# Cards probability questionWatch

#1
All the spades have been withdrawn from a pack of ordinary playing cards, leaving a pack of 39 cards (13 hearts, 13 clubs and 13 diamonds). This pack is shuffled and 5 cards are dealt. Calculate the probability that the 5 cards dealt will consist of (a) 2 hearts, 2 clubs and 1 diamond, (b) aces and kings only, (c) no ace, no king, no queen and no jack.

Please can someone show me how to answer this question, particularly part (c). Thanks.
0
2 years ago
#2
(Original post by nettogrof)
All the spades have been withdrawn from a pack of ordinary playing cards, leaving a pack of 39 cards (13 hearts, 13 clubs and 13 diamonds). This pack is shuffled and 5 cards are dealt. Calculate the probability that the 5 cards dealt will consist of (a) 2 hearts, 2 clubs and 1 diamond, (b) aces and kings only, (c) no ace, no king, no queen and no jack.

Please can someone show me how to answer this question, particularly part (c). Thanks.
Easiest method, IMO, is to work out the number of possible hands that meet each criterion, and divide be the total number of possible hands where there are no restrictions.

Since this is not the most basic of questions, you should be able to do at least some of that.
2 years ago
#3
(Original post by ghostwalker)
Easiest method, IMO, is to work out the number of possible hands that meet each criterion, and divide be the total number of possible hands where there are no restrictions.
This.

39 cards left in the pack -12 (A,K,Q,J*3 (for each remaining suit)) = 27
So we have 27 cards (2-10 (9 cards in each suit)) that can make legal hands, according to the question.
Using the nCr function on your calculator you can do 27 and 5 which gives you 80730 legal hands. Using it with 39/5 gets you 435897 hands, leaving in the A,K,Q,J cards.
That gives you the probability of 80730/435897 or 18.52%

However, this is without removing a card from the pack, but you should get the idea.
0
2 years ago
#4
(Original post by Luneth)
...
A) You've quoted the wrong person.

B) Posting full solutions straight off is contrary to the forum guidelines.

2 years ago
#5
(Original post by ghostwalker)
/
I quoted you as I agreed with what you said.

I didn't give a full solution at all.
2
2 years ago
#6
(Original post by Luneth)
I quoted you as I agreed with what you said.

I didn't give a full solution at all.
2 years ago
#7
(Original post by ghostwalker)
No problem m8.
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