Two cards are drawn from a pack of well shuffled cards. Find probability one is red card and second a queen.?

Question

I tried solving the problem by splitting the question in two parts. first card taking a red queen and second one a queen ( that is 5 cases) and the second part where the first card is the balance 24 cards (26 red less 2 red queens) and 4 queens leading to 24 X 4 96 cases. hence total 96 + 5 = 101 cases. Total sample space is 52 C 2 cases, but this does not give the right answer.

Sorry the correct question is one is a red card and other (not second) is a queen (the answer made me realise how small errors can change the question) It is obviously a question of combinations not permutations.The answer given in the book is (96 + 4)/ 52 C 2 but I think it should be (96 + 5) /52 C2 because the 5 cases I consider are DQHQ, DQCQ,DQSQ, HQCQ,HQSQ.

Answer 1

Are you asking whether the first is red and the second is a queen? I think that's what you mean.

P(1st is red and 2nd is queen)
= P({1st is red queen and 2nd is queen} or {1st is red and not queen and 2nd is queen})
= P(1st is red queen and 2nd is queen) + P(1st is red and not queen and 2nd is queen)
=(2/52)(3/51) + (24/52)(4/51)
= 102/2652

There are six cases in the first one, not five. Also, you are looking at combinations when you should be looking at permutations. QH first and QS second is different from QS first and QH second. That's why you need to look at permutations, where order matters. The number of outcomes in the sample space is 52P2, which is 2652.

edit: Given you correction, I get the same answer you did, only in a different way.

P(one red and other queen)
= P({1st R and 2nd Q} or {1st Q and 2nd R})
= P(1st R and 2nd Q) + P(1st Q and 2nd R) - P(1st RQ and 2nd RQ)
= 102/2652 + 102/2652 - (2/52)(1/51)
= 202/2652
= 101/1326

You're right about the combinations stuff. I have a feeling that the book did not count the QD and QH possibility, but I do think that counts. I don't know what to tell you.

Answer 2

Your own answer doesn't seem to quite fit your question - is the first card supposed to be a red queen, or just any red card? In the first case, there are two red queens in 52 cards, and then three queens left out of 51 cards, if you draw them as you want to. Probabilities are multiplied here, so it's 2/52 x 3/51. That comes to 6/2652, or 1/442.

In the second scenario, there are 26 red cards in 52 (half the deck), and either three or four queens left, depending upon whether or not you got one of the red queens on the first draw. If you didn't, it would be 1/2 x 3/51 = 3/102 or 1/34. If you did draw a queen first, it would be 1/2 x 4/51 = 4/102 or 2/51.

Answer 3

Do you mean the first is a red card an the second a queen? Or do you mean at least or exactly one of the two is a red card? This gives different answers.

In the case you mean the first is a red card and the second a queen. this is the answer:
the first is a red card : 1/2 chance
The second is a queen: 1/13 chance
total: 1/26 chance

Taking into account the fact that the first one could also be a queen gives the same answer, since:
the first is a red queen (1/13 chance) so the chance of the second beeing a queen is 3/51
the first is not a queen (12/13 change) then the chance of the second beeing a queen is 4/51
the total chance of the second being a queen is 1/13*3/51+12/13*4/51=1/13.

Answer 4

Probability of the 1st card being red but not a red queen is 24/52 because there are 26 red cards in a pack of 52
Now given that you have drawn one card from the pack of 52, there are only 51 cards to choose from.
Out of these 51 we have to get a queen.
Probability of 2nd card being a queen is 4/51

Now consider the case that when u drew the 1st card it was a red queen
Probability of that happening is 2/52
If you have chosen a red queen in the 1st draw then you have one queen less in remaining 51 cards
In this case the probability of choosing a queen is 3/51

P(red card on 1st draw and a queen on second) = P(choosing a red card in 1st draw which is not a queen) x P(choosing a queen in 2nd draw) + P(choosing a red queen in the 1st draw)x P(of choosing a queen in the 2nd draw)

P = 24/52 x 4/51 + 2/52 x 3/51 = 1/26

Answer 5

Probability of drawing red queen on first draw is 1/26

Probability of drawing a red card that is not a queen is 6/13

Probability of drawing a queen after drawing a red queen is 3/51

Probability of drawing a queen after drawing a red non-queen is 4/51

So the probability of drawing a red card first, and a queen second:

(1/26)*(3/51)+(6/13)*(4/51)=9/442, which is approximately 0.0203

Answer 6

King probability = 4/fifty two (a million/13) b/c there are 4 kings in deck of fifty two enjoying cards crimson Card probability = 26/fifty two (a million/2) b/c there are 26 crimson enjoying cards King or crimson Card probability = 28/fifty two (7/13) b/c there are 26 crimson enjoying cards and 2 kings that are actually not crimson (26+2 = 28) wish i substitute into of help!

Answer 7

your number of cases is off by one. above you say "first card taking a red queen and second one a queen (that is 5 cases)." Actually, this is 6 cases:
QH - QD
QH - QC
QH - QS
QD - QH
QD - QC
QD - QS

Answer 8

1/26