For problems 11-14, 3 consecutive cards are drawn from a deck of 52 cards. Find
ID: 3175291 • Letter: F
Question
For problems 11-14, 3 consecutive cards are drawn from a deck of 52 cards. Find the probability of each of the following: Exactly 2 red cards are drawn with replacement. The possibilities are RRB, RBR, or BRR. In each case the probability is (1/2)^3 = 1/8, so P(RRB, RBE, or BRR) = 1/8 + 1/8 + 1/8 = 3/8 3 face cards are drawn without replacement. There are 12 face cards in the deck, so the probability is 12/52 mid dot 11/51 mid dot 10/50 = 11/1105 = 0.0100 3 black cards are drawn with replacement. There are 26 black cards, so the probability is (26/52)^3 = 1/8 No even numbered cards are drawn without replacement. There are 32 non-even cards, so the probability is 32/52 mid dot 31/51 mid dot 30/50 = 248/1105Explanation / Answer
The questions are already answered . but the explanations are as follows
with replacement mean that once card is drawn , it is kept back in the deck
now out of a pack of 52 cards , 26 are red and 26 are black , so proabability of getting a red is 26/52 = 12
and for black is 26/52 = 1/2
so if 3 cards are drwan with replacement , every time the probability is 1/2 for 2 red cards
1/2 * 1/2 * 1/2 = 1/8
next 3 face cards are drwan without replacement
so total face cards are 12
now 1 st draw 12/52
onc drwan 1 card is removed from the deck
so next face can be drwan with probability 11/51
3rd face with 10/50 , so total probability is 12/52 * 11/51 * 10/50
3 black cards with replacement
again 1 black card is picked with probabilty is 26/52 = 1/2
hence every time the card is kept back in the deck so we have
1/2 * 1/2 * 1/2 = 1/8
no even cards , without replacement
we have 20 even cards (2,4,6,8,10) , so total cards are 5*4 = 20
now non even cards are thus 52-20 = 32
hence the probability is
32/52 * 31/51 * 30/50 , as the cards are drawn without replacement
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