Question

A bag of fruit contains 3 apples and 2 oranges and 1 banana and 4 pears.Gerald will randomly selected two pieces of fruit one at a time from the bag and not put is back. What is the probability that the first piece of fruit Gerald selects will be a banana and the second piece of fruit will be a pear??

Answer 1

The probability that Gerald will first select a banana and then a pear from the bag without replacement is 2/45.

Step-by-step explanation:

To determine the probability that Gerald selects a banana first and then a pear without replacement, we have to consider the total number of possible outcomes for each draw and the favorable outcomes for the event.

For the first draw, the total number of fruits is 10 (3 apples + 2 oranges + 1 banana + 4 pears). The favorable outcome of drawing a banana is 1 since there's only one banana.

The probability of drawing a banana on the first draw is therefore 1/10. After drawing the banana, there are 9 fruits left in the bag with 4 pears among them.

The probability of then drawing a pear is 4/9. To find the total probability of both events happening in sequence (a banana first and then a pear), multiply the two probabilities:

P(banana first and pear second) = P(banana first) × P(pear second)
= (1/10) × (4/9)
= 4/90
= 2/45.

The simplification process shows that the probability Gerald will first select a banana and then a pear is 2/45.

Answer 2

1/4 would be the probability