Mr. Abu has a box of 7 blue marbles and 3 red marbles. 2 marbles were selected randomly without replacement. What is the possibility that,

Question

(1) Both marbles were blue

(2) both marbles were red.

(3) The first marble was blue and the second marble was red.

Answer 1

Hello Issahaku,

1. Use the multiplication rule. The probability of the first marble being blue is 7/10. To get both blues, we must assume the first draw was a blue. That means we have one fewer blue marble and one fewer marble overall for the second draw, so the probability of the second marble being blue given the first one is blue is 6/9. Now, to the multiplication rule:

P = (7/10)*(6/9) = 0.467

2.This is essentially the same procedure. Probability of first marble being red is 3/10. If we want both reds, we have to assume the first draw was a red. That means we have one fewer red and one fewer total marbles, so probability of the second marble being red is 2/9. Multiplication Rule:

P= (3/10)*(2/9)

P= 0.067

#3. Recall that the probability of the first marble being blue is 7/10. With a marble removed, the probability of the second marble being red is 3/9. Use the Multiplication Rule:

P = (7/10)*(3/9)

P = 0.233

I hope this helps.