Question

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

Answer 1

Answer:

:

`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`.