Laura has a deck of standard playing cards with 13 of the 52

My question may sound incorrect. Why can't I think of the problem as geometric distribution in terms of first success ( heart) in k (10) trials. Please tell where I am going wrong

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Thank you Ian,
You and Bunuel make math problems sound so simple. Kudos.

and then deals 10 cards off the top of the deck, what is the probability that the 10th card dealt is a heart?



WHAT does this statement implies that 10 cards are withdrawn with replacement or without replacement ??

If there is no replacement then how is the (P) that the 10th card is 13/52 ??

there are many cases here to be considered here if there is no replacement such as:

H- Denotes heart X-may be any diamond, spade or club.

1. HXXXXXXXXH
2. HHXXXXXXXH
3. HHHXXXXXXH
.
.
.
.
.
9. HHHHHHHHHH
10. XXXXXXXXXH

All cases from 1 to 10 will have different probabilities for heart to be at the 10th place and it will take hell lot of time to calculate all of them.

For according to me the above solution by Ian is only valid if cards are replaced (Every card has a 13/52 = 1/4 chance of being a heart; it doesn't matter if it's the top card in the deck or the tenth card in the deck.)If that's the case that brings back me to my original question ---- how do we determine that the cards are replaced or not ?? based on the question given ....

When we have a case with replacement it's always clearly mentioned in the question. We are told that "Laura deals 10 cards off the top of the deck", which means that there is no replacement whatsoever.

As for the question, concept behind it is discussed in the topics given in my previous post.

Hi! I went through all the similar problems provided by Bunuel and was able to understand and solve them. However, I am still not able to get why this is not an arrangement problem. Why do we need to ignore the number 10? Bunuel/Karishma, kindly elaborate.

I solved the question using reverse probability: 1- P(10th is not heart) = 1- (3*13)/52 = 1/4.

I was confused as well,but the solution is something like this.
Let's assume the question is to find the probability of drawing second card as heart instead of 10th?

How we will approach this problem is:
A) find the probability that first drawn Is heart and second drawn is heart too i.e. (13/52*12/51)
B) find the probability that first drawn is not a heart and second drawn is a heart(39/52*13/51)

Adding the two will give you 1/4.

The above example is to give a sense of what is happening in this question.

Since we don't know the outcome we can't assume weather the first draw is a heart or not a heart. We added the outcome of two situations.

I see a lot of people saying how can the probability of drawing 10th card as a heart be 1/4 when all the cards from 1 to 9 can be heart. Yes! Certainly it can be but its one of the cases. Adding all such cases will give you 1/4.

And had we known the result of previous outcomes, indeed the question would be different.

As Bunuel said. May be laura knows the outcome but we don't.

Great question.
Thanks guys!

Regards
M

Posted from my mobile device

Posted from my mobile device

For anyone looking at this problem and thinking "I need to consider what the first 9 cards are", perhaps I can offer an easy explanation of why that's unnecessary (if you did not think that, there's no reason to read any further!).

In a deck of cards, we have 13 hearts, 13 spades, 13 clubs and 13 diamonds, out of 52 cards, so 1/4 of the deck is hearts. Consider these questions:

If a magician spreads out a deck of cards, and asks you to choose one at random, what's the probability you pick a heart? It's 13/52 = 1/4, since you're just picking one card at random.

Now if the magician fans the deck of cards, and asks you to choose one, and you happen to pick the 10th card, what is the probability you picked a heart? Again, it's 1/4, for the same reason as before; you're still picking a random card from the deck. Whether you pick the 1st card, the 10th card, the 37th card or the last card, each has a 1/4 probability of being a heart.

But that's the same question as the one in the OP above - Laura is just picking the tenth card from the deck. Unless you have some information about the top nine cards, the tenth card is just as random as the first one.

My question may sound incorrect. Why can't I think of the problem as geometric distribution in terms of first success ( heart) in k (10) trials. Please tell where I am going wrong

Posted from my mobile device

You don't need to know what a "geometric distribution" is for the GMAT, but I think you're trying to answer a more complicated question than the one asked. If you're thinking in terms of geometric distributions here, you're trying to answer a question like "what is the probability, if Laura picks cards one by one from the top of the deck, that she doesn't get a heart until she picks the tenth card?" The problem here is simpler - we don't need to even consider the first nine cards. It's just asking for the probability the tenth card is a heart, and nothing more.

The answer is 1/4 Option A

When we shuffle and deal the 19 cards each card getting chosen would have a probability of 1/52.

And each card would have a probability of 1/4 being a heart.

The first 9 cards can be of any suite this probability = 1
The probability of the tenth card being a heart = 1/4

Thus Probability of 10th card being a heart = 1*1/4

Posted from my mobile device

Even I got it qrong first, but it's important to note that question doesn't mention about whethere first, second, third... etc was a heart or not. So replacement should not be taken into consideration.
So simply, withdrawal of a card at any point with hearts = 13/52

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