We shuffle a standard deck of cards obtaining a permutation(that is random over
ID: 2953219 • Letter: W
Question
We shuffle a standard deck of cards obtaining a permutation(that is random over all
52! possible permutations). Find the probability of thefollowing events:
i. The first two cards include at least oneace.
ii. The first ve cards include at least oneace.
iii. The first two cards are a pair of the samerank.
iv. The first ve cards are all diamonds.
v. The first ve cards form a full house (three of a rankand two of another rank).
(b) Shuffle the deck again and pick 5 cards randomly out of52. Compute the probability of
events (ii), (iv), (v) above, for these new 5 cards. Do theanswers in (b) agree or disagree
with the answers in (a) and why?
Explanation / Answer
a) i P(ace) = 4/52 = 1/13 P(not ace) = 12/13 P(neither of first two cards areaces) = 12/13 * 12/13 = 144/169 P(at least one of the first twocards are aces) = 1 - 144/169 = 25/169. ii P(neither of first five cards are aces) = 12/13 ^5 ˜0.67. P(at least one of the first five cards are aces)˜ 1 - 0.67 ˜ 0.33. iii Given the rank of the first card, there are then 12 cards leftin the deck of 51 that will match it. Thus the answer is 12/51 or4/17. iv P(first card diamond) = 13/52 = 1/4 P(second card diamond | first card diamond) =12/51 = 4/17 etc. Thus the required probability is given by 13/52 * 12/51 * 11/50 *10/49 * 9/48 ˜ 0.0005. v In a similar way to iii or iv, P(three of a kind) = 4/52 * 3/51 *2/50 ˜ 0.0002 and P(two of a different kind) = 4/52 * 3/51 ˜ 0.004525 As the order doesn't matter, we also multiply by 5! to representthe different arrangements possible. Thus P(full house) = P(three of a kind) * P(two of a differentkind) *5! ˜ 0.0002 * 0.004525 * 120 ˜ 9.828 x 10-5. b) The position in the pack from where the cards are taken makes nodifference, so the answers should be the same as in part a.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.