JEE Main 2026 — Mathematics Probability & Statistics

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

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

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

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

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

For each pair, the number of possible sequences of heads and tails is given by $^{h+t}C_{h}$h+tCh​, and the probability of any specific sequence is $\left(\dfrac{1}{2}\right)^{h+t}$(21​)h+t.

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

For $h = 3, t = 0$h=3,t=0, the probability is $^{3}C_{3} \left(\dfrac{1}{2}\right)^{3} = 1 \times \dfrac{1}{8} = \dfrac{8}{64}$3C3​(21​)3=1×81​=648​.

For $h = 2, t = 2$h=2,t=2, the probability is $^{4}C_{2} \left(\dfrac{1}{2}\right)^{4} = 6 \times \dfrac{1}{16} = \dfrac{24}{64}$4C2​(21​)4=6×161​=6424​.

For $h = 1, t = 4$h=1,t=4, the probability is $^{5}C_{1} \left(\dfrac{1}{2}\right)^{5} = 5 \times \dfrac{1}{32} = \dfrac{10}{64}$5C1​(21​)5=5×321​=6410​.

For $h = 0, t = 6$h=0,t=6, the probability is $^{6}C_{0} \left(\dfrac{1}{2}\right)^{6} = 1 \times \dfrac{1}{64} = \dfrac{1}{64}$6C0​(21​)6=1×641​=641​.

Total probability $= \dfrac{8}{64} + \dfrac{24}{64} + \dfrac{10}{64} + \dfrac{1}{64} = \dfrac{43}{64}$=648​+6424​+6410​+641​=6443​.

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

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

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

Answer: $107$107