Suppose that `90%` of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed ?

Answer

N/a

Explanation

A person can be either right handed or left handed.
Given that 90% of people are right handed.
Therefore, `p=90/100=9/10`
and `q=1-p=1-9/10=1/10`,
`n=10`
Clearly `X` is a binomial distribution with
`n=10,p=9/10` and `q=1/10`
`:.P(X=r)=.^(n)C_(r)p^(r)q^(n-r)`
`=.^(10)C_(r)(9/10)^(r)(1/10)^(10-r)`
Therefore, required probability `=P(Xle6)`
`=P(X=0)+P(X=1)+P(X=2)+P(X=3)+`
`P(X=r)+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)+10pq^(9)+(10xx9)/(1xx2)p^(2)q^(8)+(10xx9xx8)/(1xx2xx3)p^(3)q^(7)`
`+(10xx9xx8xx7)/(1xx2xx3xx4)p^(4)q^(6)+(10xx9xx8xx7xx6)/(1xx2xx3xx4xx5)p^(5)q^(5)`
`+(10xx9xx8xx7)/(1xx2xx3xx4)p^(6)q^(4)`
`=q^(10)+10pq^(9)+45p^(2)q^(8)+120p^(3)q^(7)+210p^(4)q^(6)`
`+252p^(5)q^(5)+210p^(6)q^(4)`
`=(1/10)^(10)+10(9/10)(1/10)^(2)+45(9/10)^(2)(1/10)^(8)`
`+120(9/10)^(3)(1/10)^(7)+210(9/10)^(4)(1/10)^(6)`
`+252(9/10)^(5)(1/10)^(5)+210(9/10)^(6)(1/10)^(4)`
`1+90+45xx9^(2)+120xx9^(3)`
`=(210xx9^(4)+252xx9^(5)210xx9^(6))/(10^(10))`