Is there a simpler way to solve this problem?

[John Salerno]

Hello everyone. I've got another probability puzzle that I actually did solve, but I feel like I may have taken a longer route than necessary to arrive at the solution. I'm wondering if there is a simpler, more elegant way to solve it than the "backwards" approach I took.

Basically, the problem is that you have 40 cards remaining in a deck, with all four aces still in the deck. If you draw two cards, what is the probability that exactly one card will be an ace?

I didn't know how to solve it directly, so I thought it would be easier to find the probability of *both* cards being an ace and the probability of *neither* card being an ace, and subtract those amounts from 1:

1 - ( (4/40) * (3/39) ) - ( (36/40) * (35/39) )

This works, and I understand why it does, but it feels like more of a brute force method. Is there a way to directly solve for the probability of only one card being an ace?

Thanks!

[Prometheus]

I can't think of an easier way. Subtracting the probability of an event not occurring from 1 is a pretty standard way of finding the probability of an event so kudos for thinking it up.

[John Salerno]

4 hours ago, Prometheus said:

I can't think of an easier way. Subtracting the probability of an event not occurring from 1 is a pretty standard way of finding the probability of an event so kudos for thinking it up.

Thanks! Just wanted to make sure I was learning how to do it the proper way, and not taking detours like you mentioned in the other thread!

Edited November 19, 2017 by John Salerno

[Country Boy]

If there are 40 cards in the deck and 4 of them are aces, the probability the first card drawn is an ace is 4/40= 1/10.  There are then 39 cards left in the deck, 3 of them aces.  The probability the second card drawn is not an ace is (39- 3)/39= 36/39= 12/13.  The probability of drawing "ace, non-ace" in that order is (1/10)(12/13)= 12/130= 6/65.  Similarly, the probability the first card drawn is not an ace is 36/40= 9/10 and the probability the second card is  an ace is 4/39 so the probability of drawing "non-ace, ace" in that orderis (9/10)(4/39)= (3/5)(2/13)= 12/130= 6/65.  The probability of drawing one ace out  of the two cards is 2(6/65)= 12/65.

I am not sure that using your other method, while correct, would be any simpler.

If there are 40 cards in the deck and four of them are aces, there is 4/40= 1/10 probability the first card is an ace.  There are then 39 cards in the deck, 3 of them aces.  The probability the second card drawn is an ace is 3/39= 1/13.  The probability both cards are aces is (1/10)(1/13)= 1/130.  If there are 40 cards in the deck and four of them are aces, 36 cards are not aces so there is 36/40= 9/10 probability the first card is not an ace.  There are then 39 cards in the deck, 35 of then not aces.  The probability the second card drawn is not an ace is 35/39. The probability the cards are two non-aces is (9/10)(35/39)= 21/26.  The probability the are not "two aces" nor "two non-aces" is 1/130+ 21/26= 3/390+315/390= 318/390= 159/195= 53/65.   The probability the cards drawn are one ace and one non-ace is 1- 53/65= (65- 53)/65= 12/65 as before.

I found the second method much harder than the first!

[phyti]

For clarity, I would prefer the events defined with the corresponding ratios.

a=ace, b=not ace, a=4, b=36, e(x)=number of events (x)

e(aa)=4*3=12

e(ab)=4*36=144

e(ba)=36*4=144

e(bb)=36*35=1260

total events=1560

 

p(ab or ba)=288/1560=12/65=.185