The following questions are about a game in which you draw r

The following questions are about a game in which you draw randomly from a bag containing 4 red balls. 4 blue balls, 3 green balls anti 3 yellow balls. The ball is not replaced after each draw, and each ball in the bag is equally likely to be selected. If you draw two balls from the bag without replacing them, what Is the probability of drawing two red balls? Starting over with all 14 balls back in the bag, what Is the probability of drawing two balls that are the same color?

Solution

a) The probability of choosing one red ball out of 14 balls= 4/14.

Now in the bag the number of balls is 13 and the number of red balls is 3.The probability of choosing one red ball out of 13 balls and now the number of red balls is 3 is 3/13

so the total probability of choosing 2 red balls is(4/14) *(3/13)

=6/91.

b) The probability of choosing 2 balls with same color implies (probability of choosing 2 red balls) or( probability of choosing 2 blue balls) or ( probability of choosing 2 green balls) or (probability of choosing 2 yellow balls)

probablility of choosing 2 red balls=(4/14) *(3/13) =(12/182)

probability of choosing 2 blue balls =(4/14)*(3/13) = (12/182)

probability of choosing 3 green balls= (3/14)*(2/13) = (6/182)

probability of choosing 2 yellow balls = (3/14)*(2/13) = (6/182)

total probability = (12/182) +(12/182) + (6/182) + (6/182) = (18/91).