card problem

Using a standard deck of cards, you deal 3 cards. From those 3 cards you choose one card. If the chosen card is a heart, what is the probability that it is the only heart in the hand?
I thought that it would be the number of ways to choose one heart multiplied by the number of ways to choose 2 cards not hearts divided by the sum of the ways to choose 3, 2, and 1 heart.
(13C1*39C2)/(13C3 + 13C2*39C1 + 13C1*39C2)
I get about 0.743 but the answer in the back of the book says 0.581
I've also tried the number of ways to choose 1 heart and 2 not hearts divided by the number of ways to choose 3 cards out of 52.
(13C1*39C2)/(52C3) which is about 0.436

sl8on said:

Using a standard deck of cards, you deal 3 cards. From those 3 cards you choose one card. If the chosen card is a heart, what is the probability that it is the only heart in the hand?
I thought that it would be the number of ways to choose one heart multiplied by the number of ways to choose 2 cards not hearts divided by the sum of the ways to choose 3, 2, and 1 heart.
(13C1*39C2)/(13C3 + 13C2*39C1 + 13C1*39C2)
I get about 0.743 but the answer in the back of the book says 0.581
I've also tried the number of ways to choose 1 heart and 2 not hearts divided by the number of ways to choose 3 cards out of 52.
(13C1*39C2)/(52C3) which is about 0.436

Click to expand...

Reduce your sample space. There are 51 cards left, 39 of which are non-hearts. The probability you have no additional hearts is 39/51*38/50.

Another way to think about it:

(39 choose 2)/(51 choose 2)

Hello, sl8on!

This is a very messy problem.
It requires Bayes' Theorem and a lot of work . . .

Using a standard deck of cards, you deal 3 cards.
From those 3 cards you choose one card.
If the chosen card is a heart, what is the probability that it is the only heart in the hand?[
Answer: \(\displaystyle 0.581\)

Click to expand...

This is conditional probability.

You have a hand of 3 cards.
Given that you randomly choose a card from your hand and it is a Heart,
. . what is the probability that it is the only Heart?

\(\displaystyle P(\text{only }\heartsuit\,|\,\text{choose a }\heartsuit) \;=\;\dfrac{P(\text{only }\heartsuit\,\wedge\,\text{choose a }\heartsuit)}{P(\text{choose a }\heartsuit)} \) .[1]


We know your hand contains at least one Heart.,

There are \(\displaystyle {52\choose3} = 22,\!100\) possible 3-card hands.

There are \(\displaystyle {39\choose3} = 9,\!139\) hands with no Hearts.

Hence, there are: \(\displaystyle 22,\!100 - 9,\!139 \:=\:12,\!169\) hands with some Hearts.
. . This is our denominator for this problem.


There are 3 types of hands.

(1) One Heart, two Others: .\(\displaystyle {13\choose1}{39\choose2} \,=\,9633\) hands.

(2) Two Hearts, one Other: .\(\displaystyle {13\choose2}{39\choose1} \,=\,3042\) hands.

(3)Three Hearts: .\(\displaystyle {13\choose3} \,=\,286\) hands.


The probability of having a Type (1) hand and choosing a Heart is:
. . \(\displaystyle \frac{9,633}{12,961}\cdot\frac{1}{3} \:=\:\frac{3,211}{12,961}\) .This is the numerator of [1]

The probability of having a Type (2) hand and choosing a Heart is:
. . \(\displaystyle \frac{3,042}{12,961}\cdot\frac{2}{3} \:=\:\frac{2,028}{12,961}\)

The probability of having a Type (3) hand and choosing a Heart is:
. . \(\displaystyle \frac{286}{12,961}\cdot\frac{3}{3} \:=\:\frac{286}{12,961}\)

Hence: .\(\displaystyle P(\text{choose a }\heartsuit) \:=\:\frac{3,211}{12,961} + \frac{2,028}{12,961} + \frac{286}{12,961} \:=\:\frac{5,525}{12,961}\)
. . This is the denominator of [1].


Therefore: .\(\displaystyle P(\text{only }\heartsuit\,|\,\text{chose a }\heartsuit) \;=\;\dfrac{\frac{3,211}{12,961}}{\frac{5,525}{12,961}} \;=\;\frac{3,211}{5,525} \;=\;0.581\,176\,471 \)

Whew! .I need a nap . . .

Personally, I prefer daon2's method.

Hey! Thank you all for your help.

daon2 I really like the simplicity of your answer, but don't quite follow how to reduce the sample space. Is it because you already know that you drew a heart that you only need to look at the probability of the other two cards not being hearts?

soroban, this problem did come from the section using Bayes' Theorem, so the teacher probably intended for us to use Bayes' Theorem. I've noticed that the denominators for both parts of the fractions are the same in every problem I've worked. Is this always true? If it is, then my original attempt only needed to multiply each part by the probability of choosing a heart each time. That is the numerator would be the number of ways to choose one heart times the number of ways to choose 2 not hearts times the probability of choosing the heart; and the denominator would be the sum of the ways to choose 3, 2, and 1 heart, each multiplied by the probability of choosing a heart from that combination.
(13C1*39C2*1/3)/(13C3*1 + 13C2*39C1*2/3 + 13C1*39C2*1/3)

Again, thank you all,
Sean

sl8on said:

daon2 I really like the simplicity of your answer, but don't quite follow how to reduce the sample space. Is it because you already know that you drew a heart that you only need to look at the probability of the other two cards not being hearts?

Click to expand...

Correct, though perhaps mine is not a(n obvious) rigorous solution. Reducing your sample space is equivalent to calculating a conditional probability. You take your given information, and collect the subset of events which satisfies it to form your new sample space. Then use the counting principle.

Think of your reduced sample space as all sets of the form: {Known Heart} U {Two unknown cards}. It is then easy to see there are (1 choose 1)*(51 choose 2) elements in the sample space.

How many satisfy the event in question? They are all events of the form: {Known heart} U {Two non-Hearts}. So: (1 choose 1)*(39 choose 2).