Question

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards. What is the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind)? (Round your answer to four decimal places.)

Answer 1

Answer:

0.0083

Step-by-step explanation:

The gambler will have full house if he is dealt two kings or ace and a king.Now, there are 47 cards left in the deck and two which are aces and three are king.

The probability of these event are $\frac{3C_2}{47C_2}$

and  $\frac{3C_1\times 2C_1}{47C_2}$  respectively. So, the probability of a full house is given as:

$\frac{3C_2}{47C_2}+\frac{3C_1\times 2C_1}{47C_2}$

=0.0083

Answer 2

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

• If he is given with two kings.
• If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  $\frac{^{3}C_2 }{^{47}C_2}$   {∵ one king is already there}

              =  $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$           {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

              =  $\frac{3}{1081}$  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

                                                                   =  $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

                                                                   =  $\frac{6}{1081}$  =  0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

                                                                                    =  $\frac{9}{1081}$ = 0.0083.