PERMUTATIONS HELP: When drawing 7 cards from a deck, what is the probability of drawing...

Question

a.) 3 face cards? - answer 15.03%

b.) at most 5 spaces? - answer 99.95%

For a, I am doing this but this isn't right:

[(12P3)(40P4)] / 52P7

Answer 1

We draw 7 cards from a deck of 52, with replacement. (I'm assuming with replacement since the answers are closer to what you have.)

There are 12 face cards in a deck of 52 so the probability of picking one is 12/52 = 3/13. The chance of *not* picking one is 1-3/13 = 10/13.

Let's examine the case where you pick 3 faces in a row and the 4 others are not faces. The chance of this happening is:

(3/13)^3*(10/13)^4

However, this is only one such case. The set of combinations you are looking for is 7C3. The number of total objects is 7, and you have three items that can move among those 7 objects.

So the final probability is:

7C3*(3/13)^3*(10/13)^4 = 0.1506 = 15.06%

There are 12 spade cards in a deck of 52. So the probability of picking a spade is 1/4. The probability of not picking one is 1 - 1/4 = 3/4.

What we can do here is eliminate the cases where we'd pick 6 or 7 spades.

The probability of NOT picking 7 spades is trivial to calculate. It is (3/4)^7. There is only one configuration.

We can calculate the probability of picking 6 spades in a row, with the last card not being a spade as:

(1/4)^6*(3/4)

However, there are 7C6 such combinations. If we combine everything together, we get:

1 - (7C6*(1/4)^6*(3/4) + (3/4)^7) = 1 - 7C6*(1/4)^6*(3/4) - (3/4)^7 = 0.9987 = 99.87%

Answer 2

7 cards from 52 card deck, with replacement

probability of exactly 3 face cards =

7C3(3/13)^3(10/13)^4 = about 15.06%

without replacement a little less,= about 10.86%=

7C3(12/52)(11/51)(10/50)(37/49)(36/48)(35/47)(34/46)

at most 5 spades, with replacement

= 1 -P(6S) - P(7S)

= 1- 7(1/4)^6(3/4) - (1/4)^7

=1- 21/4^7 - 1/4^7

= 1 - 22/4^7

= about 1- .0013

= .9987

= 99.87%

without replacement, a little less

= 1- 7(13/52)(12/51)(11/50)(10/49)(9/48)(8/47)(39/46) - (13/52)(12/51)(11/50)(10/49)(9/48)(8/47)(7/46)

=1- .00288- 8.429x10^-5

= .99712 - .00008420

= .9962

= 99.62%

the calulations get a little tedious, so possible error(s) along the way

another problem

at most 5 aces = 1 - probability of 6 or 7 aces

= 1 - P(6A) - P(7A)

without replacement

= 1- 0-0

=100%

it's impossible to get more than 4 aces. this part would be a trick question. the deck only has 4 aces

You could have worked these problems with permutations or combinations, but permutations are just unnecessarily more tedious