Question

A fair die is thrown three times. What is the probability of getting exactly two 6s?

• A. \(\frac{5}{72}\)
• B. \(\frac{1}{18}\)
• C. \(\frac{1}{12}\)
• D. \(\frac{7}{72}\)

Hint:

Think about how many ways you can pick which single throw is the "non-six" among the three throws, then count outcomes for each face.

Answer 1

The Correct Option is A

Step 1: Total outcomes when a fair die is thrown 3 times $= 6^3 = 216$.
Step 2: "Exactly two 6s" means exactly one of the three throws is NOT a 6. Choose which throw is the non-6 throw: this can happen in 3 ways (1st, 2nd, or 3rd throw).
Step 3: For each such arrangement, the two chosen throws must show 6 (1 way each), and the remaining throw must show any of the 5 non-6 faces.
Step 4: Number of favourable outcomes $= 3 \times 1 \times 1 \times 5 = 15$.
Step 5: Probability $=\dfrac{\text{favourable}}{\text{total}} = \dfrac{15}{216} = \dfrac{5}{72}$.
\[\boxed{\dfrac{5}{72}}\]