1. ## 2 questions need 2 answers

1. acard is drawn from adeck of cards .suppose that A and B represent the following two events [/font]
A =the cards drawn is an ace. 4/52
B = the card drawn is adiamond.13/52
Are A and B are disjoint?
Two events A and b are disjoint (mutually exclusive) if A∩B=Ø
ii. Compute the probability that at least one of the two events A or B occur .

2. abox contains one white ,three black and four green balls .
i. what is the probability of drawing a white ball followed by agreen ball without returning the first ball to the box .

2. Originally Posted by compufatwa
1. acard is drawn from adeck of cards .suppose that A and B represent the following two events [/font]
A =the cards drawn is an ace. 4/52
B = the card drawn is adiamond.13/52
Are A and B are disjoint?
Two events A and b are disjoint (mutually exclusive) if A∩B=Ø
ii. Compute the probability that at least one of the two events A or B occur .

2. abox contains one white ,three black and four green balls .
i. what is the probability of drawing a white ball followed by agreen ball without returning the first ball to the box .
1. A and B are not disjoint, because the Ace of diamonds is in both sets. The probability of getting either an ace or a diamond is 16/52 (which is 4/13) because there are 13 diamonds, and 3 aces which are not diamonds, this gives you 16 cards that are in AUB (A unioned with B)

2. There are 8 balls in the box, you need to draw a white one, there is only 1 white one, so you have 1/8 of getting a white ball in 1 pick. Now there are 7 balls in the box, you need a green one, 4 of the 7 are green, so you have a 4/7 chance of getting a green ball on the second pick. So 1/8 * 4/7 = 4/56 = 1/14
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