Maths Q8 – JEE-MAIN 2026

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 :

Clue 1 of 3: Identify Combinations

Determine all possible non-negative integer pairs of (Heads, Tails) that sum up to exactly 30 points, given the point values for each outcome.

🧠 Full Solution Path

Step 1: Determine Combinations of Heads and Tails✦ Active

Let $H$ be the number of heads and $T$ be the number of tails. The points obtained are $10 H + 5 T$ . We are given that the total points are 30. So, $10 H + 5 T = 30$ . Dividing by 5, we get $2 H + T = 6$ . We need to find non-negative integer solutions for $(H, T)$ :

H = 0 ⟹ T = 6 (N = 6 throws)

H = 1 ⟹ T = 4 (N = 5 throws)

H = 2 ⟹ T = 2 (N = 4 throws)

H = 3 ⟹ T = 0 (N = 3 throws)

Step 2: Calculate Probability for Each Combination○ Expand

For a fair coin, the probability of a head is $P (H) = 1 / 2$ and a tail is $P (T) = 1 / 2$ . The probability of getting $H$ heads in $N$ throws is $(\binom{N}{H}) (\frac{1}{2})^{N}$ .

P (H = 0, T = 6) = (\binom{6}{0}) {(\frac{1}{2})}^{6} = 1 \cdot \frac{1}{64} = \frac{1}{64}

P (H = 1, T = 4) = (\binom{5}{1}) {(\frac{1}{2})}^{5} = 5 \cdot \frac{1}{32} = \frac{5}{32}

P (H = 2, T = 2) = (\binom{4}{2}) {(\frac{1}{2})}^{4} = 6 \cdot \frac{1}{16} = \frac{6}{16}

P (H = 3, T = 0) = (\binom{3}{3}) {(\frac{1}{2})}^{3} = 1 \cdot \frac{1}{8} = \frac{1}{8}

💡 Teacher's Secret Hint

Remember that the number of throws $N$ changes for each combination of heads and tails.

Step 3: Sum Probabilities and Find m+n○ Expand

The total probability of getting exactly 30 points is the sum of these individual probabilities:

P = \frac{1}{64} + \frac{5}{32} + \frac{6}{16} + \frac{1}{8}

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

P = \frac{1 + 10 + 24 + 8}{64} = \frac{43}{64}

We have $\frac{m}{n} = \frac{43}{64}$ . Since 43 is a prime number and 64 is $2^{6}$ , their greatest common divisor is $gcd (43, 64) = 1$ . Thus, $m = 43$ and $n = 64$ . The required value is $m + n$ :

m + n = 43 + 64 = 107

💡 Teacher's Secret Hint

Always simplify the fraction to its lowest terms before identifying $m$ and $n$ to ensure $gcd (m, n) = 1$ .

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Maths Question 8 – JEE-MAIN 2026

Clue 1 of 3: Identify Combinations

🧠 Full Solution Path