Worksheet: Binomial Distribution Problems
Problem
1. A manufacturer of electronics components produces precision resistors designed to have a tolerance of ±1%.
From quality-control testing, the manufacturer knows that about one resistor in six is actually within just 0.3%
of its nominal value. A customer needs two of these more precise resistors. What is the probability of finding
exactly two such resistors among the first four tested?
2. Paula moves to an area with a different telephone exchange. Telephone numbers in the new exchange start
with 753, and all combinations of the four remaining digits are equally likely.
a) Calculate the probability that the last four digits in Paula’s new telephone number are even.
b) What is the expected number of even digits in her new telephone number?
3. The Choco-Latie Candies company makes candy-coated chocolates, 40% of which are red. The production
line mixes the candies randomly and packages ten per box.
a) What is the probability that less than four candies in a given box are red?
b) What is the probability that at least four candies in a given box are red?
c) Describe a second way of finding the answer to part b).
4. Prepare a table and a graph for a binomial probability distribution with n = 5 and p = .
5. One type of jet engine has a 0.0001 probability of failure while in flight. For a jet that has four of these engines, what is the probability of at least two of them failing?
6. Suppose that 65% of the families in a town own computers, If ten families are surveyed at random,
a) what is the probability that at least five own computers?
b) what is the expected number of families that own computers?
7. Ninety percent of a country’s population are right-handed.
a) What is the probability that exactly 29 people in a group of 30 are right-handed?
b) What is the expected number of right-handed people in a group of 30?
c) Design a simulation to show that the expectation calculated in part b) is accurate.
8. Suppose that Bayanisthol, a new drug, is effective for 65% of the participants in clinical trials. If a group of
fifteen patients take this new drug,
a) what is the expected number of patients for whom the drug will be effective?
b) what is the probability that the drug will be effective for less than half of them?
9. Jason knows that his favourite player on the Raptors basketball team scores on 83% of his free-throw attempts. Since 10 0.83 = 8.3, Jason expects that in ten attempts this player will score eight times.
a) Is Jason’s reasoning correct? Explain why or why not.
b) Is the player more likely to score exactly eight times or not to score exactly eight times?
10. A student writes a five question multiple-choice quiz. Each question has four possible responses. The student
guesses at random for each question. Calculate the probability for each possible score on the test from 0 to 5.
11. There are 10 members on a committee. The probability of any member attending a randomly chosen meeting
is 0.9. The committee cannot do business if more than 3 members are absent. What is the probability that 7 or
more members will be present on a given date?
12. A school fills each of its Grade 9 mathematics classes with 22 students. Assume that the likelihood of a male
or female being given a place in a class is equal. Design a simulation that could be used to model the distribution of males and females in these classes.
13. A small math class consists of 16 students. What is the probability that the difference in the number of male
and female students in the class is greater than 4?
14. A baseball player has a batting average of 0.350. Compare the expected value for his number of hits in a game
with 6 at bats to the probability of the number of hits he is most likely to get.
15. A soccer linesman will make the correct call for a possible offside pass 90% of the time. What is the probability that he will make 2 or fewer incorrect calls in a game in which he sees 32 passes?
Worksheet: Binomial Distribution Problems
Answer Section
PROBLEM
1. ANS:
PTS: 1
2. ANS:
REF: Applications OBJ: Section 7.2
TOP: Binomial probabilities
a)
b) Here, n = 4 and p = 0.4, so the expected number of even digits is
E(X) = 4 × 0.4
= 0.16
PTS: 1
REF: Applications OBJ: Section 7.2
TOP: Binomial probabilities | Expected value
3. ANS:
a)
= 0.3823
b) 1 – 0.3823 = 0.6177
c) The probability that at least four candies in a given box are red can also be found by adding the individual
probabilities for exactly four, five, six, seven, eight, nine, and ten red candies.
PTS: 1
REF: Applications | Communication
TOP: Binomial probabilities
4. ANS:
The probabilities are given by the formula
P(x) = 5Cx(0.5)x(0.5)5–x
= Cx(0.5)5
Number of Successes, x
0
1
2
3
4
5
OBJ: Section 7.2
Probability, P(x)
0.031 25
0.156 25
0.312 5
0.312 5
0.156 25
0.031 25
PTS: 1
REF: Knowledge & Understanding
TOP: Binomial probabilities
5. ANS:
OBJ: Section 7.2
PTS: 1
REF: Applications OBJ: Section 7.2
6. ANS:
a) Here, p = 0.65 and n = 10.
