Question

On a lunch counter, there are 3 oranges, 5 apples, and 2 bananas. If 3 pieces of fruit are selected, find the probability that 1 orange, 1 apple, and 1 banana are selected (does not have to be in that order).

Answer 1

Answer:

$Probability = \frac{1}{24}$

Step-by-step explanation:

Given

$Oranges = 3$

$Apples = 5$

$Banana = 2$

$Total = 10$

Required

Determine the probability of 1 orange, 1 apple and 1 banana

Since, order is not important:

$Probability = P(Orange) * P(Apple) * P(Banana)$

$Probability = \frac{3}{10} * \frac{5}{9} * \frac{2}{8}$

The difference in the numerator is as a result of picking the fruit without replacement

$Probability = \frac{30}{720}$

$Probability = \frac{1}{24}$