Question

2) At COS, 58% of all students identify as female. If 12 COS students are selected at random, findthe probability that less than 5 identify as female.

Answer 1

Let p denotes the probability of COS students identify as female.

We are given

`\begin{gathered} p=0.58 \\ n=12 \end{gathered}`

We want to find

`p(X<5)`

This is a binomial distribution

Note: The Binomial distribution formula

`p(X=x)=\text{ }^nC_xp^x(1-p)^(n-x)`

For the probability of less than 5

`p(X\lt5)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)`

we will find them one after the other

`\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^(n-x) \\ p(X=0)=\text{ }^(12)C_0*(0.58)^0*(0.42)^(12) \\ p(X=0)=1*1*0.00003012946949 \\ p(X=0)=0.00003013 \end{gathered}`

p(X = 1)

`\begin{gathered} p(X=x)=^\text{ }^nC_xp^x(1-p)^(n-x) \\ p(X=1)=\text{ }^(12){}C_1(0.58)^1(0.42)^(11) \\ p(X=1)=12*0.58*0.00007173683211 \\ p(X=1)=0.0004992883515 \\ p(X=1)=0.0004993 \end{gathered}`

p(X = 2)

`\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^(n-x) \\ p(X=2)=\text{ }^(12)C_2(0.58)^2(0.42)^(10) \\ p(X=2)=66*0.3364*(1.7080*10^(-4)) \\ p(X=2)=0.003792 \end{gathered}`

p(X = 3)

`\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^(n-x) \\ p(X=3)=\text{ }^(12)C_3(0.58)^3(0.42)^9 \\ p(X=3)=0.01746 \end{gathered}`

p(X = 4)

`\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^(n-x) \\ p(X=4)=\text{ }^(12)C_4(0.58)^4(0.42)^8 \\ p(X=4)=0.05423 \end{gathered}`

Thus, we will need to add and the answer will be

`\begin{gathered} p(X\lt5)= p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4) \\ p(X\lt5)=0.07601143 \\ p(X\lt5)=0.0760\text{ }(to\text{ four decimal places}) \end{gathered}`