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Method 1: We correct your attempt. As you observed, the probability that at least one ace is drawn is $$\Pr(\text{at least one ace is drawn}) = \frac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51} = \frac{33}{221}$$ If we draw two aces, they must be of different suits. This event can occur in $4 \cdot 3$ ways. If we draw one of the four aces with the first draw, there are $51 - 3 - 12 = 36$ cards left in the deck which are not aces and are of a different suit than the ace. Hence, the number of ways of drawing an ace and then drawing a non-ace of a different suit is $4 \cdot 36$. If we draw one of the $48$ non-aces with the first draw, there are $3$ aces left in the deck with a different suit than the non-ace we drew first. Hence, the number of ways of drawing a non-ace and then drawing an ace of a different suit is $48 \cdot 3$. Hence, the probability that the two cards are of different suits and at least one ace is drawn is $$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51} = \frac{25}{221}$$ Hence, the probability that the two cards are of different suits given that at least one ace is drawn is \begin{align*} \Pr(\text{of different suits} \mid & \text{at least one ace is drawn})\\ \qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\ \qquad & = \frac{\dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51}}{\dfrac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51}}\\ \qquad & = \frac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}\\ \qquad & = \frac{25}{33} \end{align*} Method 2: We work with combinations to avoid considering the order in which the cards are drawn, which simplifies the calculations. The probability of drawing at least one ace when two cards are drawn is $$\Pr(\text{at least one ace}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}$$ since we must either select two of the four aces or one of the four aces and one of the $48$ non-aces while selecting two of the $52$ cards in the deck. The probability that the two cards are of different suits and at least one ace is drawn is $$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}} = \frac{25}{221}$$ since either two aces are drawn or one ace and one of the $36$ non-aces that are of a different suit than the ace are drawn when two cards are drawn from the deck of $52$ cards. The probability of drawing two cards of different suits given that at least one ace has been selected is \begin{align*} \Pr(\text{of different suits} & \mid \text{at least one ace is drawn})\\ \qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\ \qquad & = \frac{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}}}{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}}\\ \qquad & = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}\\ \qquad & = \frac{25}{33} \end{align*} Note: Observe that in both methods, after we cancel the common denominators, the numerator reduces to the number of cases in which at least one ace is drawn and cards are of different suits and the denominator reduces to the number of cases in which at least one ace is drawn. This is because the sample space for the conditional probability that the cards are of different suits given at least one ace is drawn is the set of cases in which at least one ace is drawn.
If two cards are drawn from a pack without replacement, what is the probability of getting A. An ace and a king B. Two aces ?
In Mathematics 4 Answers Available Asked by Nana mohammed on 9th July, 2020  |
If two cards are drawn from a pack without replacement, what is the probability of getting two aces
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