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(5 marks) Consider randomly selecting 5-cards from a standard 52-card deck: Let the random variable, X, denote the number of aces in your hand. Given you have at le...

Question

(5 marks) Consider randomly selecting 5-cards from a standard 52-card deck: Let the random variable, X, denote the number of aces in your hand. Given you have at least one ace, what is the probability you have two aces?

(5 marks) Consider randomly selecting 5-cards from a standard 52-card deck: Let the random variable, X, denote the number of aces in your hand. Given you have at least one ace, what is the probability you have two aces?



Answers

What is the probability that a five-card poker hand contains at least one ace?

Okay. What is the probability that a five card poker hand content exactly one X so the number or put the number of possible outcomes, he's going to be equal two out of 52 cards. We're gonna choose life of them. So the reason why it's 52 truths five instead of 50 to pick five is because the ordering off the cards do not matter. So since the ordering doesn't matter, we don't use it. We used shoes. Now the number off favorable outcomes, favorable outcomes. Well, in this case, we want a We want exactly one ace. So you imagine we already take one ace out of the bed. Didn't we have four spots left to feel? So it Since one car is taking out off the deck, there's 51 card left and off the 51 card. We want a feeling four spaces. So then the probability is 51 choose four divided by 15 to choose by. So then this is going to be equal to 303,243. You bought it by 10,829

Well Why did the number of aces in the five curves? Then we know that wife knows that hyper geometric ridges for ocean with capital and vehicles 52 R equals full, can smile and eagles flock. Then we can compute the probability that Y equals four and Y equals three and this year's the number. And then we want to compute the conditional probability that is a probability that hand contains all four is is if it is known that it contains and needs three aces. So there's conditional probability vehicles. The probability of Y equals fall over the probability of Y equals three plus the probability of what he calls for which has 48 over 48 plus two times 48 times 47 And verticals 1/95. Its approximate value is .010 five Street

So we want to know if we have a deck of 52 cards. What is the probability that we pick up on a CE? Well, we have 52 cards and each card has equal probability of being chosen since the cards are selected at random. Therefore each card has a probability of one over 52. Well, we have four aces. Therefore we have, ah, four over 52 probability that we will select an ace at random. Four over 52 is equal to one over 13 and that's our probability that we pick up a a sz.

Okay, so we want the probability that second is a C given first was not a C. And so then we've taken one court out. So we'll have 51 on the bottom and then since will still have all four aces. Where will be on the top? I can't be divided by anything were simplified, so that's all right.


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