Multiple choice

Seven unbiased coins are tossed $128$ times. In how many throws would you find at least three heads?

  1. $99$
  2. $102$
  3. $103$
  4. $104$
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A Correct answer
Explanation

Probability of at least 3 heads in 7 tosses is 1 - P(0, 1, or 2 heads). P(X=k) = C(7,k) * (0.5)^7. P(0)=1/128, P(1)=7/128, P(2)=21/128. Sum = 29/128. P(at least 3) = 1 - 29/128 = 99/128. In 128 throws, expected count is 99.

AI explanation

The probability of getting exactly k heads when seven unbiased coins are tossed is found using the binomial distribution formula, with the probability of heads being 1/2. We calculate the probabilities of getting zero, one, and two heads as 1/128, 7/128, and 21/128, respectively. The probability of getting at least three heads is 1 minus the sum of these probabilities, yielding 1 minus 29/128, which equals 99/128. Multiplying this probability by the 128 total throws gives an expected 99 throws with at least three heads.

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