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Question

108

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A) 136/345	B) 17/87
C) 316/435	D) 158/435

Answer: C) 316/435

Explanation:

$n (s) = {}^{30}C_{2}$

Let A be the event of getting two oranges and

B be the event of getting two non-defective fruits.

and $(A \cap B)$ be the event of getting two non-defective oranges

$∴ P (A) = \frac{{}^{20}C_{2}}{{}^{30}C_{2}}, P (B) = \frac{{}^{22}C_{2}}{{}^{30}C_{2}} a n d P (A \cap B) = \frac{{}^{15}C_{2}}{{}^{30}C_{2}}$

$∴ P (A \cup B) = P (A) + P (B) - P (A \cap B)$

= $\frac{{}^{20}C_{2}}{{}^{30}C_{2}} + \frac{{}^{22}C_{2}}{{}^{30}C_{2}} - \frac{{}^{15}C_{2}}{{}^{30}C_{2}} = \frac{316}{435}$

Subject: Probability

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