# help on a probability problem

• Feb 7th 2010, 09:19 AM
ilc
help on a probability problem
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:

a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------

a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?

b) (26/52)x(23/52) = 650/2704

C) (4/52)x(3/52) = 12/2704

d) (26/52)x(23/52) = 2/2704

how would i do e?
• Feb 7th 2010, 09:28 AM
e^(i*pi)
Quote:

Originally Posted by ilc
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:

a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------

a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?

b) (26/52)x(23/52) = 650/2704

C) (4/52)x(3/52) = 12/2704

d) (26/52)x(23/52) = 2/2704

how would i do e?

Where did you get 23/52 from in part B? Surely removing 1 red card would mean that 25 remain?

Is the first card replaced? I have assumed it is not

All your second terms should have a denominator of 51.

For example in B:

$\displaystyle \frac{26}{52} \times \frac{25}{51} = \frac{25}{102}$

And in C:

$\displaystyle \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$

Then part E would be

$\displaystyle \frac{25}{102} + \frac{1}{221}$
• Feb 7th 2010, 09:37 AM
ilc
yes, that should be 25...that was a typo. thanks!
• Feb 7th 2010, 09:42 AM
Quote:

Originally Posted by ilc
Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:

a) both cards are clubs
b) both cards are red
c) both cards are queens
d) both cards are red queens
e) both cards are queens or both cards are red
-----------

a) 13 clubs out of 52 = 13/52
there will be 12 clubs left after picking the first one so: 12/52
so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?

b) (26/52)x(23/52) = 650/2704

C) (4/52)x(3/52) = 12/2704

d) (26/52)x(23/52) = 2/2704

how would i do e?

Hi ilc,

in making these calculations, you are replacing the first card.
This is incorrect as you are choosing 2 from 52,
hence after picking the first card, you have 51 left.

Therefore redo parts a), b) and c).

In part d), you must count the number of red queens.

For part e),

You can calculate the probabilities of getting 2 red cards and add to the probability of getting 2 queens....

but you must subtract the probability of getting 2 red queens as this is a conditional probability question. They will have already been accounted for.
There is overlap since some queens are red.
• Feb 7th 2010, 10:04 AM
ilc
how would i get the probability of the two red queens?

2/53 x 1/51 = 2/2652 correct?
• Feb 7th 2010, 10:13 AM
You only need to know how a pack is organised.

There are 26 red and 26 black.

There are 4 queens, 2 are red and 2 are black.

Therefore 2 of the 26 red cards are red queens.

Therefore there are 24 red cards that are not queens.

There are another 2 queens which are black.

If both cards are queens or both are red, then we must include the probability of also getting 2 black queens.
Also, the queens can be a black and red.
However if we include the 2 red queens, we must be aware that these are already counted,
since the 2 red queens have been already included in 2 red cards total.

Hence we can add the probabilities of 2 red cards to the probability of getting 2 queens,
but we must subtract the probability of getting 2 red queens.
• Feb 7th 2010, 10:33 AM
$\displaystyle \frac{2}{52}\ \frac{1}{51}$
Or choose both red queens at once....$\displaystyle \frac{\binom{2}{2}}{\binom{52}{2}}$