Question

A basket contains 5 apples and 7 oranges, and another basket contains 4 apples and 8 oranges. If one fruit is picked out at random from each basket, then the probability of getting one apple and one orange is

• A. $\frac{1}{6}$
• B. $\frac{7}{18}$
• C. $\frac{17}{36}$
• D. $\frac{19}{36}$

Hint:

For probability of mutually exclusive events (e.g., one apple and one orange), sum the probabilities of all favorable cases.

Answer 1

The Correct Option is C

Basket 1: 5 apples, 7 oranges (total 12). Basket 2: 4 apples, 8 oranges (total 12). We need the probability of picking one apple and one orange. Case 1: Apple from Basket 1, orange from Basket 2: \[ P(\text{A}_1, \text{O}_2) = \frac{5}{12} \cdot \frac{8}{12} = \frac{40}{144} = \frac{5}{18} \] Case 2: Orange from Basket 1, apple from Basket 2: \[ P(\text{O}_1, \text{A}_2) = \frac{7}{12} \cdot \frac{4}{12} = \frac{28}{144} = \frac{7}{36} \] Total probability: \[ \frac{5}{18} + \frac{7}{36} = \frac{10 + 7}{36} = \frac{17}{36} \] Option (3) is correct. Options (1), (2), and (4) do not match the computed probability.