Multiple choice

An unbiased coin is tossed $10$ times. By using binomial distribution, find the probability of getting at least $3$ heads.

  1. $\dfrac{110}{128}$.
  2. $\dfrac{51}{63}$.
  3. $\dfrac{121}{128}$.
  4. $\dfrac{150}{176}$.
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C Correct answer
Explanation

Binomial distribution: P(X >= 3) = 1 - [P(0) + P(1) + P(2)]. P(k) = 10Ck * (0.5)^10. P(0) = 1/1024, P(1) = 10/1024, P(2) = 45/1024. Sum = 56/1024. P(X >= 3) = 1 - 56/1024 = 968/1024 = 121/128.

AI explanation

Using the binomial distribution, the probability of getting at least 3 heads is 1 minus the sum of the probabilities of getting exactly 0, 1, and 2 heads. The probability of a head on each toss is 1/2, so the calculation is 1 minus the quantity (1/1024 plus 10/1024 plus 45/1024). This simplifies to 1 minus 56/1024, giving a final probability of 121/128.

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