Question

Consider a different experiment: EXPERIMENT: A bag contains 3 red balls, 4 blue balls, 4 green balls and 3 yellow balls. One ball is drawn at random and then returned to the bag. Then a ball is drawn again, possibly the same one or a different one.

(a) Specify the sample space (set of outcomes) and probability distribution of the outcomes.
(b) What is the probability that both balls drawn were red balls?
(c) Starting over with all 14 balls back in the bag, what is the probability that both draws were balls of different colors?

Answer 1

a) Sample space = RR, RB, RG, RY, BB, BR, BG, BY, GG, GR, GB, GY, YY, YR, YB, YG .

probability distribution of outcomes

= (3/14*3/14), (3/14*4/14), (3/14*4/14), (3/14*3/14), (4/14 *4/14 ), (4/14*3/14), (4/14 * 4/14), (4/14 * 3/14 ), ( 4/14 *4/14 ), (4/14*3/14), ( 4/14 *4/14 ),

(4/14*3/14), (3/14 *3/14 ), (3/14*3/14), (3/14 *4/14 ), ( 3/14 * 4/14 )

b) 0.0459

c) ( 3/14 * 4/14 ) + ( 3/14 * 4/14 ) + ( 3/14 * 3/14 ) + ------ ( 3/14 * 4/14 )

Explanation:

Total number of balls = 3 + 4 + 4 + 3 = 14 balls

3 Red balls = 3/14

4 blue balls = 4/14

4 green balls = 4/14

3 yellow balls = 3/14

a) Determine the sample space and probability distribution of outcomes

since ball is drawn with replacement

Sample space = RR, RB, RG, RY, BB, BR, BG, BY, GG, GR, GB, GY, YY, YR, YB, YG .

i.e. The sample space has 16 possible combination

probability distribution of outcomes

= (3/14*3/14), (3/14*4/14), (3/14*4/14), (3/14*3/14), (4/14 *4/14 ), (4/14*3/14), (4/14 * 4/14), (4/14 * 3/14 ), ( 4/14 *4/14 ), (4/14*3/14), ( 4/14 *4/14 ),

(4/14*3/14), (3/14 *3/14 ), (3/14*3/14), (3/14 *4/14 ), ( 3/14 * 4/14 )

b) Determine P( R and R )

P ( R and R ) = 3/14 * 3/14

= 9/196 = 0.0459

C) Determine probability of drawing two different balls

P( R and B ) + P ( R and G ) + P ( R and Y ) + ------ P ( Y and G )

= ( 3/14 * 4/14 ) + ( 3/14 * 4/14 ) + ( 3/14 * 3/14 ) + ------ ( 3/14 * 4/14 )