Multiple choice

A coin is tossed $n$ times. The probability of getting at least one head is greater than that of getting at least two tails by $\displaystyle \frac {5}{32}$. Then $n$ is

  1. $5$
  2. $10$
  3. $15$

  4. none of these
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A Correct answer
Explanation

P(at least 1 head) = 1 - (1/2)^n. P(at least 2 tails) = 1 - P(0 tails) - P(1 tail) = 1 - (1/2)^n - n(1/2)^n. Difference = (1 - (1/2)^n) - (1 - (1/2)^n - n(1/2)^n) = n(1/2)^n = 5/32. For n=5, 5(1/32) = 5/32.

AI explanation

The probability of getting at least one head is 1 minus the probability of getting all tails, which equals 1 minus (1/2)^n. Getting at least two tails is the complement of getting zero tails or one tail, meaning its probability is 1 minus ((1/2)^n plus n times (1/2)^n). The difference between these two probabilities is given by the equation ((n+1) times (1/2)^n) minus (1/2)^n equals 5/32. Simplifying this yields n times (1/2)^n equals 5/32, or n times 2^n equals 32. The only value of n that satisfies this equation is 5.

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