Multiple choice

Six coins are tossed simultaneously. The probability of getting at least $4$ heads is

  1. $\dfrac { 11 } { 64 }$
  2. $\dfrac { 11 } { 32 }$
  3. $\dfrac { 15 } { 44 }$
  4. $\dfrac { 21 } { 32 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total outcomes = 2^6 = 64. Favorable outcomes (at least 4 heads) = 4 heads (6C4=15) + 5 heads (6C5=6) + 6 heads (6C6=1) = 15 + 6 + 1 = 22. Probability = 22/64 = 11/32.

AI explanation

Using the binomial distribution formula, the probability of getting exactly r heads in n tosses is nCr * (1/2)^n. The probability of getting at least 4 heads includes the probabilities of getting exactly 4, exactly 5, and exactly 6 heads. The calculations are 6C4 * (1/2)^6 + 6C5 * (1/2)^6 + 6C6 * (1/2)^6, which equals 15/64 + 6/64 + 1/64 = 22/64 = 11/32.

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