Probability Help

[757nowhere7]

Here is the problem that I have.
In a deck of cards there are 13 “types” of cards and each type has 4 of its same kind. What is the probability of getting a “pair”, or 2 of the same kind getting dealt 2 cards?

Here is what I have come up, please let me know if this is correct or what I should do:

P(A) = 0/52 (since first card dealt doesnt matter.
P(B) = 3/51 (since the I have already got one card)

so 0/52+3/51 = 3/51 ?

 

[correctlogic]

these are my thoughts

• i suppose P(A) is the probability when taking a card to be any type , thus it is equal to 1 not 0
• getting a pair = A AND B happening together

setting R as the probability getting a pair
P(R)=P(A["U" inverse]B)=P(A)*P(B)=1*3/51=3/51
hope helped

 

[Ethan3141]

I find adding context makes it easier to express the problem, also since all cases are identical this can be done without loss of generality.

What is the probability of drawing a king out of a deck of cards?
What is the probability of drawing another king out of that same deck of cards?

Multiplying these probabilities together will get you the answer you desire.

In regards to what you have posted. I agree with P(B) however not P(A).

 

[Steven G]

I find adding context makes it easier to express the problem, also since all cases are identical this can be done without loss of generality.

What is the probability of drawing a king out of a deck of cards?
What is the probability of drawing another king out of that same deck of cards?

Multiplying these probabilities together will get you the answer you desire.

In regards to what you have posted. I agree with P(B) however not P(A).

Actually the desired answer is 13*(probability of drawing a king out of a deck of cards)*(the probability of drawing another king out of that same deck of cards)

 

[Steven G]

Here is the problem that I have.
In a deck of cards there are 13 “types” of cards and each type has 4 of its same kind. What is the probability of getting a “pair”, or 2 of the same kind getting dealt 2 cards?

Here is what I have come up, please let me know if this is correct or what I should do:

P(A) = 0/52 (since first card dealt doesnt matter.
P(B) = 3/51 (since the I have already got one card)

so 0/52+3/51 = 3/51 ?

P(A) is not zero! Since the card does not matter you want the probability of getting any card. Well with 100% certainty if you pick a card from a deck you will get a card from the deck!

More importantly: You pick a card from the AND THEN you pick another card. AND means to multiply the probabilities NOT add then.

Here is the standard way of doing this. You 1st want to pick a denomination AND THEN you want to pick 2 of the 4 cards of that value. This can be done in ₁₃C₁*₄C₂ ways. So the probability is ₁₃C₁*₄C₂/₅₂C₂