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Two cards are drawn from a standard deck of cards at the same time. Find:

a) Probability of drawing the King of hearts and a red card
b) Probability of drawing the King of hearts and a black card

Progress

a) I get a red card and a king hearts: P(R or K). There are 26 reds card out of 52 cards, so P(R) = 26/52 = 1/2. There are 4 kings out of 52 cards, so P(K) = 4/52 = 1/13. There is 1 card which is a red card and a king of hearts so $$P(\text{H and K}) = \frac{1}{13}\times \frac12 = \frac{1}{26}$$

But I don't know if this is right.

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  • $\begingroup$ yup , I tried it .. a) I get A red card and a king hearts: P(R or K) There are 26 reds card out of 52 cards, so P(R) = 26/52 = 1/2 There are 4 kings out of 52 cards, so P(K) = 4/52 = 1/13 There is 1 card which is a red card and a king of hearts so P(H and K) = (1/13) x (1/2) = 1/26 $\endgroup$ Aug 21 '14 at 1:32
  • $\begingroup$ But i don't know right or wrong ? $\endgroup$ Aug 21 '14 at 1:33
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One way is by a counting procedure. There are $\binom{52}{2}$ two-card hands. They are all equally likely.

We now count the favourables. There are $25$ hands that have the King of $\heartsuit$ and an additional red card. For there are $25$ red cards that are not the King of $\heartsuit$.

Thus the required probability is $\frac{25}{\binom{52}{2}}$.

For King of $\heartsuit$ and a black card, it's your turn.

Another way: Imagine drawing the cards one at a time (it makes no difference to the probability). We will be happy if (i) we draw the King of $\heartsuit$, and then another red card, or (ii) if we draw a red card other than the King of $\heartsuit$, and then the King of $\heartsuit$.

We find the probability of (1). The probability the first card is the King of $\heartsuit$ is $\frac{1}{52}$. Given that this happened, the probability the next card is red is $\frac{25}{51}$. So the probability of (i) is $\frac{1}{52}\cdot \frac{25}{51}$.

The probability of (ii) is the same. Add.

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  • $\begingroup$ Hey sir , I thought there are 4 kings out of 52 cards, so P(K) = 4/52 = 1/13 , i can find out King of hearts from here ? Right or wrong ? $\endgroup$ Aug 21 '14 at 1:57
  • $\begingroup$ You could say that the probability the first card drawn is a King is $4/52$, and given that it is a King, the probability it is the Heart King is $\frac{1}{4}$, for a probability of $(1/13)(1/4)=1/52$. But it is much quicker to say there is exactly $1$ King of Hearts, so the probability is $1/52$. $\endgroup$ Aug 21 '14 at 2:06
  • $\begingroup$ Oh .. I understood .. Thank sir so much :) $\endgroup$ Aug 21 '14 at 2:09
  • $\begingroup$ Hey sir , can you tell me concepts involved about it ? $\endgroup$ Aug 21 '14 at 2:14
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    $\begingroup$ I am not sure what you are referring to. But if an experiment can result in $N$ equally likely outcomes, and if $H$ of those will make us happy, then the probability we will be happy is $\frac{H}{N}$. $\endgroup$ Aug 21 '14 at 2:24

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