Question

Three ships A, B, C sail from England to India. If the odds in favour of their safe arrival are 2:5, 3:7 and 6:11 respectively, then the probability that the exactly two ships arrive safely is

• A. \(\frac{24}{119}\)
• B. \(\frac{24}{1190}\)
• C. \(\frac{66}{1190}\)
• D. \(\frac{84}{1190}\)

Hint:

Convert odds to probabilities and add the three two-safe cases.

Answer 1

The Correct Option is A

Step 1: Complement idea:
P(0 safe) = $\tfrac57\cdot\tfrac7{10}\cdot\tfrac{11}{17} = \tfrac{385}{1190}$. P(3 safe) = $\tfrac27\cdot\tfrac3{10}\cdot\tfrac6{17} = \tfrac{36}{1190}$.

Step 2: P(exactly 1 safe):
$\tfrac{2\cdot7\cdot11 + 5\cdot3\cdot11 + 5\cdot7\cdot6}{1190} = \tfrac{154 + 165 + 210}{1190} = \tfrac{529}{1190}$.

Step 3: Exactly 2:
$1 - \tfrac{385 + 36 + 529}{1190} = \tfrac{1190 - 950}{1190} = \tfrac{240}{1190} = \tfrac{24}{119}$, option (A).

Final Answer:
The probability is 24/119. \[ \boxed{\text{(A) }\dfrac{24}{119}} \]