Question

A box contains a total of 12 crayons: 2 red, 3 green, 3 blue, 1 yellow, 2 purple, and 1 brown. Without looking, Frieda picks two crayons from the box. What is the probability that both will be blue?

Answer 1

Probability Question.

Red (R) = 2 Prob(R) = 2/12 = 1/6

Green (G) = 3 Prob(G) = 3/12 = 1/4

Blue (B) = 3 Prob(B) = 3/12 = 1/4

Yellow (Y) = 1 Prob(Y) = 1/12

Purple (P) = 2 Prob(P) = 2/12 = 1/6

Brown (Br) = 1 Prob(Br) = 1/12

TOTAL = 12

`\text{Probability =}\frac{no\text{ of required outcomes}}{no\text{ of total outcomes}}`

The probability of picking two blue crayons without looking is:

The first crayon being blue is : Prob(first Blue) = 3/12

The second crayon being blue is : Prob( second Blue) = 2/11 , since the first first is no longer in the box, that is, after the first selection of the first blue crayon, we were left with 2 Blue in the box and the total crayon has also dropped to 11 crayons.

`\begin{gathered} \text{Prob(BB) = Prob(first B) }*\text{ Prob(second B) } \\ \text{ = }(3)/(12)*(2)/(11)=(1)/(22) \end{gathered}`

Hence, the correct answer is 1/22