Question

One out of 9 ships is likely to sink when they are set on sail. When 6 ships are set on sail, the probability that exactly 3 of them will not arrive safely is:

• A. \( 1 - \frac{1}{9^6} \)
• B. \( {}^6C_3 \cdot \frac{8^3}{9^6} \)
• C. \( \frac{25 \times 8^3}{9^5} \)
• D. \( {}^6C_3 \cdot \frac{8}{9^6} \)

Hint:

Use binomial distribution formula: \( {}^nC_r p^r q^{n-r} \) where \( p \) is failure and \( q \) is success.

Answer 1

The Correct Option is B

Probability a ship sinks = \( \frac{1}{9} \), hence the probability it arrives safely = \( \frac{8}{9} \). We are to find the probability that exactly 3 ships out of 6 do not arrive safely. This is a binomial probability: \[ P(X = 3) = {}^6C_3 \cdot \left(\frac{1}{9}\right)^3 \cdot \left(\frac{8}{9}\right)^3 = {}^6C_3 \cdot \frac{8^3}{9^6} \]