Probability questions with 52 cards deck

Forums: Probability
Email this Topic • Print this Page

Hi,

I am given some questions and I need help.

(1)You are dealt two cards from a pack of 52 cards. What is the conditional probability that
(a) both of them are aces if at least one of them is the ace of spades?
(b) both of them are aces if at least one of them is an ace?

a) I know that the probability of being dealt 2 aces out of 2 cards is : (4x3)/(52x52)=1/221
there are 6 possible ways to arrange the aces and 3 to have Ace of spades. So should it be just 1/221/2 ?
This is the conditional probability of being dealt 2 aces out of 2 cards, right?
P(A1A2)=P(A1)P(A2|A1)=P(A2|A1)=P(A1A2)/P(A1) , but what should I change to get the condition right?
b)Isnt it P(A2|A1)/P(A1)?
And I don't have idea about the 2nd one:
(2) Consider a pack of 52 cards.
(a), What is the probability that there is a king right after the first ace?
(b), What is the probability that there is an ace right after the first ace?
(c) What is the probability that the 1st ace is the 30-th card?
(d) If the 1st ace is the 30-th card, what is the conditional probability the next card is the ace of spades?
(e) If the 1st ace is the 30-th card, what is the conditional probability the next card is
the jack of diamonds?
Note that you have to give these probabilities without looking at any cards from the pack.

a) if the Ace is the first card it gives 4/52 and the king is right after the ace is 4/51, but what when they are not 1st and 2nd cards
b) is almost the same 4/52*3/51 ?
c) I have no idea what is the correct approach, but I think since there are 22 cards left we should have 4/52 - 4/22?
I have no Idea on the last 2.

Thanks a lot!

Topic Stats
Top Replies
Link to this Topic

Type: Question • Score: 0 • Views: 4,271 • Replies: 3

No top replies

@gl0ck,

P(A|B) = P(B|A)*P(A)/P(B)

1a) A = both cards are aces, B = one card is the ace of spades
P(B|A) = 1/2
P(A) = C(4,2)/C(52,2) = 1/221
P(B) = C(51,1)/C(52,2) = 1/26

1b) A = both cards are aces, B = at least one card is an ace
P(B|A) = 1
P(A) = C(4,2)/C(52,2) = 1/221
P(B) = 1 - P(neither card is an ace) = 1 - C(48,2)/C(52,2) = 33/221

But if you look at http://www.stat.ucla.edu/~rosario/classes/081/100a-2a/100aHW2Soln.pdf, you'll see a different approach to 1a and 1b.

2a) Without knowing what came before the first ace, I'd say 4/51.

2b) Without knowing what came before the first ace, I'd say 3/51.

2c) P(first 29 are not an ace) * P(30th is an ace) = C(48,29)/C(52,29) * 4/23

2d) 1/4 of the time the 30th card will be the ace of spades, 3/4 of the time it will be a different ace. Therefore,
(1/4)*0 + (3/4)*1/22 = 3/88

2e) From 2d we know that P(31st is ace of spades) = 3/88. The same argument applies to P(31st is ace of clubs), P(31st of ace of hearts), and P(31st is ace of diamonds).
Therefore, the probability of the 31st card being an ace is 12/88. That leaves 76/88 for the rest of the cards.
Therefore, the probability of any specific non-ace being the 31st card is (76/88)/48.

But if you look at http://www3.nd.edu/~cstanton/courses/math30530/lecture08.pdf, you'll see the "proper" way to do 2d and 2e.

Probability questions with 52 cards deck

Related Topics

Quick Links

My Account

able2know