# Thread: [SOLVED] Word Problem - Probability

1. ## [SOLVED] Word Problem - Probability

Hello Forum, I need help with the following,

Five cards are chosen at random from an ordinary deck of playing cards. In how many ways can the cards be chosen under each of the following conditions?

a. All are hearts. b. All are the same suit.

c. Exactly three are kings. d. Two or more are aces.

2. Originally Posted by AlgebraicallyChallenged
Hello Forum, I need help with the following,

Five cards are chosen at random from an ordinary deck of playing cards. In how many ways can the cards be chosen under each of the following conditions?

a. All are hearts. b. All are the same suit.

c. Exactly three are kings. d. Two or more are aces.
In each case you should think about where the bits of each answer have come from:

(a) $\displaystyle ^{13}C_5 \cdot ^{39}C_0$.

(b) 4 times (a).

(c) $\displaystyle ^{4}C_3 \cdot ^{48}C_1$.

(d) $\displaystyle ^{52}C_5 -$ (number of ways of getting less than 2 aces).

3. Hello, AlgebraicallyChallenged!

Since you seem to be unable to do any of these,
. . I'll give you a walk-through ... with baby steps.

I assume you know about Combinations.

Five cards are chosen at random from an ordinary deck of playing cards.
In how many ways can the cards be chosen under each of the following conditions?

a. All are Hearts.
We want 5 of the 13 Hearts.

There are: .$\displaystyle C(13,5) \;=\;_{13}C_5 \;=\;{13\choose5} \;=\;\frac{13!}{5!\,8!} \;=\;1287$ ways to get Five Hearts.

b. All are the same suit ( a "flush").

There are 4 choices for the suit.

And there are: .$\displaystyle C(13,5) \:=\:1287$ to get 5 cards of that suit.

Therefore, there are: .$\displaystyle 4 \times 1287 \:=\:5148$ flushes.

c. Exactly three are kings.

There are 4 Kings ... and 48 non-Kings.

We want 3 of the 4 Kings: .$\displaystyle C(4,3) = 4$ ways.

We want 2 of the 48 non-Kings: .$\displaystyle C(48,2) \:=\: 1128$ ways.

Therefore, there are: .$\displaystyle 4 \times 1128 \:=\:4512$ ways to get Three Kings.

d. Two or more are aces.
This can be done by the "complement" approach; I'll do it head-on.

There are 4 Aces and 48 Others.

There are three cases to consider . . .

Two Aces
There are $\displaystyle C(4,2) = 6$ ways to choose the two Aces.
There are $\displaystyle C(48,3) = 17,\!296$ ways to get three Others.
. . Hence, there are: .$\displaystyle 6 \times 17,\!296 \:=\:103,\!776$ ways to get Two Aces.

Three Aces
There are $\displaystyle C(4,3) = 4$ ways to choose the three Aces.
There are $\displaystyle C(48,2) = 1128$ ways to get two Others.
. . Hence, there are: .$\displaystyle 4 \times 1128 \:=\:4512$ ways to get Three Aces.

Four Aces
There ae $\displaystyle C(4,4) \ 1$ way to choose the four Aces.
There are $\displaystyle C(48,1) = 48$ ways to get one Others.
. . Hence, there are: .$\displaystyle 1 \times 48 \:=\:48$ ways to get Four Aces.

Therefore, there are: .$\displaystyle 103,\!776 + 4512 + 48 \:=\:108,\!336$ ways to get two or more Aces.