving binomial...

  8. Binomial distribution (video)

    In the problem around 6:12 how does 5!/4!=5? • ( 5 votes)

  9. Calculating binomial probability (practice)

    Calculating binomial probability. 70 % of a certain species of tomato live after transplanting from pot to garden. Najib transplants 3 of these tomato plants. Assume that the plants live independently of each other. Let X = the number of tomato plants that live. What is the probability that exactly 2 of the 3 tomato plants live?

  10. PDF Binomial distribution questions: formal word problems

    Binomial distribution questions: formal word problems For the following questions, write the information given in a formal way before solving the problem, something like: = number of . . . out of 12, so ∼ B(12, 0.2). Then P(X 6 3) = . . . Thirty percent of students at a large university are known to be short-sighted.

  11. Binomial Distribution: Formula, What it is, How to use it

    Step 1: Identify 'n' from the problem. Using our example question, n (the number of randomly selected items) is 9. Step 2: Identify 'X' from the problem. X (the number you are asked to find the probability for) is 6. Step 3: Work the first part of the formula. The first part of the formula is. n! / (n - X)!

  12. ThatTutorGuy

    This video covers some really important tips for how to solve binomial probability distribution word problems, starting with how to know which problems you should be doing this to in the first place!

  13. Binomial Probability Distribution Word Problem Example 2

    The probability that a motorcycle will change lanes while making a turn is 80%. Suppose a random sample of 16 motorcycle are observed making turns at Fordham...

  14. Binomial Probability Word Problem

    3 Answers Sorted by: 1 (a) Let G G be the event that the whole lot is good (zero defectives), and let P P be the event there were no defectives in the sample of 2 2. We are asked for the conditional probability Pr(G|P) Pr ( G | P), the probability that the whole lot is good given the information that the sample of 2 2 had no defectives.

  15. Using the Binomial Formula to Solve a Basic Word Problem

    Practice Using the Binomial Formula to Solve a Basic Word Problem with practice problems and explanations. Get instant feedback, extra help and step-by-step explanations. Boost your Algebra grade ...

  16. Binomial Distribution Examples, Problems and Formula

    Conditions for using the formula. 3 examples of the binomial distribution problems and solutions. Many real life and business situations are a pass-fail type. For example, if you flip a coin, you either get heads or tails. You either will win or lose a backgammon game. There are only two possible outcomes - success and failure, win and lose.

  17. How to Solve a Binomial Distribution Word Problem. [HD]

    Suppose a local bus service accepted 12 reservations for a commuter bus with 10 seats. Seven of the ten reservations went to regular commuters who will show ...

  18. Binomial Distribution

    The binomial distribution in probability theory gives only two possible outcomes such as success or failure. Visit BYJU'S to learn the mean, variance, properties and solved examples. ... Solve the following problems based on binomial distribution: The mean and variance of the binomial variate X are 8 and 4 respectively. Find P(X<3).

  19. Probability Problem Solver

    Problem Solver Subjects. Our math problem solver that lets you input a wide variety of probability math problems and it will provide a step by step answer. This math solver excels at math word problems as well as a wide range of math subjects. Here are example math problems within each subject that can be input into the calculator and solved.

  20. Binomial probability example (video)

    For example (S=Success, M=Miss): I see it as SSMMMM, SMSMMM, SMMSMM, SMMMSM, SMMMMS all being the same thing, which I would consider a permutation and not a combination. But is it actually: [S1, S2, M3, M4, M5, M6,] [S1, M2, S3, M4, M5, M6,] [S1 ,M2, M3, S4, M5, M6,]

  21. Binomial Probability

    Learn how to use the binomial probability theorem to calculate probabilities in this free math video tutorial by Mario's Math Tutoring. We go through a coupl...

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  23. Word Problems Calculator

    How do you solve word problems? To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform.