The lost card.

[Queen09]

Out of a pack of 52 cards one is lost; from the remaining of the pack, two cards are drawn and are found to be spades. Find the chance that the missing card is a spade.

 

a. 11/50

b. 11/49

c. 10/49

d 10/50

 

OA to follow.

[Shooter]

11/50

 

Let the one lost is spade and the other two that are taken out are also spades. We are left over with 11 spades in the pack.

[Itsmyturn]

IMO,

 

Card was lost before the 2 other cards were picked so the chance that the card is a spade will be 13/52 = 1/4

 

I know it is not one of the options

[Queen09]

Itsmyturn

hint for you ...

 

"two cards are drawn and are found to be spades" is just a distraction.

It is as good as a pack of cards with 50 cards, with 11 spades.

[Shooter]

I am on a second thought now:

 

Queen,

 

If 3 cards have been removed then the denominator should consist of (52-3) = 49 right?

[sanjeev_sharma]

IMO it is C = 10/49,

 

since we are left with 10 spades if one of the spade is missiong and two other are drawn from the pack. The remaing cards in the pack = 51-2 =49. So the probabilty = 10/49

[Makumajon]

At the point when the card was lost, the card could be any one of the 52 cards, and the probability that it was spade is 13/52.

 

Now, we have drawn another 2 cards and looked at them. Hmm, 2 spades! SO, our prediction about the lost card changes and becomes stronger: we are certain that the lost card is any one of the [remaining] (52-2) or 50 cards,[among which there are also (13-2) or 11 spades].

 

Therefore, P=11/50

[Queen09]

Makumajon

You explained it so well.. wish we could borrow your brain for one day ;)

[Makumajon]

Thanks. Just pray I can contribute to this forum for many days. I have not planned to sit for GMAT, but I enjoy the great discussions and learning here. Because I love to explore the beauty of mathematics, especially in the area of number theory, and the lofty grammar rules. :)

[Shooter]

Nice Explanation!

[Storm]

Hi,

I had a Question:

Lets take a simple problem first to illustrate what I exactly mean:

We roll a Dice. We get an even number. WHat is the probability that it is prime.

 

Initially, the sample set is (1,2,3,4,5,6). Therefore the prob of getting a prime number is 3/6 = 1/2

 

However, once we get an even number, the prob changes to 1/3 (2,4,6). Only 2 is prime. Here, the key is that the prob changed because the number of sample outcomes became less (3 as opposed to 6 before).

 

In the problem being discussed in the thread, I don't understand why finding 2 spades would change anything. Look at it another way. 1 card is lost. What is the prob of finding 2 (or even 3 spades). It obviously is 1 because even if the lost card is a spade, we are sure to find 3 spades in the pack. Essentially, this extra statement has changed nothing. Its something like saying we roll a STANDARD dice and we get a natural number between 1 and 10(Don't we already know that as it can range from 1-6). Does it change anything. No. Then why should our finding 2 spades (which is a given as only 1 card is lost) changing the probability. Yes, had it said 13 cards are found to be spades, then the prob would go down to 0 as a spade was certainly not stolen. Could somebody explain if I am missing somethin here. In light of this I do think that the prob should still be 1/4. I know, its not an option, but that still doesn't mean anything.

 

Normally such cases are calculated as follows P(F|E) * P(E)= P(E n F)

[Queen09]

See..

I think I can reframe this question :

 

Imagine a deck of 50 cards (13 diamonds, 13 hearts , 13 clubs, 11 spades).

What is the probablity of picking one spade?

 

I think it's the language of Q that is confusing.

[Storm]

After you have 11 spades its fine. However, the card was lost before you found 2 spades. If your Q is we find 2 spades and then we lose a card, what is the prob that a pade was lost, yes it'd be 11/50. No doubt. But to find 2 spades makes no diff (acc to me. I really am not sure if I am right here.I am sticking my neck out!!) to a card that has already been lost. In fact I'd be surprised if you didnt find 12 spades in the deck coz only 1 card was lost.

[Makumajon]

Of course, finding two spades changes the probability from 1/4. Because the problem asks to find the probability after you have drawn two spades, not before. So, you are calculating a dependent probability, not a free probability. As the successive events unfold, you would get new probabilities for an outcome because you are calculating probability of that outcome based on new circumstances/evidences. And the formula

P(F|E) * P(E)= P(E n F)is nothing but the formula for a dependent probability.

[Storm]

Guys, its just that I don't take anything on face value. Obvious things are not very 'obvious' to me. So, I decided to go in for a more rigorous approach.

 

Let Event E = Finding 2 spades after 1 card is lost.

Event F = The event that the lost card is a spade.

 

Then event E n F = event that after pulling 2 spades from the pack, the lost card was a spade.

 

now P(F|E) = P(EnF)/P(E)

 

P(EnF) = 1/4 * 12c2/51c2 (the 2 probabilities are independent)

 

P(E) = (1/4 * 12c2/51c2) + (3/4 * 13c2/51c2)

 

essentially states that prob of finding 2 spades after losing a spade + prob of finding 2 spades after losing a non spade. Events mutually exclusive and therefore are added.

 

now just divide p(EnF)/P(E) and we do indeed get 11/50.

 

I can lay my fears and doubts to rest!!

[bose]

like makumajon's expln