Multiple choice

What is the least number of times a fair coin needs to be flipped so that the probability of getting at least two tails exceeds 90%?

  1. 4

  2. 5

  3. 6

  4. 7

  5. 8
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The probability of getting at least two tails in n flips is 1 - P(0 tails) - P(1 tail). P(0 tails) = (1/2)^n, and P(1 tail) = n * (1/2)^n. We need 1 - (1+n)/2^n > 0.9, which simplifies to (1+n)/2^n < 0.1. Testing values, for n=6, 7/64 = 0.109; for n=7, 8/128 = 0.0625, which is less than 0.1.

AI explanation

Let n be the number of coin flips; the probability of getting at most one tail is the sum of zero tails and one tail. Using the binomial probability formula, this sum is (1/2 to the power of n) plus n multiplied by (1/2 to the power of n). For the probability of at least two tails to exceed 90 percent, the probability of at most one tail must be less than 10 percent. Testing n equals 6 gives 7/64, which is about 10.94 percent, and testing n equals 7 gives 8/128, which equals 6.25 percent. Therefore, the least number of flips needed is 7.

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