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Coin Probability of Exactly 30 Points: m + n = 107 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
A man throws a fair coin repeatedly. He gets 10 points for each head and 5 points for each tail. If the probability that he gets exactly 30 points is mn\dfrac{m}{n}, gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to
A5353
B5555
C107107correct
D105105
Solution
Step 1: To total exactly 30 points, the (heads, tails) combinations solving 10h+5t=3010h+5t=30 are: 66 tails; 11 head + 44 tails; 22 heads + 22 tails; 33 heads. The probability sums the arrangements of each case:
P=(12)6+5!1! 4!(12)5+4!2! 2!(12)4+(12)3.P=\left(\tfrac12\right)^6+\frac{5!}{1!\,4!}\left(\tfrac12\right)^5+\frac{4!}{2!\,2!}\left(\tfrac12\right)^4+\left(\tfrac12\right)^3.
 Step 2: Evaluate each term:
P=164+532+38+18.P=\frac{1}{64}+\frac{5}{32}+\frac{3}{8}+\frac{1}{8}.
 Step 3: Common denominator 6464:
P=1+10+24+864=4364=mn.P=\frac{1+10+24+8}{64}=\frac{43}{64}=\frac{m}{n}.
 (Cross-check by the recursion qk=12qk−1+12qk−2q_k=\tfrac12 q_{k-1}+\tfrac12 q_{k-2} for reaching a multiple of 5: q6=4364q_6=\tfrac{43}{64}.) Step 4: With m=43, n=64m=43,\ n=64,
m+n=43+64=107.m+n=43+64=107.
 Correct answer: (3)
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