Question

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
• D. 105

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Let's simplify the points system by dividing everything by 5.
So, Heads (H) gives 2 points, and Tails (T) gives 1 point.
We now need the probability of obtaining exactly 6 points in this scaled system.
Step 2: Key Formula or Approach:
The man can reach exactly 6 points in multiple sequences of coin tosses.
Let's list all possible combinations of Heads and Tails that sum to exactly 6 points.
- \(0\) Heads, \(6\) Tails: \(0(2) + 6(1) = 6\)
- \(1\) Head, \(4\) Tails: \(1(2) + 4(1) = 6\)
- \(2\) Heads, \(2\) Tails: \(2(2) + 2(1) = 6\)
- \(3\) Heads, \(0\) Tails: \(3(2) + 0(1) = 6\)
Step 3: Detailed Explanation:
Now, we calculate the probability for each valid combination using permutations.
Case 1: 6 Tails in a total of 6 tosses.
The number of ways is \(\frac{6!}{6!} = 1\).
The probability is \(1 \times \left(\frac{1}{2}\right)^6 = \frac{1}{64}\).
Case 2: 1 Head, 4 Tails in a total of 5 tosses.
The number of ways is \(\frac{5!}{1! 4!} = 5\).
The probability is \(5 \times \left(\frac{1}{2}\right)^5 = \frac{5}{32} = \frac{10}{64}\).
Case 3: 2 Heads, 2 Tails in a total of 4 tosses.
The number of ways is \(\frac{4!}{2! 2!} = 6\).
The probability is \(6 \times \left(\frac{1}{2}\right)^4 = \frac{6}{16} = \frac{24}{64}\).
Case 4: 3 Heads, 0 Tails in a total of 3 tosses.
The number of ways is \(\frac{3!}{3!} = 1\).
The probability is \(1 \times \left(\frac{1}{2}\right)^3 = \frac{1}{8} = \frac{8}{64}\).
The total probability of scoring exactly 30 points (or scaled 6 points) is the sum of these probabilities.
\[ P = \frac{1}{64} + \frac{10}{64} + \frac{24}{64} + \frac{8}{64} = \frac{43}{64} \] We are given that this probability is \(\frac{m}{n} = \frac{43}{64}\).
Since 43 is a prime number, \(\gcd(43, 64) = 1\) is satisfied.
Therefore, we deduce that \(m = 43\) and \(n = 64\).
Finally, we calculate \(m + n = 43 + 64 = 107\).
Step 4: Final Answer:
The value of \(m + n\) is \(107\).