A coin is tossed 3 times. The probability of getting exactly 2 heads

A coin is tossed 3 times. The probability of getting exactly 2 heads is

[amp_mcq option1=”$\frac{1}{3}$” option2=”$\frac{3}{8}$” option3=”$\frac{1}{2}$” option4=”$\frac{5}{8}$” correct=”option2″]

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UPSC CAPF – 2022

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When a fair coin is tossed, there are two possible outcomes: Heads (H) or Tails (T). When tossed 3 times, the total number of possible outcomes is 2³ = 8. The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. We are interested in the outcomes with exactly 2 heads. These are {HHT, HTH, THH}. There are 3 favorable outcomes. The probability of getting exactly 2 heads is the number of favorable outcomes divided by the total number of outcomes: 3/8.

– List all possible outcomes (sample space) for the given number of trials.
– Identify the outcomes that satisfy the condition (favorable outcomes).
– Probability = (Number of favorable outcomes) / (Total number of outcomes).
– For independent events like coin tosses, the total number of outcomes for ‘n’ trials is 2^n.

This type of problem can also be solved using the binomial probability formula: P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where n is the number of trials (3), k is the number of successes (2 heads), p is the probability of success on a single trial (0.5 for heads), and C(n, k) is the binomial coefficient “n choose k”.
C(3, 2) = 3! / (2! * 1!) = 3.
P(exactly 2 heads) = C(3, 2) * (0.5)² * (0.5)¹ = 3 * 0.25 * 0.5 = 3 * 0.125 = 0.375 = 3/8.

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