JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

A man throws a fair coin repeatedly. He gets $10$ points for each head he throws and $5$ points for each tail he throws. If the probability that he gets exactly $30$ points is $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $m + n$ is equal to :

A
$53$
B
$55$
C
$107$
D
$105$

Solution

$P = (6 tail) + (4 tail + 1 head) + (2 tail + 2 head) + (3 head)$

$P = (\frac{1}{2})^{6} + \frac{5 !}{1 ! 4 !} (\frac{1}{2})^{5} + \frac{4 !}{2 ! 2 !} (\frac{1}{2})^{4} + (\frac{1}{2})^{3}$

$P = \frac{1}{64} + \frac{5}{32} + \frac{3}{8} + \frac{1}{8}$

$P = \frac{1 + 10 + 24 + 8}{64}$

$P = \frac{43}{64} = \frac{m}{n}$

$m + n = 43 + 64 = 107$

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