﻿ Binomial Probability Problems - Page 3 | Problems & Solutions

# Binomial Probability Problems - Page 3

Binomial Probability Problems
• Page 3
21.
A survey was conducted in a class of 50 about the games the students play. Each student had to select only one particular game or no game. The results of the survey are: 20 play football, 15 play basketball and 10 play baseball. What is the probability that a student selected does not play any game?
 a. $\frac{3}{10}$ b. Zero c. $\frac{9}{10}$ d. $\frac{1}{10}$

#### Solution:

Probability of event E : P(E) = Number of outcomes in E / Total number of outcomes in the sample space
P(E) = n(E)n(S)

Total number of students in the class n(S) = 50

Number of students who do not play any games = Total number of students in the class - (Number of students who play football + Number of students who play basketball + Number of students who play baseball)

Number of students who do not play any games = 50 - (20 + 15 + 10)

Number of students who do not play any games = 50 - 45

Number of students who do not play any games n(E) = 5

P(E) = 5 / 50
[Substitute n(E) = 5 and n(S) = 50 in P(E).]

P(E) = 1 / 10
[Simplify.]

Probability that a selected student does not play any game is 1 / 10 .

22.
The distribution of the ages of the members of a colony is as follows :
 Age Frequency 0 - 20 35 21 - 30 15 31 - 40 20 41 - 50 25 51 - 60 15 61 - up 10
If a person is selected at random, then find the probability that his/her age is less than 30 years.
 a. $\frac{5}{12}$ b. $\frac{1}{8}$ c. $\frac{7}{12}$ d. $\frac{7}{24}$

#### Solution:

Given a frequency distribution, the probability of an event being in a given class is
P(E) = frequency for the classtotal frequencies in the distribution = f / n

Total frequencies in the distribution = 35 + 15 + 20 + 25 + 15 + 10

n = 120

Frequency for the class 0 - 30 = Frequency for the class 0 - 20 + Frequency for the class 21 - 30

f = 35 + 15

f = 50

P(E) = 50 / 120
[Substitute the values of f and n in P(E).]

P(E) = 5 / 12
[Simplify.]

Probability that a person selected is less than 30 years is 5 / 12 .

23.
The distribution of the ages of the members of a colony is as follows :
 Age Frequency 0 - 20 35 21 - 30 15 31 - 40 20 41 - 50 25 51 - 60 15 61 - up 10
If a person is selected at random, then find the probability that his/her age is over 41 years and under 60.
 a. $\frac{1}{8}$ b. $\frac{1}{3}$ c. $\frac{5}{12}$ d. $\frac{5}{24}$

#### Solution:

Given a frequency distribution, the probability of an event being in a given class is
P(E) = frequency for the classtotal frequencies in the distribution = f / n

Total frequencies in the distribution = 35 + 15 + 20 + 25 + 15 + 10

n = 120

Frequency for the class 41 - 60 = Frequency for the class 41 - 50 + Frequency for the class 51 - 60

f = 25 + 15 = 40

P(E) = 40 / 120
[Substitute the values of f and n in P(E).]

P(E) = 1 / 3
[Simplify.]

Probability that a person selected is over 41 and under 60 is 1 / 3 .

24.
Specify the nature of probability that the following statement denotes:
The probability that it will rain tomorrow is 83%( Weather Department Forecast).
 a. Cannot be determined b. Empirical probability c. Classical probability d. Subjective probability

#### Solution:

Empirical probability relies on actual experience to determine the likelihood of outcomes.

The probability statement given is based on professional information and experience.

So, the statement denotes empirical probability.

25.
Specify the nature of probability that the following statement denotes:
The probability of drawing an Ace from a pack of 52 cards is $\frac{1}{13}$.
 a. Subjective probability b. Classical probability c. Cannot be determined d. Empirical probability

#### Solution:

Classical probability assumes that all outcomes in the sample space are equally likely to occur.

In the probability statement each outcome has equal probability to occur.

So, the statement denotes classical probability.

26.
What type of probability does the statement denote?
The probability that a student will get 60% and above is 0.23.
 a. Subjective probability b. Classical probability c. Empirical probability d. Cannot be determined

#### Solution:

Empirical probability relies on actual experience to determine the likelihood of outcomes.

Probability statement given is based on professional information and experience.

So, the statement denotes empirical probability.

27.
Specify the nature of probability that the following statement denotes:
The probability that a new motel venture in the State will succeed is 25%.
 a. Empirical probability b. Classical probability c. Cannot be determined d. Subjective probability

#### Solution:

Empirical probability relies on actual experience to determine the likelihood of outcomes.

The probability statement is based on professional information and experience.

So, the statement denotes empirical probability.

28.
Identify the nature of probability in the statement:
The probability that a patient suffers from viral fever is 0.7 (Hospital Diagnosis).
 a. Subjective probability b. Cannot be determined c. Classical probability d. Empirical probability

#### Solution:

Empirical probability relies on actual experience to determine the likelihood of outcomes.

Probability is based on professional information and experience.

So, the statement denotes empirical probability.

29.
The Sample Space (S) derived from drawing 2 students from a group of 2 Boys B1, B2 and 3 Girls G1, G2, G3 is:
 a. { B1B2, G1G2, G1G3, G2G3 } b. { B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3, G1G2, G1G3, G2G3 } c. { B1G1, B1G2, B1G3, B2G1, B2G2, B2G3 } d. { B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3 }

#### Solution:

Sample space is the set of all possible combinations.

The sample space is: { B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3, G1G2, G1G3, G2G3 } .

30.
A coin and a die are rolled. The sample space(S) of getting a Head(H) on the coin and an even number on the die is:
 a. S = { (H, 2) (H, 4) (H, 6) (T, 2) (T, 4) (T, 6) } b. S = { (H, 2) (H, 4) (H, 6) } c. S = { (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (H, 6) } d. S = { (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) }

#### Solution:

Sample space is the set of all the possible combinations.

The sample space is: S = { (H, 2) (H, 4) (H, 6) }.