Question

Suppose that $90\%$ of people are right handed. What is the probability that alomost 6 of random sample of 10 people are right handed?

Answer 1

A person can be either right handed or left handed. It is given that $90\%$ of the people are right handed.

Therefore $p=\frac{90}{100}=\frac{9}{10}andq=1-p=1-\frac{9}{10}=\frac{1}{10},n=10$

Clearly, X has a binomial distribution with n=10,

$p=\frac{9}{10}andq=\frac{1}{10}\therefore P(X=r){=}^n{C}_r{p}^r{q}^{n-r}{=}^{10}{C}_r{\left(\frac{9}{10}\right)}^r{\left(\frac{9}{10}\right)}^{10-r}$

$Requiredprobability=P(X\le 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6){=}^{10}{C}_0{p}^0{q}^{10}{+}^{10}{C}_1{p}^1{q}^9{+}^{10}{C}_2{p}^2{q}^8{+}^{10}{C}_3{p}^3{q}^7{+}^{10}{C}_4{p}^4{q}^6{+}^{10}{C}_5{p}^5{q}^5{+}^{10}{C}_6{p}^6{q}^4=q^{10}+10p{q}^9+\frac{10\times 9}{1\times 2}{p}^2{q}^8+\frac{10\times 9\times 8}{1\times 2\times 3}{p}^3{q}^7+\frac{10\times 9\times 8\times 7}{1\times 2\times 3\times 4}{p}^4{q}^6+\frac{10\times 9\times 8\times 7\times 6}{1\times 2\times 3\times 4\times 5}{p}^5{q}^5+\frac{10\times 9\times 8\times 7}{1\times 2\times 3\times 4}{p}^6{q}^4$

$=q^{10}+10p{q}^9+45p^2{q}^8+120p^3{q}^7+210p^4{q}^6+252p^5{q}^5+210p^{+}{q}^4={\left(\frac{1}{10}\right)}^{10}+10\left(\frac{9}{10}\right){\left(\frac{1}{10}\right)}^9+45{\left(\frac{9}{10}\right)}^2{\left(\frac{1}{10}\right)}^8+120{\left(\frac{9}{10}\right)}^3{\left(\frac{1}{10}\right)}^7+210{\left(\frac{9}{10}\right)}^4{\left(\frac{1}{10}\right)}^6+252{\left(\frac{9}{10}\right)}^5{\left(\frac{1}{10}\right)}^5+210{\left(\frac{9}{10}\right)}^6{\left(\frac{1}{10}\right)}^4=\frac{1+90+45\times 9^2+120\times 9^3+210\times 9^4+252\times 9^5+210\times 9^6}{{10}^{10}}$

Alternatively,
Required probability

=$P(X\le 6)$=1-P(more than 6 are right handed) $(7\le X\le 10)$

$=1-{\sum }_{r=7}^{10}{}^{10}{C}_r{\left(\frac{9}{10}\right)}^r{\left(\frac{1}{10}\right)}^{10-r}$