DM6.3: Problem 2 (1 pt) A card is drawn from a standard deck of 52 cards, the va
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DM6.3: Problem 2 (1 pt) A card is drawn from a standard deck of 52 cards, the value noted, and the card replaced. This is repeated 10 times. What is the probability that: a) Exactly 7 of the cards are diamonds? b) At least two of the cards are diamonds? c) At most 7 of the cards are diamonds? Note: You can earn partial credit on this problem. Preview A Submit Answers You have attempted this problem 2 times. Your overall recorded score is 096. You have unlimited attempts remaining ons Email instructor WeBWorK © 1996-201 1 The yieBWork Proiect ers?Explanation / Answer
Please note nCx = n! / [(n-x)!*x!]
In a deck of 52 Cards there are 13 diamonds and 39 other cards (Spades, clubs and Hearts)
In a draw of 10 cards, the total outcomes = 52C10 = 15820024220
(a) Exactly 7 diamonds means Out of 13 diamonds we get 7, and from the remaining 39 we get 3
13C7 * 39C3 = 1716 * 9139 = 15682524
Therefore the required probability = 15682524/15820024220 = 0.00099
(b) At Least 2 cards are diamonds. Means P(2 diamonds and 8 others + 3 diamonds and 7 others + ...+ 10 diamonds and 0 others). This is a long process. We know that the sum of probabilities = 1
Therefore At least 2 diamonds = 1 - P(1 diamond and 9 others + 0 diamonds and 10 others)
1 diamond and 9 others = 13C1 * 39C9 = 13 * 211915132 = 2754896716
0 diamonds and 10 others = 13C1 * 39C10 = 1 * 635745396 = 635745396
P( 1 diamond and 9 others + 0 diamonds and 10 others) = (2754896716+635745396)/15820024220 = 0.214
Therefore P( At least 2 diamonds) = 1 - 0.214 = 0.786
(c) At Most 7 = 1 - P(8 diamonds and 2 others + 9 diamonds and 1 another + 10 diamonds and 0 others)
8 diamond and 2 others = 13C8 * 39C2 = 1287*741 = 953667
9 diamond and 1 others = 13C9 * 39C1 = 715 * 39 = 27885
10 diamond and 0 others = 13C10 * 39C0 = 286 * 1 = 286
There P(At most 7) = 1 - (953667 + 27885 + 286)/15820024220 = 0.999
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