# Math Help - Answer Check - Probability

1. ## Answer Check - Probability

Q. Two cards are drawn from a deck of 52 playing cards. Find the probability of each of the following events occurring:

a) Both cards are clubs (3 marks)

N(S) = (52,2)
= 1326

C (13,2) = 78
P(E) = 78/1326
=39/663
=13/221
Therefore the probability that both cards are clubs is 13/221

b) Both cards are red (3 marks)

(36, 2) = 630
P (E) = 630/1326
=315/663
=105/221
Therefore, the probability that both are red is 105/221.

c) Both cards are queens (3 marks)

(4,2) = 6
P(E) = 6/1326
= 1/221
Therefore, the probability that both cards are queens is 1/221

d) Both cards are red queens (3 marks)

(2,2) = 1
P(E)= 1/1326
Therefore, the probability that both are red queens is 1/221.

e) Both cards are queens or both cards are red.

(32,2) + (4,2)
= 78 + 6
= 84
P(E) = 84/1326
= 42/663
= 14/221
Therefore the probability that both cards are queens or both are red is 14/221.

Please check and let me know if I am correct. THANK YOU

2. Originally Posted by cnmath16
Q. Two cards are drawn from a deck of 52 playing cards. Find the probability of each of the following events occurring:

a) Both cards are clubs (3 marks)

N(S) = (52,2)
= 1326

C (13,2) = 78
P(E) = 78/1326
=39/663
=13/221
Therefore the probability that both cards are clubs is 13/221

You have a 1 in 4 chance of picking a club to start with. Once you have picked a club, then there are 51 cards left in the pack, 12 of which are clubs. So you have a 12/51 chance. The odds of both occuring is $\frac{1}{4} \times \frac{12}{51}$.

b) Both cards are red (3 marks)

(36, 2) = 630
P (E) = 630/1326
=315/663
=105/221
Therefore, the probability that both are red is 105/221.

1/2 chance of picking a red card. Then there will be 25 red cards in a pack if 51. So chance of picking another one is 25/51. Total probability of both occuring: $\frac{1}{2} \times \frac{25}{51}$

c) Both cards are queens (3 marks)

(4,2) = 6
P(E) = 6/1326
= 1/221
Therefore, the probability that both cards are queens is 1/221

4 Queens. So 4/52 chance at first. Then 3/51. Total probability $\frac{4}{52} \times \frac{3}{51}$. You are correct here.

[b]
d) Both cards are red queens (3 marks)

(2,2) = 1
P(E)= 1/1326
Therefore, the probability that both are red queens is 1/221.

2 red queens. Chance of the first being a red queen is 2/52. Chance of 2nd being a red queen is 1/51. Total probability 2/52 x 1/51

e) Both cards are queens or both cards are red.

(32,2) + (4,2)
= 78 + 6
= 84
P(E) = 84/1326
= 42/663
= 14/221
Therefore the probability that both cards are queens or both are red is 14/221.

Please check and let me know if I am correct. THANK YOU
Yes.

3. Originally Posted by cnmath16
Q. Two cards are drawn from a deck of 52 playing cards. Find the probability of each of the following events occurring:

a) Both cards are clubs (3 marks)

N(S) = (52,2)
= 1326

C (13,2) = 78
P(E) = 78/1326
=39/663
=13/221
Therefore the probability that both cards are clubs is 13/221

b) Both cards are red (3 marks)

(36, 2) = 630
P (E) = 630/1326
=315/663
=105/221
Therefore, the probability that both are red is 105/221.

c) Both cards are queens (3 marks)

(4,2) = 6
P(E) = 6/1326
= 1/221
Therefore, the probability that both cards are queens is 1/221

d) Both cards are red queens (3 marks)

(2,2) = 1
P(E)= 1/1326
Therefore, the probability that both are red queens is 1/221.

e) Both cards are queens or both cards are red.

(32,2) + (4,2)
= 78 + 6
= 84
P(E) = 84/1326
= 42/663
= 14/221
Therefore the probability that both cards are queens or both are red is 14/221.

Please check and let me know if I am correct. THANK YOU
You and Mush are both correct for (a), (c) and (d) (different methods but both methods are valid and give the same correct answer).

-----------------------------------------------------------------------------

You're wrong for (b). Doing it your way, it should be (26, 2) = 325. NOT (36, 2). Then you'll get the same correct answer as Mush.

---------------------------------------------------------------------------

For (e) you have to be careful not to double count since two red Queens also gets counted as two red cards ....

So it's Pr(2 red cards) + Pr(2 Queens) - Pr(2 red Queens). That is, add (b) and (c) and subtract (d).

Both you and Mush have double counted.

4. Hello, cnmath16!

Two cards are drawn from a deck of 52 playing cards.
Find the probability of each of the following events occurring:

a) Both cards are Clubs (3 marks)

N(S) = (52,2) = 1326

C (13,2) = 78

P(E) = 78/1326 = 13/221
= 1/17 . . . Yes!

b) Both cards are Red (3 marks)

(36, 2) = 630 . . . no

We want: . $C(26,2) \:=\: 325$

Therefore: . $P(\text{both Red}) \:=\:\frac{325}{1326} \:=\:\frac{25}{51}$

c) Both cards are Queens (3 marks)

(4,2) = 6

P(E) = 6/1326 = 1/221
. . . Correct!

d) Both cards are Red Queens (3 marks)

(2,2) = 1

P(E)= 1/1326
. . . Yes!

e) Both cards are Queens or both cards are Red.

There are $C(4,2) = 6$ ways to get two Queens.
There are $C(26,2) = 325$ ways to get two Red cards.
. . But there is 1 way to get two Red Queens.

Hence, there are: . $6 + 325 - 1 \:=\:330$ ways to get two Queens or two Reds.

. . $P(E) \:=\:\frac{330}{1326} \:=\:\frac{55}{221}$