# i) Find the probability of getting an even number on the first die or a total of 8 in a single throw of two dice. ii) Two cards are drawn at random from a pack of 52 cards.What is the probability that the cards are either both aces or both black cards. (1) There are 36 items in the sample space -- all possible throws of 2 dice. The probability that the first die is even is 1/2 as 18 of the possible 36 throws have an even first die.

The probability of throwing an 8 has an event space of (2,6),(3,5),(4,4),(5,3),(6,2)...

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(1) There are 36 items in the sample space -- all possible throws of 2 dice. The probability that the first die is even is 1/2 as 18 of the possible 36 throws have an even first die.

The probability of throwing an 8 has an event space of (2,6),(3,5),(4,4),(5,3),(6,2) ; so the probability of throwing an 8 is 5/36. (The number of things in the event space divided by the size of the sample space.)

The probability of (even first die) or (8) is the sum of the probabilities less the number of items in the intersection.

Thus the probability is 1/2+5/36-3/36=20/36=5/9

(2) The probability of the first card being an ace is 4/52 or 1/13. The probability of the second card being an ace (assuming the first is an ace) is 3/51 or 1/17. (There are 3 aces left, and 51 cards to choose from.) So the probability that there are two aces is 1/13*1/17=1/221

The probability that the first card is black is 1/2. The probability that the second card is black (assuming the first card is black) is 25/51. So the probability that both cards are black is 1/2*25/51=25/102.

The probability of (2 aces) or (2 black cards) is the sum of the probabilities minus the intersection. Thus the probability is 1/221+25/102-1/1326. (There are 12 ways to pull two aces, 2 of which are all black. Thus the probability that you have selected 2 black aces is 2/2652=1/1326; there are a total of 52*51=2652 different 2 card selections.)

So the probability is 1/221+25/102-1/1326=55/221

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