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 :

Explanation

Using Binomial distribution,
`P(Xge2)=1-P(X=0)-P(x=1)`
`=1-((1)/(2))^(n)-[overset(n)C_(1).((1)/(2)).((1)/(2))^(n-1)]`
`=1-(1)/(2^(n))-overset(n)""C_(1).(1)/(2^(n))=1-((1+n)/(2^n))`
Given, `P(Xge2)ge0.96`
`:. 1-((n+1))/(2^(n))ge(24)/(25)`
`rArr (n+1)/(2^(n))le(1)/(25)`
`:. n=8`