Chapter 12.7, Problem 3CYU

To determine

To find : Probability of atleast $5$ of the $25$ are left handed.

Answer to Problem 3CYU

  $\text{13}\text{.31\%}$

Explanation of Solution

Given information :

We know that, $11\%$ of the population is left handed.

Formula Used:

  ${}^n{c}_r=\frac{n!}{r!(n-r)!}$

Calculation:

Probability of atleast $\text{5}=1-$ Probability of at most $\text{4}$

  $\begin{array}{l}=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)]\\ $ =1-{[}^{25}{c}_0\times {(0.11)}^0\times {(0.89)}^{25}+{}^{25}{c}_1\times {(0.11)}^1\times {(0.89)}^{24}$$ +{}^{25}{c}_2{(0.11)}^2{(0.89)}^{23}+{}^{25}{c}_3{(0.11)}^3{(0.89)}^{22}$$ +{}^{25}{c}_4{(0.11)}^4\times {(0.89)}^{21}]$=13.31\%\\ \text{As}P(x=r)={}^n{c}_r{p}^r{q}^{n-r}\\ p=0.11\text{and}q=0.89\end{array}$