Question

The probability of getting a total of 5 at least once in three tosses of a pair of fair dice is:

• A. \(1 - \left(\dfrac{8}{9}\right)^3\)
• B. \(1 - \left(\dfrac{3}{7}\right)^3\)
• C. \(\left(\dfrac{8}{9}\right)^3\)
• D. \(\left(\dfrac{3}{7}\right)^3\)

Hint:

Use complementary probability for "at least once" problems: \(1 - P(\text{not happening})^n\)

Answer 1

The Correct Option is A

There are \(6 \times 6 = 36\) outcomes when two dice are tossed.
Favorable outcomes for sum = 5:
(1,4), (2,3), (3,2), (4,1) → Total = 4 outcomes.
Probability of sum 5 in one trial = \( \dfrac{4}{36} = \dfrac{1}{9} \)
Probability of NOT getting 5 in one trial = \(1 - \dfrac{1}{9} = \dfrac{8}{9}\)
Probability of NOT getting 5 in all 3 tosses = \( \left( \dfrac{8}{9} \right)^3 \)
So, probability of getting it at least once = \[ 1 - \left( \dfrac{8}{9} \right)^3 \]