1. A person has purchased 5 of 1,000 tickets sold in acertain raffle. To determi

Question

1. A person has purchased 5 of 1,000 tickets sold in acertain raffle. To determine the 10th prizewinners, 10 tickets aredrawn at random and without replacement. a) Compute the probability that this person wins exactly oneprize. b)Compute the probability that he wins at least oneprize. Hint: first compute the probability that he does not win aprize.
2. A bowl contains 20 chips, of which 9 are red, 8 are white,3 are blue. Six chips are taken at random and without replacement. Find the probability that a) each of the 6 chips is red. b) there are 3 red, 2 white, and 1 blue chip among the sixchips selected. c) none of the 6 chips selected is blue; and d) there is at least 1 white and at least 1 blue chip amongthe 6 selected. Hint: Consider the complement of at least 1 white and at least1 blue.
3. Compute the probability of being dealt, at random withoutreplacement, a 13-card bridge hand consisting of a) 4 spades, 4 hearts, 3 diamonds, and 2 clubs; b) 13 cards of the same suit.
a) Compute the probability that this person wins exactly oneprize. b)Compute the probability that he wins at least oneprize. Hint: first compute the probability that he does not win aprize.
2. A bowl contains 20 chips, of which 9 are red, 8 are white,3 are blue. Six chips are taken at random and without replacement. Find the probability that a) each of the 6 chips is red. b) there are 3 red, 2 white, and 1 blue chip among the sixchips selected. c) none of the 6 chips selected is blue; and d) there is at least 1 white and at least 1 blue chip amongthe 6 selected. Hint: Consider the complement of at least 1 white and at least1 blue.
3. Compute the probability of being dealt, at random withoutreplacement, a 13-card bridge hand consisting of a) 4 spades, 4 hearts, 3 diamonds, and 2 clubs; b) 13 cards of the same suit.

Explanation / Answer

For distinugishable widgets without replacement, where orderdoes not matter, use the (n choose k) = n!/(n!*(n-k)!) formulawhere n is total number of widgets and k is the number ofspots.
Total number of 13(spots) card hands from a standard52(widgets) card deck = 52 choose 13
a) 4 spades, 4 hearts, 3 diamonds, 2 clubs = (13 choose 4) *(13 choose 4) * (13 choose 3) * (13 choose 2) Divide the above number by (52 choose 13) to get probabilityof such a hand

b) 13 cards of the same suit = 4 * (13 choose 13) Divide the above number by (52 choose 13) to get probabilityof such a hand
I do not have a calculator on me so I am leaving thecalculations up to you.
Total number of 13(spots) card hands from a standard52(widgets) card deck = 52 choose 13
a) 4 spades, 4 hearts, 3 diamonds, 2 clubs = (13 choose 4) *(13 choose 4) * (13 choose 3) * (13 choose 2) Divide the above number by (52 choose 13) to get probabilityof such a hand

b) 13 cards of the same suit = 4 * (13 choose 13) Divide the above number by (52 choose 13) to get probabilityof such a hand
I do not have a calculator on me so I am leaving thecalculations up to you. Total number of 13(spots) card hands from a standard52(widgets) card deck = 52 choose 13
a) 4 spades, 4 hearts, 3 diamonds, 2 clubs = (13 choose 4) *(13 choose 4) * (13 choose 3) * (13 choose 2) Divide the above number by (52 choose 13) to get probabilityof such a hand

b) 13 cards of the same suit = 4 * (13 choose 13) Divide the above number by (52 choose 13) to get probabilityof such a hand
I do not have a calculator on me so I am leaving thecalculations up to you.

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