Probability (deck of cards problem)

[Hill]

Two cards are drawn from a shuffled deck. What is the probability that both are red? If at least one is red, what is the probability that both are red? If one is a red ace, what is the probability that both are red?

Now below is my attempt at a solution:

P(Both Red) = (26/52) * (25/51)

P(Atleast 1 is red) = P(Red,Not Red) + P(Not Red, Red) + P(Both Red)
=(26/52) * (26/51) + (26/52) * (26/51) + (26/52) * (25/51)

If atleast 1 is red then Probability that both are red can be achieved in this way:

P(Both Red|Atleast 1 red)

I hope I am correct upto this but it's the third part of the question I am unable to do. How do we get the probability that both are red given one is a red ace?

 

[Cubist]

Hi, welcome to the forum! I agree with the work that you've shown so far.

Concerning the last question that you're stuck on, I can think of two approaches to solve it:- Bayes' theorem or combinatorics. I assume that you'll want to use Bayes' theorem since this seems more "in line" with your other work. Have you studied this? Here's how to set it up...

Event A is both cards are red
Event B is one red ace

Bayes' theorem
P(A|B) = P(B|A) * P(A) / P(B)

Can you work out P(A), P(B) and P(B|A) ?

 

[pka]

Two cards are drawn from a shuffled deck. What is the probability that both are red? If at least one is red, what is the probability that both are red? If one is a red ace, what is the probability that both are red?

the probability that both are red is $\mathcal{P}(BR)=\mathcal{P}(R_1\cap R_2)=\dfrac{\dbinom{26}{2}}{\dbinom{52}{2}}$
That is the same probability that both is black. The complement is the probability that at least one is red.

 

[Cubist]

I can think of two approaches to solve it:- Bayes' theorem or combinatorics

...by picturing the Venn diagram, I also worked out the answer using probabilities but without using Bayes' theorem

P1 = P( one red ace AND other card a non-ace red )
P2 = P( one red ace AND other card black )

Ans = P1 / (P1+P2)