A fair coin is tossed three times and the outcomes are noted. What is the probability of getting exactly two heads ?
[amp_mcq option1=”\(\frac{2}{3}\)” option2=”\(\frac{1}{2}\)” option3=”\(\frac{5}{8}\)” option4=”\(\frac{3}{8}\)” correct=”option4″]
This question was previously asked in
UPSC CAPF – 2024
Each toss has two possible outcomes: Heads (H) or Tails (T).
Since the tosses are independent, the total number of possible outcomes for three tosses is 2 * 2 * 2 = 2³ = 8.
The sample space (set of all possible outcomes) is:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
We are looking for the probability of getting exactly two heads.
The outcomes with exactly two heads are:
HHT
HTH
THH
There are 3 favourable outcomes.
The probability of an event is defined as (Number of favourable outcomes) / (Total number of possible outcomes).
Probability of getting exactly two heads = (Number of outcomes with exactly two heads) / (Total number of outcomes)
Probability = 3 / 8.
In this case, n=3 (three tosses), k=2 (exactly two heads), and p=0.5 (probability of getting a head with a fair coin).
P(X=2) = C(3, 2) * (0.5)² * (1-0.5)³⁻²
P(X=2) = C(3, 2) * (0.5)² * (0.5)¹
P(X=2) = C(3, 2) * (0.5)³
C(3, 2) = 3! / (2! * (3-2)!) = 3! / (2! * 1!) = (3 * 2 * 1) / ((2 * 1) * 1) = 3.
(0.5)³ = 0.5 * 0.5 * 0.5 = 0.125 = 1/8.
P(X=2) = 3 * (1/8) = 3/8.
This confirms the result obtained by listing the outcomes.