In card games we assume a well shuffled deck, where the odds of any card in a de
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Question
In card games we assume a well shuffled deck, where the odds of any card in a deck of n cards has a probability of 1/n of being drawn. The following problems deal with two card hand from a standard 52 card deck, so the denominator will always be 522, which is 52 times 51. Again, give the final answer as "1 in x", where x is a whole number. The probability of drawing two aces in a two card hand. ___________ The probability of drawing any pair in a two card hand. ___________ The probability of drawing two spades in a two card hand. ___________ The probability of drawing two cards of the same suit in a two card hand. ___________ If no one is a group is born on February 29, the probability of no one in a group of n people sharing a birthday is 365n/365n. It was mentioned in class that n = 23 the smallest number where the probability of no shared birthday is less than 50% at approximately 0.4927.... Find the values smallest values of n where the odds of no shared birthday fall under the given thresholds. The smallest n such that the p(no shared birthday) lessthanorequalto 60% n = _______ The smallest n such that the p(no shared birthday) lessthanorequalto 70% n = _______ The smallest n such that the p(no shared birthday) lessthanorequalto 80% n = _______ *The smallest n such that the p(no shared birthday) lessthanorequalto 90% n = _______Explanation / Answer
Solution for the first problem
Concept Back-up
A standard 52-card deck, consists of 4 suits, Club, Spade, Diamond and Heart and in each suit there are 13 cards – 9 ‘number cards, bearing numbers 2, 3, ….., 10’ and 4 ‘face cards, Jack, Queen, King and Ace’
A pair would imply: (2, 2), (3, 3), ……,(10, 10), (J, J), ….., (A, A).
Q1 Part (i)
Probability of two aces:
The first card can be any one of the four aces in 4 ways. Once the first card is drawn, there are only 3 ace cards left and hence the second card can also be an ace only in 3 ways. So, the probability =(4 x 3)/(52 x 51) = 1/221. ANSWER: 1 in 221
Q1 Part (ii)
Probability of any pair:
The pair can be (2, 2), (3, 3), ……,(10, 10), (J, J), ….., (A, A) – 13 possibilities.
Now, for any pair, say (2, 2), it can be 2 of Club and 2 of Spade, 2 of Club and 2 of Diamond, 2 of Club and 2 of Heart, 2 of Spade and 2 of Diamond, 2 of Spade and 2 of Heart or 2 of Diamond and 2 of Heart – 6 possibilities and in each of these, the order can be reversed in 2 ways. Thus, in total there are (13 x 6 x 2) possibilities and hence the probability = (13 x 6 x 2)/ (52 x 51)
= 1/17. ANSWER: 1 in 17
Q1 Part (iii)
Probability of two spades:
The first card can be any one of the 13 spade cards in 13 ways. Once the first card is drawn, there are only 12 spade cards left and hence the second card can also be a spade only in 12 ways. So, the probability =(13 x 12)/(52 x 51) = 1/17. ANSWER: 1 in 17.
Q1 Part (iv)
Probability of same suit:
The same suit can be (C, C), (S, S), (D, D) or (H, H) – 4 possibilities.
Now, for any suit pair, say (C, C), the first can be any one of the 13 cards and then the second has to be only any one of the remaining 12 cards of the same suit.
Thus, in total there are (4 x 13 x 12) possibilities and hence the probability = (4 x 13 x 12)/ (52 x 51) = 4/17. ANSWER: 4 in 17
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