Sheldon Ross Chapter-2 Question-15b

Question

If it is assumed that all $\binom{52}{5}$ poker hands are equally likely, what is the probability of being dealt two pairs? (This occurs when the cards have denominations a, a, b, b, c, where a, b, and c are all distinct.)

 

my approach is:
selecting a denomination=$\binom{13}{1}$ ways and selecting two suits=$\binom{4}{2}$

so selection of first pair=$\binom{13}{1}*\binom{4}{2}$

Similarly, for the other pair, ways of selecting the second pair=$\binom{12}{1}*\binom{4}{2}$

and the remaining one card can be chosen in $\binom{11}{1}*\binom{4}{1}$ ways

So, probability=$\frac{\binom{13}{1}*\binom{4}{2}*\binom{12}{1}*\binom{4}{2}*\binom{11}{1}*\binom{4}{1}}{\binom{52}{5}}$

I’m getting answer 0.095 but in the book answer is given 0.0475

where am I going wrong?

Comments

I agree
I was confused with my own solution
Arjun sir cleared it

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yes.

Actually among 13 denomination, we choose 2 cards at a time.

Now for pairing, among 4 same type cards, we chooses 2 cards for each of that two denominations.

Now select one odd pair and from 4 suit select one suit

right?

Nice question

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yes

Answer 1

I've done in this way.

For selecting 2 pairs, first, we choose 2 types of cards out of possible 13. So, we get 13C2. Now, for each pair, we have to select 2 colours out of 4. So, now, we get 13C2 * 4C2 * 4C2,  as final for selecting the pairs.

But, now for selecting the last card, we will select 1 card from rest 11 numbers, which will be 11C1, but it can be of any 4 colours, so for single card, the number of combinations are 11C1 * 4C1.

Finally, answer :

13C2 * 4C2 * 4C2 * 11C1 * 4C1 / 52C5 = 0.0475