binomial experiment practice problems

Free Mathematics Tutorials

Math Problems
Algebra Questions and Problems
Graphs of Functions, Equations, and Algebra
Free Math Worksheets to Download
Analytical Tutorials
Solving Equation and Inequalities
Online Math Calculators and Solvers
Free Graph Paper
Math Software
The Applications of Mathematics in Physics and Engineering
Exercises de Mathematiques Utilisant les Applets
Calculus Tutorials and Problems
Calculus Questions With Answers
Free Calculus Worksheets to Download
Geometry Tutorials and Problems
Online Geometry Calculators and Solvers
Free Geometry Worksheets to Download
Trigonometry Tutorials and Problems for Self Tests
Free Trigonometry Questions with Answers
Free Trigonometry Worksheets to Download
Elementary Statistics and Probability Tutorials and Problems
Mathematics pages in French
About the author
Primary Math
Middle School Math
High School Math
Free Practice for SAT, ACT and Compass Math tests

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.