Probability Question

[goelmohit2002]

Hi All,

Can someone please tell how to solve the following question:

"Laura has a deck of standard playing cards with 13 of the 52 cards designated as a "heart". If Laura shuffles the deck thoroughly and then deals 10 cards off the top of the deck, what is the probability that the 10th card dealt is a heart ? "

Ans = 1/4

[supratims]

I think probability of any card chosen at random will be 13/52 = 1/4 only.
Provided, its not conditional like first card is so and so, two cards are so and so etc.

[goelmohit2002]

I think probability of any card chosen at random will be 13/52 = 1/4 only.
Provided, its not conditional like first card is so and so, two cards are so and so etc.

Thanks. But....why it will be 13/52....after 9 cards are drawn...there are 43 cards left....

Moroeover isn't it conditional only e.g.

after drawing 9 cards we may have so many cases..so we have to add all these sort of probabilities....isn't it ?

13 spade, 13 clubs, 13 diamonds, 4 hearts.
12 spade, 13 clubs, 13 diamonds, 5 hearts.
11 spade, 13 clubs, 13 diamonds, 6 hearts.

and so on....on...

I hope my doubt is clear....

[adam.roffman]

I agree with the 1/4. We know the deck is thoroughly shuffled, so let me ask this: say all but 4 cards were dealt. What's the probability that the next card is a heart? Taking it to conclusion, say all cards were dealt but one (and no looking through the pile!), then what would be the probability that the final card is a heart?

[goelmohit2002]

I agree with the 1/4. We know the deck is thoroughly shuffled, so let me ask this: say all but 4 cards were dealt. What's the probability that the next card is a heart? Taking it to conclusion, say all cards were dealt but one (and no looking through the pile!), then what would be the probability that the final card is a heart?

Thanks. But can someone please tell how to prove the same mathematically ?

[ankitagarwal1010]

I will try to explain the answer:

The probability would be:
atleast 1 heart + atleast 2 hearts ... all 10 hearts
P(r) = p(1) + p(2) + .. + p(10) = 1 - P(with no hearts)
Hence,
P(with no hearts) = all 3 suits / all 4 suits
P(with no hearts) = 39/52
P(r) = 1- (3/4) = 1/4
Hope that answers the question.

Can someone please verify this :)

[goelmohit2002]

I will try to explain the answer:

The probability would be:
atleast 1 heart + atleast 2 hearts ... all 10 hearts
P(r) = p(1) + p(2) + .. + p(10) = 1 - P(with no hearts)
Hence,
P(with no hearts) = all 3 suits / all 4 suits
P(with no hearts) = 39/52
P(r) = 1- (3/4) = 1/4
Hope that answers the question.

Can someone please verify this :)

Hi Ankita,

Can you please tell why are you doing the atleast one heart + atleast 2 hearts + .....

Please note that question asks about only 10th card to be heart....

Probably I am missing something here that you want to say....can you please elaborate a bit ?

Thanks
Mohit

[ananj143]

Hi All,

Can someone please tell how to solve the following question:

"Laura has a deck of standard playing cards with 13 of the 52 cards designated as a "heart". If Laura shuffles the deck thoroughly and then deals 10 cards off the top of the deck, what is the probability that the 10th card dealt is a heart ? "

Ans = 1/4

I will explain this prob with two cards

P( that the first card and second Card =Heart) =13/52*12/51

P(first card is non heart and second card is heart ) = 39/52 *13/51

When you add this the probability = 1/4

This is kind of a counterintuitive problem where 10th or nth will have the same probability

[goelmohit2002]

Hi All,

Can someone please tell how to solve the following question:

"Laura has a deck of standard playing cards with 13 of the 52 cards designated as a "heart". If Laura shuffles the deck thoroughly and then deals 10 cards off the top of the deck, what is the probability that the 10th card dealt is a heart ? "

Ans = 1/4

I will explain this prob with two cards

P( that the first card and second Card =Heart) =13/52*12/51

P(first card is non heart and second card is heart ) = 39/52 *13/51

When you add this the probability = 1/4

This is kind of a counterintuitive problem where 10th or nth will have the same probability

Thanks Ananj.

[Ben Ku]

Hi,

Let me reply by posing a different question.

Suppose I toss a coin 50 times. What is the probability that the 15th toss is heads?

Well the answer is 1/2 because the probability of the 15th toss does not depend on what the other tosses are.

Likewise, it doesn't matter which card we're picking. It could be the first, the tenth, or the 50th card. Each one has a 1/4 chance of being hearts.

Hope that helps.

[aaanishaagarwal]

13 12 11 10 9 8 7 6 5 4 0.00
52 51 50 49 48 47 46 45 43 42
13 12 11 10 9 8 7 6 5 0.00
51 50 49 48 47 46 45 43 42
13 12 11 10 9 8 7 6 0.00
50 49 48 47 46 45 43 42
13 12 11 10 9 8 7 0.00
49 48 47 46 45 44 43
13 12 11 10 9 8 0.00
48 47 46 45 44 43
13 12 11 10 9 0.00
47 46 45 44 43
13 12 11 10 0.00
46 45 44 43
13 12 11 0.02
45 44 43
13 12 0.08
44 43
13 0.30
43


This gives an answer 0.41 (Roundabout 13/52)

[jnelson0612]

Is there a question here?

[vikash.121186]

Can someone please explain to me as to what would have been the answer to the above question if the wordings were as shown below :

"Laura has a deck of standard playing cards with 13 of the 52 cards designated as a "heart". If Laura shuffles the deck thoroughly and then deals 14 cards off the top of the deck, what is the probability that the 14th card dealt is a heart ?

I am a little curious about what would be the solution for this case.

I think it should be (52C13 - 1)/52C14 but not sure. Can someone please help with this?

[tim]

since we know nothing about the first 13 cards, this problem is no different from asking what the probability is that the 14th card is a heart. this is no different from asking what the probability is that any given card is a heart, which is 1/4..