Question

Trevor has a bag of jelly beans. There are 10 red, 15 green, 5 orange, and 5 purple jelly beans in the bag. What is the probability that he randomly chooses 1 orange jellybean, eats it, and then chooses another orange jellybean and eats it

Answer 1

`(2)/(119)`

Explanation:

We have been given that Trevor has a bag of jelly beans. There are 10 red, 15 green, 5 orange, and 5 purple jelly beans in the bag.

The probability of choosing an orange jelly will be total number of orange jellies over all jellies.

`\text{P(Orange)}=(5)/(10+15+5+5)`

`\text{P(Orange)}=(5)/(35)`

Now, he again chooses an orange jelly and eats it. Since he already ate 1 orange jelly, so orange jelly left is 4 and total jellies would be 34.

`\text{P(Orange)}=(4)/(34)`

Using multiplication rule of probability, we will get:

`P(A\cap B)=P(A)\cdot P(B|A)`

`P(\text{Orange}\cap\text{Orange})= (5)/(35)* (4)/(34)`

`P(\text{Orange}\cap\text{Orange})=(1)/(7)* (2)/(17)`

`P(\text{Orange}\cap\text{Orange})= (2)/(119)`

Therefore, the probability that Trevor randomly chooses 1 orange jellybean, eats it, and then chooses another orange jellybean and eats it would be
`(2)/(119)`.