Given three identical bags each containing 10 balls with different colour compositions.

Question

Given three identical bags each containing 10 balls with different colour compositions. A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $p$; if the ball is Green, the probability that it is from Bag III is $q$. Find $\left(\dfrac1p+\dfrac1q\right)$.

*(Options as recorded)*

Accepted Answer

Using the bag compositions implied by the working (Bag I: 3 red, Bag II: 4 red, Bag III: 5 red out of 10; and correspondingly for green):

$$p(B_1/R) = \frac{\frac13\times\frac{3}{10}}{\frac13\times\frac{3}{10}+\frac13\times\frac{4}{10}+\frac13\times\frac{5}{10}} = \frac{1}{4} \implies \frac1p = 4$$

$$p(B_3/G) = \frac{\frac13\times\frac{4}{10}}{\frac13\times\frac{5}{10}+\frac13\times\frac{3}{10}+\frac13\times\frac{4}{10}} = \frac13 \implies \frac1q = 3$$

$$\frac1p+\frac1q = 4+3=7$$

Solutions · 1