TOP: Binomial probabilities
This probability can also be calculated using the binomcdf( function on a graphing calculator, the
BINOMDIST function in a spreadsheet, or the binomialProbability function in Fathom™.
b)
PTS: 1
REF: Applications OBJ: Section 7.2
TOP: Binomial probabilities | Expected value
7. ANS:
a) Here, p = 0.90 and n = 30.
This probability can also be calculated using the binompdf( function on a graphing calculator, the
BINOMDIST function in a spreadsheet, or the binomialProbability function in Fathom™.
b) np = 30 0.90
=27
c) Answers may vary. The following method could be used with a graphing calculator.
Check that list L1 is clear.
Let 1 represent a right-handed person.
Enter randBin(1,9/10,30) L1.
Enter sum(L1).
Record this result.
Use this process to generate ten or more sets of random numbers and calculate the average of the sums.
PTS: 1
OBJ: Section 7.2
8. ANS:
REF: Applications | Communication | Thinking/Inquiry/Problem Solving
TOP: Binomial probabilities | Expected value | Simulations
a)
b)
This
probability can also be calculated using the binomcdf( function on a graphing calculator, the BINOMDIST function in a spreadsheet, or the binomialProbability function in Fathom™.
PTS: 1
REF: Applications OBJ: Section 7.2
TOP: Binomial probabilities | Expected value
9. ANS:
a) Jason is correct that the expected number of baskets scored is np = 8.3. However, this value is the expected
average of many sets of trials. For a single set of trials, Jason can predict only that the player is likely to
score about eight times.
b)
The player is more likely not to score exactly 8 times.
PTS: 1
OBJ: Section 7.2
10. ANS:
REF: Communication | Thinking/Inquiry/Problem Solving
TOP: Binomial probabilities | Expected value
PTS: 1
REF: Application OBJ: 5.3 Binomial Distributions
STA: CP2.05
TOP: Probability Distributions and Predictions
11. ANS:
The total probability is the sum of the probabilities for having 7, 8, 9 or 10 members present.
The total probability is 0.9872.
PTS: 1
REF: Application OBJ: 5.3 Binomial Distributions
STA: CP2.05
TOP: Probability Distributions and Predictions
12. ANS:
A simple simulation could involve using a coin and assigning one face to represent a male student and the
other face to represent a female student. The coin would be tossed 22 times to simulate filling one class. Successive repetitions could provide as many sample classes as needed to model the probability distribution.
PTS: 1
REF: Thinking/Inquiry/PS
OBJ: 5.3 Binomial Distributions
STA: CPV.03
TOP: Probability Distributions and Predictions
13. ANS:
There are five cases where the gender split is 4 or less. These are 8M/8F, 9M/7F, 9F/7M, 10M/6F, and
10F/6M. Subtract the sum of these probabilities from 1 to obtain the requested probability. Define X as the
random variable representing the number of males in the class.
P(X = 6, 7, 8, 9, 10) = 0.1964 + 2(0.1746) + 2(0.1222) = 0.79
This represents the probability that the gender split is 4 or less. The requested probability is 0.21.
PTS: 1
REF: Application OBJ: 5.3 Binomial Distributions
STA: CP2.05
TOP: Probability Distributions and Predictions
14. ANS:
The expected value for a binomial distribution is E(X) = np. In this case, we have 6 trials and a probability of
success of 0.350. The expected number of hits is 2.1.
The actual probability of getting r hits in 6 at bats is given by
. This gives
P(X = 1) = 0.2437, P(X = 2) = 0.328, and P(X = 3) = 0.235.
The most likely number of hits is closest to the expected value, while the next most likely numbers of hits are
on either side of the expected value.
PTS: 1
REF: Application OBJ: 5.3 Binomial Distributions
STA: CP2.05
TOP: Probability Distributions and Predictions
15. ANS:
We will simply evaluate the probabilities for 0, 1, and 2 incorrect calls and find their sum.
The total probability of 2 or fewer incorrect calls is 0.3667 or 36.7%.
PTS: 1
STA: CP2.05
REF: Application OBJ: 5.3 Binomial Distributions
TOP: Probability Distributions and Predictions