(1 point) You are to pick 4 cards randomly from a condensed deck of cards that c
ID: 3307233 • Letter: #
Question
(1 point) You are to pick 4 cards randomly from a condensed deck of cards that contains four suits - , ,4and following denomations: Ace, 2, 3, 4, 5, 6,7, 8, 9, and 10. There are no face-cards in this deck. What is the probability that you will get Part (a) a three-of-a-kind? P(3 of a kind) .015 lli (use four decimals) Part (b) What is the probability that all four cards are of the same suite? For example, all four cards are Vs? P(all the same suit)= 0.0092 ll (use four decimals) Part (c) What is the probability you get one-pair? P(one pair)= .4412 (use four decimals) Part (d) What is the probability you get two aces and two 10s? PC2-aces and 2-10s)= 0.0004 EEE (use four decimals) Part (e) What is the probability you get two aces and two Os? P(2-aces and 2-0)= 0.0055 (use four decimals)Explanation / Answer
Total number of ways to choose 4 cards out of 10 = 40C4 = 91390.
(a) The three of a kind can come in 10 ways and among them in 4C3 = 4 ways.
The remaining card can come in 36 ways
=> Number of ways to get three of a kind = 10 * 4 * 36 = 1440.
Probability = 1440 / 91390 = 0.0158.
(b) There are 10C4 ways to select four cards among the same suit from each suit.
= 4 * 10C4 = 4 * 210 = 840
Probability = 840 / 91390 = 0.0092.
(c) The kind of the same pair can come in 10 ways and there are 4C2 options to make a pair
=> Number of ways to obtain a pair = 10 * 6 = 60
The other two numbers can come in 36 * 32 / 2 = 576 ways
Total number of ways to obtain pair = 60 * 576 = 34560
Probability = 34560/91390 = 0.3782.
(d) The two aces can come in 4C2 = 6 ways.
The two 10s can also come in 6 ways.
Probability = 6*6/91390 = 0.0004.
(e) Beware of the king of hearts!
The two hearts can come in 9C2 ways (ace is not allowed) and there are 3C2 aces (heart is allowed)
Total combinations = 45 * 3 = 135
Probability = 135 / 91390 = 0.0015
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