QJEEAdv15Mab958: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Mab958 Solution: The minimum number of times a fair coin needs to be tossed, so…

Let the coin is tossed $n$ times.

$\because p$ (at least 2 heads) $=1-[p$ (one heads) $+p$ (No heads) $\}$

As we know, by binominal probability theorem the probability of getting $r$ success in $n$ trials with $p$ being the probability of success and $q$ be the probability of failure, is given by ${ }^n \mathrm{C}_r(p)^r(q)^{n-r}$ .

Let head be considered as the success and tail be the failure probability of getting head in a toss $=p=\frac{1}{2}$ and probability of getting tail in a toss $=q=\frac{1}{2}$

$\begin{aligned}& \therefore \mathrm{P}(\text { one head })={ }^n \mathrm{C}_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{n-1} \\& ={ }^n \mathrm{C}_1\left(\frac{1}{2}\right)^n \\& \Rightarrow \mathrm{P} \text { (one head) }=n\left(\frac{1}{2}\right)^n \\& \text { Similarly, } \mathrm{P}(\text {No heads})={ }^n \mathrm{C}_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^n \\& \Rightarrow \mathrm{P}(\text { No heads })=\left(\frac{1}{2}\right)^n \quad\left\{\because n_{c_o}=1\right\} \\& \therefore \mathrm{P} \text { (at least } 2 \text { heads) }=1-\left\{n\left(\frac{1}{2}\right)^n+\left(\frac{1}{2}\right)^n\right\}\end{aligned}$

$\begin{aligned}& \qquad=1-\frac{(n+1)}{2^n} \\& \because P(\text { at least } 2 \text { heads }) \geq 0.96 \\& \Rightarrow 1-\frac{(n+1)}{2^n} \geq 0.96 \\& \Rightarrow 0.04 \geq \frac{n+1}{2^n} \\& \Rightarrow \frac{1}{25} \geq \frac{n+1}{2^n} \\& \Rightarrow \frac{2^n}{n+1} \geq 25 \\& \Rightarrow n \geq 8\end{aligned}$

So, the minimum number of times a fair coin to be tossed is 8.

Hint:

(i) P (at least 2 heads) $=1-\{\mathrm{P}$ (one head} + $\mathrm{P}$ (No heads)}

(ii) Use binomial probability theorem and simplify it.