1. In an experiment, a ball is drawn from an urn containing 13 purple balls and 10 red balls. If the ball is purple, two coins are tossed. Otherwise three coins are tossed.
a) How many elements of the sample space will have a purple ball ?
b) How many elements of the sample space are there altogether ?
2. A 5-card poker hand is dealt from a well shuffled regular 52-card playing card deck. Find the probability that the hand is a Four of a kind (4 cards of the same face value).
3.
maths problems?
2016-09-27 8:57 am
回答 (1)
2016-09-27 5:40 pm
1a) Just count:
1 Purple ball and two heads
2 Purple ball, one heads and one tails
3 Purple ball and two tails
That gives us 3 elements with a purple ball (not all with the same probability of occurring).
1b) There are 4 elements of the sample space with red balls, because a toss of three coins can result in
three heads, three tails, two heads and one tails, or one heads and two tails.
So there are 3 + 4 = 7 elements in the sample space altogether.
2. There are 52C5 = 52! / (47! 5!)
= 52 * 51 * 50 * 49 * 48 / (5 * 4 * 3 * 2 * 1)
= 52 * 51 * 5 * 49 * 4
= 2,598,960 possible 5-card hands.
[If you need any help with combinations formulas, their use, calculation, or derivation,, see
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
for an extremely thorough and clear tutorial.]
A hand containing four of a kind has four cards of one rank (13 possibilities)
combined with any one of the remaining 48 cards, so there are
13 * 48 = 624 such hands possible.
Thus, the probability is
624 / 2,598,960
= 1 / 4,165
1 Purple ball and two heads
2 Purple ball, one heads and one tails
3 Purple ball and two tails
That gives us 3 elements with a purple ball (not all with the same probability of occurring).
1b) There are 4 elements of the sample space with red balls, because a toss of three coins can result in
three heads, three tails, two heads and one tails, or one heads and two tails.
So there are 3 + 4 = 7 elements in the sample space altogether.
2. There are 52C5 = 52! / (47! 5!)
= 52 * 51 * 50 * 49 * 48 / (5 * 4 * 3 * 2 * 1)
= 52 * 51 * 5 * 49 * 4
= 2,598,960 possible 5-card hands.
[If you need any help with combinations formulas, their use, calculation, or derivation,, see
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
for an extremely thorough and clear tutorial.]
A hand containing four of a kind has four cards of one rank (13 possibilities)
combined with any one of the remaining 48 cards, so there are
13 * 48 = 624 such hands possible.
Thus, the probability is
624 / 2,598,960
= 1 / 4,165
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