Question

51​% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan. Find the probability that the number who consider themselves baseball fans is​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

Answer 1

Answer:

a) 24.56% probability that the number who consider themselves baseball fans is exactly five.

b) 40.18% probability that the number who consider themselves baseball fans is​ at least six.

c) 15.60% probability that the number who consider themselves baseball fans is​ less than four.

Step-by-step explanation:

For each men, there are only two possible outcomes. Either they are professional baseball fans, or they are not. The probability of a man being a professional baseball fan is independent of any other men. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

51​% of men consider themselves professional baseball fans.

This means that $p = 0.51$

10 men

This means that $n = 10$

Find the probability that the number who consider themselves baseball fans is​

(a) exactly​ five,

This is P(X = 5).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.51)^{5}.(0.49)^{5} = 0.2456$

24.56% probability that the number who consider themselves baseball fans is exactly five.

(b) at least​ six, and

Either less than six are fans, or at least six are. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$. So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$​

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.51)^{0}.(0.49)^{10} = 0.0008$

$P(X = 1) = C_{10,1}.(0.51)^{1}.(0.49)^{9} = 0.0083$

$P(X = 2) = C_{10,2}.(0.51)^{2}.(0.49)^{8} = 0.0389$

$P(X = 3) = C_{10,3}.(0.51)^{3}.(0.49)^{7} = 0.1080$

$P(X = 4) = C_{10,4}.(0.51)^{4}.(0.49)^{6} = 0.1966$

$P(X = 5) = C_{10,5}.(0.51)^{5}.(0.49)^{5} = 0.2456$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0008 + 0.0083 + 0.0389 + 0.1080 + 0.1966 + 0.2456 = 0.5982$​

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.5982 = 0.4018$

40.18% probability that the number who consider themselves baseball fans is​ at least six.

(c) less than four.

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.51)^{0}.(0.49)^{10} = 0.0008$

$P(X = 1) = C_{10,1}.(0.51)^{1}.(0.49)^{9} = 0.0083$

$P(X = 2) = C_{10,2}.(0.51)^{2}.(0.49)^{8} = 0.0389$

$P(X = 3) = C_{10,3}.(0.51)^{3}.(0.49)^{7} = 0.1080$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0008 + 0.0083 + 0.0389 + 0.1080 = 0.1560$

15.60% probability that the number who consider themselves baseball fans is​ less than four.