The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is :
Correct Answer – C
Let the coin be tossed n times and let X-denote the number of heads in n tosses of the coin. Then,
`therefore P(X=r)= .^(n)C_(r )((1)/(2))^(r )((1)/(2))^(n-r) = .^(n)C_(r )((1)/(2))^(n), r=0,1,2,..,n`
It is given that
`P(X ge 2) ge 0.96`
`implies underset(r=2)overset(n)(sum)P(X=r) ge 0.96`
`implies underset(r=2)overset(n)(sum) .^(n)C_(r )((1)/(2))^(n) ge 0.96`
`implies (1)/(2^(n))( underset(r=2)overset(n)(sum) .^(n)C_(r )) ge 0.96`
`implies (1)/(2^(n)) (2^(n) – .^(n)C_(0) – .^(n)C_(1)) ge 0.96`
`implies 1-(n+1)/(2^(n)) ge 0.96`
`implies (n+1) le (0.04) 2^(n)`
`implies n=8, 9, 10,`..
`p=1/2,q=1-1/2=1/2`
`P(r)=nC_r*p^r*q^(n-r)`
r success in n trials
Required probability=1-P(1)-P(0)
`P(1)=nC_1*(1/2)^11*(1/2)^(n-1)`
`P(1)=nC_1*(1/2)^n`
`P(0)=nC_0*(1/2)^0*(1/2)^n`
`=(1/2)^n`
`P=1-n*1/(2^n)-1/(2^n)`
`=1-1/(2^n)(n+1)>=0.96`
`=11-0.96>=1/(2^n)*(n+1)`
`0.04>=(n+1)/(2^n)`
`2^n/(n+1)>=25`
Which is possible when `n>=8`
`n=8`.