A die is thrown 4 times and getting 3 is considered a success. The probability of 2 successes is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{25}{216}\)
The number of successes X follows a binomial distribution with
\( \displaystyle n = 4,\quad p = \frac{1}{6},\quad q = \frac{5}{6} \)
\( \displaystyle P(X = 2) = {}^{4}C_{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 \)
\( \displaystyle = 6 \times \frac{1}{36} \times \frac{25}{36} \)
\( \displaystyle = \frac{150}{1296} = \frac{25}{216} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{25}{1296}\) — Incorrect. This is \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2\) alone, the probability of one particular order; the factor \({}^{4}C_{2} = 6\) for the positions of the two successes is missing.
Option (2) \(\displaystyle\frac{1250}{72}\) — Incorrect. It is greater than 1, so it cannot be a probability.
Option (4) \(\displaystyle\frac{25}{72}\) — Incorrect. \(\frac{25}{72} = \frac{450}{1296}\), not \(\frac{150}{1296}\).