JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Probability

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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}, gcd (m, n) = 1

, then

m + n

is equal to :

(A)53
(B)55
(C)107✓ Correct
(D)105

Answer: (C)

Solution

A head adds

10

points and a tail adds

5

points, so every step adds a multiple of

5

; let

a_{k}

be the probability the running total ever equals

5 k

. A total of

5 k

is reached either from

5 (k - 1)

by a tail or from

5 (k - 2)

by a head, each with probability

\frac{1}{2}

, so

a_{k} = \frac{1}{2} a_{k - 1} + \frac{1}{2} a_{k - 2}, a_{0} = 1, a_{- 1} = 0.

This gives

a_{1} = \frac{1}{2}, a_{2} = \frac{3}{4}, a_{3} = \frac{5}{8}, a_{4} = \frac{11}{16}, a_{5} = \frac{21}{32}, a_{6} = \frac{43}{64}

. Since

30 = 5 \cdot 6

, the required probability is

\frac{43}{64}

, with

g cd (43, 64) = 1

, so

m + n = 43 + 64 = 107

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Probability

Solution

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