Two fair four-faced dice, each with faces numbered 1, 2, 3 and 4, are thrown together. Let X denote the sum of the numbers on the faces on which they land.
(A) \(P(X = 5) = \displaystyle\frac{1}{4}\)
(B) \(E(X) = 5\)
(C) \(P(X \ge 7) = \displaystyle\frac{1}{4}\)
(D) \(\mathrm{Var}(X) = \displaystyle\frac{5}{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
There are \(4 \times 4 = 16\) equally likely outcomes. Counting the pairs for each sum gives the probability distribution:
| \(X\) | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| \(P(X)\) | \(\frac{1}{16}\) | \(\frac{2}{16}\) | \(\frac{3}{16}\) | \(\frac{4}{16}\) | \(\frac{3}{16}\) | \(\frac{2}{16}\) | \(\frac{1}{16}\) |
The table is symmetric about 5, so E(X) = 5. Also
\( \displaystyle E(X^{2}) = \frac{4 + 18 + 48 + 100 + 108 + 98 + 64}{16} = \frac{440}{16} = 27.5 \)
\(\mathrm{Var}(X) = 27.5 - 5^{2} = 2.5\)
So \(P(X = 5) = \frac{1}{4}\), \(E(X) = 5\), \(P(X \ge 7) = \frac{3}{16}\) and \(\mathrm{Var}(X) = \frac{5}{2}\).
Checking the statements
(A) \(P(X = 5) = \frac{1}{4}\) — Correct. (1,4), (2,3), (3,2), (4,1) give 4 of the 16 outcomes.
(B) \(E(X) = 5\) — Correct. Each die has mean 2.5, and 2.5 + 2.5 = 5.
(C) \(P(X \ge 7) = \frac{1}{4}\) — Incorrect. Only (3,4), (4,3), (4,4) qualify, so it is \(\frac{3}{16}\).
(D) \(\mathrm{Var}(X) = \frac{5}{2}\) — Correct. \(27.5 - 25 = 2.5\).