SOLUTION: Two cards are dealt from ordinary pack; Event (A) : both cards are same suit Event (B): both cards are red (i) Which event is more likely. I have attempted this part and I

Question 601044: Two cards are dealt from ordinary pack;
Event (A) : both cards are same suit
Event (B): both cards are red
(i) Which event is more likely.
I have attempted this part and I believe I am right in saying that b is more likely with a probability of 0.24509... and A is 0.0588....
(ii) Find the probability that the two cards are either both red or both from the same suit.
(iii) Use the definition of p(A given B) to calculate the probability that the two cards are from the same suit, given that they are both red.
I am struggling on parts two and three.
Thanks

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
i)

Let's say you draw a diamond card. What is the probability of drawing another diamond card? The probability would be 12/51 = 0.23529 (assuming you can't replace the cards). This idea can be done with any suit.

So in general, the probability of drawing two cards that are of the same suit is roughly 0.23529

Now let's say you draw a red card. The probability of drawing another red card is 25/51 = 0.490196

So the probability of drawing two red cards is 0.490196

Clearly 0.490196 is larger than 0.23529, which means that event B is more likely.

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ii)

P(Both red or same suit) = P(Both red) + P(same suit) - P(Both red and same suit)

P(Both red or same suit) = 25/51 + 12/51 - 12/51

P(Both red or same suit) = 25/51

P(Both red or same suit) = 0.49019

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iii)

P(A given B) = ( P(A and B) )/P(B)

P(same suit given both red) = ( P(same suit and both red) )/P(both red)

P(same suit given both red) = ( 12/51 )/(25/51)

P(same suit given both red) = ( 12/51 )*(51/25)

P(same suit given both red) = 12/25

P(same suit given both red) = 0.48

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