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

Answer:

$\frac{2}{119}$

Step-by-step 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.

$\text{P(Orange)}=\frac{5}{10+15+5+5}$

$\text{P(Orange)}=\frac{5}{35}$

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.

$\text{P(Orange)}=\frac{4}{34}$

Using multiplication rule of probability, we will get:

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

$P(\text{Orange}\cap\text{Orange})= \frac{5}{35}\times \frac{4}{34}$

$P(\text{Orange}\cap\text{Orange})=\frac{1}{7}\times \frac{2}{17}$

$P(\text{Orange}\cap\text{Orange})= \frac{2}{119}$

Therefore, the probability that Trevor randomly chooses 1 orange jellybean, eats it, and then chooses another orange jellybean and eats it would be $\frac{2}{119}$.