JEE Main 2026 — Probability Question with Solution

Let $h$ be the number of heads and $t$ be the number of tails obtained.

The total points obtained is $10h+5t$.

For the man to get exactly $30$ points, we must have:

$10h+5t=30\Rightarrow 2h+t=6$

Since $h$ and $t$ are non-negative integers, the possible pairs of $(h,t)$ are $(3,0)$, $(2,2)$, $(1,4)$, and $(0,6)$.

For each pair, the number of possible sequences of heads and tails is given by ${}^{h+t}{C}_h$, and the probability of any specific sequence is ${\left(\frac{1}{2}\right)}^{h+t}$.

The total probability is the sum of the probabilities of all these mutually exclusive cases:

For $h=3,t=0$, the probability is ${}^3{C}_3{\left(\frac{1}{2}\right)}^3=1\times \frac{1}{8}=\frac{8}{64}$.

For $h=2,t=2$, the probability is ${}^4{C}_2{\left(\frac{1}{2}\right)}^4=6\times \frac{1}{16}=\frac{24}{64}$.

For $h=1,t=4$, the probability is ${}^5{C}_1{\left(\frac{1}{2}\right)}^5=5\times \frac{1}{32}=\frac{10}{64}$.

For $h=0,t=6$, the probability is ${}^6{C}_0{\left(\frac{1}{2}\right)}^6=1\times \frac{1}{64}=\frac{1}{64}$.

Total probability $=\frac{8}{64}+\frac{24}{64}+\frac{10}{64}+\frac{1}{64}=\frac{43}{64}$.

Thus, $m=43$ and $n=64$.

Since $\gcd (43,64)=1$, the fraction is in its simplest form.

$m+n=43+64=107$.

Answer: $107$