JEE Mains · Maths · STD 12 - 13. probability
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 \(\dfrac{m}{n}\), \(\gcd(m, n) = 1\), then \(m + n\) is equal to:
- A \(53\)
- B \(55\)
- C \(107\)
- D \(105\)
Answer & Solution
Correct Answer
(C) \(107\)
Step-by-step Solution
Detailed explanation
Let \(h\) be the number of heads and \(t\) be the number of tails obtained. The total points obtained is \(10h + 5t\). For the man to get exactly \(30\) points, we must have: \(10h + 5t = 30 \Rightarrow 2h + t = 6\) Since \(h\) and \(t\) are non-negative integers, the possible…
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