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Chapter 2: Problem 107

Find the probability that a face card is drawn on the first draw and an ace on the second in two consecutive draws, without replacement, from a standard deck of cards.

Short Answer

Expert verified
The probability of drawing a face card on the first draw and an ace on the second draw in two consecutive draws, without replacement, from a standard deck of cards is \(\frac{12}{663}\).

Step by step solution

01

Calculate probability of drawing a face card on the first draw

The probability of drawing a face card on the first draw is: \(P(Face \, Card) = \frac{Number \, of \, Face \, Cards}{Total \, number \, of \, cards}\) In a standard deck, there are 12 face cards and 52 cards in total. So the probability of drawing a face card is: \(P(Face \, Card) = \frac{12}{52}\)
02

Calculate the probability of drawing an ace on the second draw given a face card was drawn on the first draw

As the face card is not replaced, there are now 51 cards left in the deck, and still 4 aces. We can calculate the probability of drawing an ace on the second draw given a face card was drawn on the first draw: \(P(Ace \, | \, Face \, Card) = \frac{Number \, of \, Aces}{Remaining \, number \, of \, cards}\) So, the probability of drawing an ace on the second draw given a face card was drawn on the first draw is: \(P(Ace \, | \, Face \, Card) = \frac{4}{51}\)
03

Calculate the probability of both events occurring

Now we can find the total probability by multiplying the probabilities for each draw: \(P(First \, Draw \, Face \, Card \, and \, Second \, Draw \, Ace) = P(Face \, Card) \times P(Ace \, | \, Face \, Card)\) Using the probabilities we calculated in Steps 1 and 2: \(P(First \, Draw \, Face \, Card \, and \, Second \, Draw \, Ace) = \frac{12}{52} \times \frac{4}{51}\)
04

Simplify and present final probability

The probability can be further simplified as follows: \(P(First \, Draw \, Face \, Card\, and \, Second \, Draw \, Ace) = \frac{12}{52} \times \frac{4}{51} = \frac{3}{13} \times \frac{4}{51}=\frac{3 \times 4}{13\times51} = \frac{12}{663}\) The final probability of drawing a face card on the first draw and an ace on the second draw in two consecutive draws, without replacement, from a standard deck of cards is \(\frac{12}{663}\).

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Most popular questions from this chapter

Find the probability that three successive face cards are drawn in three successive draws (without replacement) from a deck of cards. Define Events \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) as follows: Event A: a face card is drawn on the first draw, Event B: a face card is drawn on the second draw. Event \(C\) : a face card is drawn on the third draw.

A box contains 4 black marbles, 3 red marbles, and 2 white marbles. What is the probability that a black marble, then a red marble, then a white marble is drawn without replacement?

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Twenty percent of the employees of a company are college graduates. Of these, \(75 \%\) are in supervisory position. Of those who did not attend college, \(20 \%\) are in supervisory positions. What is the probability that a randomly selected supervisor is a college graduate?

A coin is tossed nine times. What is the total number of possible outcomes of the nine-toss experiment? How many elements are in the subset" 6 heads and 3 tails"? What is the probability of getting exactly 6 heads and 3 tails in nine tosses of thus unbiased coin?
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