Question

Martin puts two bowls of fruit out for his friends. The bowl of grapes has $9$ green grapes and $16$ red grapes. The bowl of tangerines has $7$seeded and $3$ seedless tangerines. Martin's friend chooses a random grape and a random tangerine. What is the probability that she chooses a green grape and a seedless tangerine?

Answer 1

Finding the probability that she chooses a green grape and a seedless tangerine:

Given,

The bowel one contains $9$ green grapes and $16$ red grapes.

The bowel two contains $7$seeded and $3$ seedless tangerines.

Therefore,

The total grapes in bowl one is $9+16=25$

The total tangerines in bowel two is $7+3=10$

We know that

$\mathrm{Probability}=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$

Probability of getting green grape and seedless tangerine $=\left(\frac{\text{number of green grapes}}{\text{total number of grapes}}\right)\times \left(\frac{\text{number of seedless tangerines}}{\text{total number of tangerines}}\right)$

$$\begin{gathered} =\left(\frac{9}{25}\right)\times \left(\frac{3}{10}\right) \\ =\frac{27}{250} \end{gathered}$$

Hence, the required probability is $\frac{27}{250}$.