Q
Two cards are drawn from a pack of card , find the probability that 1st card is red and the second is an Ace, when the 1st card is.
a) Replaced ?
b) not Replaced?
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probability of red and ace ???
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Anurag@Gurome
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When replaced:
probability of the 1st card to be red = 1/2
when we replaced, we will have the same set of cards
probability for an ace = 4/52
probability of both events to happen = 1/2 * 4/52 = 1/26
when not replaced:
probability of the 1st card to be red = 26/52
but the 1st card can be red ace too..
so we will divide it into two red cases..
1st one red suit without an ace, and second red ace
the probability of 1st card to be red non-ace = 24/52
the probability of 2nd card to be ace = 4/51
the probability that the 1st card is non-ace red card and second card is ace = 24/52 * 4/51
the probability of the 1st card to be red ace = 2/52
the probability of the 2nd card to be ace = 3/51
the probability that the 1st card is ace red card and second card is ace = 2/52 * 3/51
the probability that 1st card is red and the second is an Ace = the probability that the 1st card is non-ace red card and second card is ace + the probability that the 1st card is ace red card and second card is ace
= (24/52 * 4/51) + (2/52 * 3/51)
= 1/26
probability of the 1st card to be red = 1/2
when we replaced, we will have the same set of cards
probability for an ace = 4/52
probability of both events to happen = 1/2 * 4/52 = 1/26
when not replaced:
probability of the 1st card to be red = 26/52
but the 1st card can be red ace too..
so we will divide it into two red cases..
1st one red suit without an ace, and second red ace
the probability of 1st card to be red non-ace = 24/52
the probability of 2nd card to be ace = 4/51
the probability that the 1st card is non-ace red card and second card is ace = 24/52 * 4/51
the probability of the 1st card to be red ace = 2/52
the probability of the 2nd card to be ace = 3/51
the probability that the 1st card is ace red card and second card is ace = 2/52 * 3/51
the probability that 1st card is red and the second is an Ace = the probability that the 1st card is non-ace red card and second card is ace + the probability that the 1st card is ace red card and second card is ace
= (24/52 * 4/51) + (2/52 * 3/51)
= 1/26
Last edited by Anurag@Gurome on Wed Sep 26, 2012 2:19 am, edited 1 time in total.
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Anurag@Gurome wrote:When replaced:
probability of the 1st card to be red = 1/4
when we replaced, we will have the same set of cards
probability for an ace = 4/52
probability of both events to happen = 1/4 * 4/52 = 1/52
when not replaced:
probability of the 1st card to be red = 13/52
but the 1st card can be red ace too..
so we will divide it into two red cases..
1st one red suit without an ace, and second red ace
the probability of 1st card to be red non-ace = 12/52
the probability of 2nd card to be ace = 4/51
the probability that the 1st card is non-ace red card and second card is ace = 12/52 * 4/51
the probability of the 1st card to be red ace = 1/52
the probability of the 2nd card to be ace = 3/51
the probability that the 1st card is ace red card and second card is ace = 1/52 * 3/51
the probability that 1st card is red and the second is an Ace = the probability that the 1st card is non-ace red card and second card is ace + the probability that the 1st card is ace red card and second card is ace
= 12/52 * 4/51 + 1/52 * 3/51
= 1/52
Sir
Can you explain how did you find the probability of red ?
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anuprajan5
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Anurag,Anurag@Gurome wrote:When replaced:
probability of the 1st card to be red = 1/4
when not replaced:
probability of the 1st card to be red = 13/52
I am a bit stumped with this one. There are 26 red and black cards each in a deck. Therefore, probability of the 1st card to be red should be 26/52 = 1/2
Regards
Anup
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Anurag@Gurome
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Its been long time i played cards. i was thinking of Red diamonds. Yeah the probability for a red is 1/2.Anurag,
I am a bit stumped with this one. There are 26 red and black cards each in a deck. Therefore, probability of the 1st card to be red should be 26/52 = 1/2
Regards
Anup
i will edit the ans.
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