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: Understanding the Concept:
Odds in favour \(m : n\) mean probability \(\dfrac{m}{m+n}\).

Step 2: Individual probabilities:
A: \(\dfrac27\), not safe \(\dfrac57\). B: \(\dfrac3{10}\), not safe \(\dfrac7{10}\). C: \(\dfrac6{17}\), not safe \(\dfrac{11}{17}\).

Step 3: Three cases for exactly two:
A and B safe, C not: \(\dfrac{2\cdot3\cdot11}{1190} = \dfrac{66}{1190}\).
A and C safe, B not: \(\dfrac{2\cdot7\cdot6}{1190} = \dfrac{84}{1190}\).
B and C safe, A not: \(\dfrac{5\cdot3\cdot6}{1190} = \dfrac{90}{1190}\).

Step 4: Add:
\[ \frac{66 + 84 + 90}{1190} = \frac{240}{1190} = \frac{24}{119} \]
Option (A). Options (B), (C), (D) are individual terms or arithmetic slips.

Final Answer:
The three cases add up to 240/1190 = 24/119. \[ \boxed{\text{(A) }\dfrac{24}{119}} \]