adding probability question

[kevinf2090]

this problem has been frustrating me for a few hours. can anybody help me. the probem is in a deck of 52 cards what is the probability of drawing 2 spades or 2 face cards. pleas help

 

[Schfifty Five]

P(A or B) = P(A) + P(B) - P(A and B)

Spades - P(A) 13/52

Face Cards - P(B) 12/52 (Jack, Queen, King for each suit)

P(A and B) = 3 spade cards with face cards, so subtract those b/c they would be repeating when adding up the probability.

So it's 22/52....I think. There is a chance I might be wrong though.

 

[kevinf2090]

the thing is that i know the answer is 47/442 but i just don't know how to get to it. nice try though

 

[Atomicus]

( 13/52 * 12/51 ) + ( 9/52 * 8/51 ) - (3/52 * 2/51)

 

[DrPizza]

First, how many ways are there to choose two cards?

How many ways are there to choose 2 spades?
How many ways are there to choose 2 face cards?

And, most importantly, is there any overlap between those two groups. (yes, there is... Qspades Kspades would be counted twice, so you'd have to subtract it.)

 

[chuckywang]

(13/52 * 12/51 ) + ( 9/52 * 8/51 ) - (3/52 * 2/51) = 37/442.

 

[Schfifty Five]

I think you need to specify if it's ONLY 2 spades, or ONLY 2 face cards, or 1 of each is...and how many times can you draw from the deck.

 

[DrPizza]

Oh, and don't forget, when you're chosing two cards, there are two possible orders to get them in...
Then again, if you forget, it'll all work out in the end; so long as you forget all of the time and not just some of the time 🙂

 

[Schfifty Five]

Originally posted by: Atomicus
( 13/52 * 12/51 ) + ( 9/52 * 8/51 ) - (3/52 * 2/51)

At least I got the forumla P(A) + P(B) - P(A and B) right...haha.

Just didn't get the individual #'s right inside for each prob.

 

[DrPizza]

Originally posted by: chuckywang
(13/52 * 12/51 ) + ( 9/52 * 8/51 ) - (3/52 * 2/51) = 37/442.

Or, Chucky's way...
First, he found the probability that the first card is a spade... then without replacing the card, the probability that the second card was a spade. Multiply to get the probability of 2 spades.

Then, he tried to find the probability of choosing 2 face cards. However, since he subtracted, I think he meant 12/52 * 11/51 for the probability of 2 face cards. What he subtracted was the probability of 2 spade face cards.

 

[DrPizza]

Hopefully you know there're 52 cards in a deck 😛 (I often have freshmen who don't know that)
Take a guess why after finding the probability of choosing the first spade (13/52), the probability of choosing another spade is only 12/51...

 

[chuckywang]

Originally posted by: DrPizza

Originally posted by: chuckywang
(13/52 * 12/51 ) + ( 9/52 * 8/51 ) - (3/52 * 2/51) = 37/442.

Or, Chucky's way...
First, he found the probability that the first card is a spade... then without replacing the card, the probability that the second card was a spade. Multiply to get the probability of 2 spades.

Then, he tried to find the probability of choosing 2 face cards. However, since he subtracted, I think he meant 12/52 * 11/51 for the probability of 2 face cards. What he subtracted was the probability of 2 spade face cards.

Grr...There are 12 face cards, not 9. That's the last time I blindly copy paste work from someone else's post without checking it.

(13/52 * 12/51 ) + ( 12/52 * 11/51 ) - (3/52 * 2/51) = 47/442

That is the correct answer.

 

[DrPizza]

But, of course, having some understanding of the problem is just as important as having the numbers written down..
The OP already had the answer... With that answer, you could conceivably have multiplied any two or 3 things together that give the same result and could have come up with some rational reason for it.