Question

41​% 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) $P(X=5)=0.208$

b) $P(X\geq 6)=0.1854$

c) $P(X$

Step-by-step explanation:

Given : 41​% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan.

To Find : The probability that the number who consider themselves baseball fans is​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.  

Solution :

Applying binomial theorem,

$P(X=r)=^nC_r\times p^r\times (1-p)^{n-r}$

The success p= 41%=0.41

The failure (1-p)=1-041=0.59

Number of selection n=10

a) The probability that the number who consider themselves baseball fans is​ exactly​ five,

i.e. X=5

$P(X=5)=^{10}C_5\times (0.41)^5\times (0.59)^{10-5}$

$P(X=5)=\frac{10!}{5!5!}\times (0.41)^5\times (0.59)^{5}$

$P(X=5)=252\times 0.0115\times 0.071$

$P(X=5)=0.208$

b) The probability that the number who consider themselves baseball fans is​ at least​ six,

i.e. $X\geq 6$

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

$P(X\geq 6)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)]$

$P(X\geq 6)=1-[^{10}C_0\times (0.41)^0\times (0.59)^{10-0}+^{10}C_1\times (0.41)^1\times (0.59)^{10-1}+^{10}C_2\times (0.41)^2\times (0.59)^{10-2}+^{10}C_3\times (0.41)^3\times (0.59)^{10-3}+^{10}C_4\times (0.41)^4\times (0.59)^{10-4}+^{10}C_5\times (0.41)^5\times (0.59)^{10-5}]$

$P(X\geq 6)=1-[0.0051+0.0355+0.111+0.205+0.250+0.208]$

$P(X\geq 6)=1-[0.8146]$

$P(X\geq 6)=0.1854$

c) The probability that the number who consider themselves baseball fans is​ less than four,

i.e. $X$

$P(X$

$P(X$

$P(X$

$P(X$