Question

A bag contains apples and oranges, five in all and atleast one of each, all combinations being equally likely. If one fruit is selected at random from the bag, assuming all fruits are distinguishable,the probability that it is an orange is

• A. $\frac{1}{20}$
• B. $\frac{1}{10}$
• C. $\frac{1}{5}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

There are 4 possible cases (each case equally likely, hence probability of each is $\frac{1}{4}$).
Probability of selecting an orange is:
(i) 1 apple, 4 oranges: $\frac{4}{5}$
(ii) 2 apples, 3 oranges: $\frac{3}{5}$
(iii) 3 apples, 2 oranges: $\frac{2}{5}$
(iv) 4 apple, 1 orange: $\frac{1}{5}$
Therefore, probability = $\frac{4}{5}\ast \frac{1}{4}+\frac{3}{5}\ast \frac{1}{4}+\frac{2}{5}\ast \frac{1}{4}+\frac{1}{5}\ast \frac{1}{4}=\frac{1}{2}$