Mathematics, 18.06.2020 16:57 pradojosselinep34b1k

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.)

Answers: 2

Show answers

Answers

Answer from: Jehovel

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.

Answer from: twalalgreen26

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 from: Quest

hi there, i would be interested to you with your test, may you add the questions and answer options.

- : )

Answer from: Quest

b. e = 7.5h

hope this , lovely!

Another question on Mathematics

Mathematics, 21.06.2019 13:30

Which statement is not always true for a parallelogram? (a) opposite sides are congruent. (b) diagonals bisect each other. (c) it has 4 congruent angles. (d) consecutive angles are supplementary.

Answers: 1

Answer

Mathematics, 21.06.2019 19:00

You give a $4 tip for a meal that cost $22. approximately what percent of the bill was your tip?

Answers: 1

Answer

Mathematics, 21.06.2019 22:00

(05.02)a farmer has decided to divide his land area in half in order to plant soy and corn. calculate the area of the entire area so he knows how much soil is needed. a parallelogram with a height of 6 yards and side length 9 yards. the height forms a triangle with the slanted side of the rhombus with a base of 2.5 yards. rhombus is split into a soy half and a corn half. each bag of soil covers 40 square yards. how many bags should the farmer purchase?

Answers: 3

Answer

Mathematics, 22.06.2019 01:50

If f (x) = 2x2 +5/(x-2), complete the following statement: f(6) =

Answers: 3

Answer

You know the right answer?

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two a...

Questions