Discussion Forum

 The minimum number of times a fair coin needs to be tossed, so that the probability of getting atleast two heads is atleast 0.96, is 

Answer:

Explanation:

 Using Binomial  distribution,

    $P(X\geq2)=1-P(X=0)-P(X=1)$

    =   $1-\left(\frac{1}{2}\right)^{n}-\left[^{n}C_{1}.\left(\frac{1}{2}\right).\left(\frac{1}{2}\right)^{n-1}\right]$

   =  $1-\frac{1}{2^{n}}-^{n}C_{1}.\frac{1}{2^{n}}=1-\left(\frac{1+n}{2^{n}}\right)$

   Given,   $P(X\geq2)\geq0.96$

 $\therefore$   $1-\frac{(n+1)}{2^{n}}\geq\frac{24}{25}\Rightarrow\frac{n+1}{2^{n}}\leq\frac{1}{25}$

  $\therefore$     n=8