Question

Jordan has 10 candies in a bowl. There are 3 blue candies, 2 red candies, and the rest are green. Without looking, he grabs one candy, puts it back, and then grabs another candy. What is the probability that both candies he grabs are blue?

Answer 1

Answer: $\frac{9}{100}$

Step-by-step explanation:

Given: The total number of candies in the bowl = 10

The number of blue candies = 3

The probability to grab first candy he grab is blue is given by:-

$P(B)=\frac{\text{blue candies}}{\text{total candies}}\\\\\Rightarrow\ P(B)=\frac{3}{10}$

Since, before the second candy drawn he put back the first ball in the bowl , therefore both the events of grabbing blue candy are independent.

The probability that both candies he grabs are blue is given by:-

$P(BB)=P(B)\times P(B)\\=\frac{3}{10}\times\frac{3}{10}=\frac{9}{100}$

Answer 2

Answer:  The required probability is 9%.

Step-by-step explanation:  Given that Jordan has 10 candies in a bowl, out of which 3 are blue, 2 are red and rest are green. Jordan randomly grabs one candy, puts it back, and then grabs another one.

We are to find the probability that both the candies he grabs are blue.

Let S denotes the sample space for the experiment of choosing a candy from the jar.

Then, n(S) = 10.

Also, let A an B denote the events that the first candy is blue and second candy grabbed is blue respectively.

Then, n(A) = 3  and  n(B) = 3.

Therefore, the probability that both the candies are blue is given by

$p=\dfrac{n(A)}{n(S)}\times \dfrac{n(A)}{n(S)}=\dfrac{3}{10}\times\dfrac{3}{10}=\dfrac{9}{100}=9\%.$

Thus, the required probability is 9%.