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: [tex]\frac{9}{100}[/tex]Step-by-step explanation:Given: The total number of candies in the bowl = 10The number of blue candies = 3The probability to grab first candy he grab is blue is given by:-[tex]P(B)=\frac{\text{blue candies}}{\text{total candies}}\\\\\Rightarrow\ P(B)=\frac{3}{10}[/tex]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:-[tex]P(BB)=P(B)\times P(B)\\=\frac{3}{10}\times\frac{3}{10}=\frac{9}{100}[/tex]