Probability of Getting 1 Jack when taking 5 cards from a deck

In summary: Then for the interpretation that the phrase 'considering that in the first 3 cards there is a jack' means there is at least one jack, then I need to consider the cases whereby there are 2 or 3 jacks. So would I need to set up an expression along the following lines?You can do it either by direct counting of permutations, or by conditional probability (Bayes' thm).
  • #1
Master1022
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Homework Statement
(a) What is the probability of picking up ONLY one jack when taking 5 cards from a 52 card deck
(b) Considering that in the first 3 cards there is a jack what is the probability that the jack was the first card to be extracted
Relevant Equations
Probability and combinatorics
Hi,

I was looking through probability questions and attempting it.

Question:
(a) What is the probability of picking up ONLY one jack when taking 5 cards from a 52 card deck
(b) Considering that in the first 3 cards there is a jack what is the probability that the jack was the first card to be extracted

Attempt:
For part (a), I did:
[tex] \text{probability} = \frac{\text{number of ways of picking up 5 cards with only 1 jack}}{\text{total number of ways of picking up 5 cards from 52 cards}} [/tex]
[tex] P = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 52 - 4 \\ 4 \end{pmatrix}}{\begin{pmatrix} 52 \\ 5 \end{pmatrix}} = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 48 \\ 4 \end{pmatrix}}{\begin{pmatrix} 52 \\ 5 \end{pmatrix}} [/tex]
The terms in the numerator represent:
- ##\begin{pmatrix} 4 \\ 1 \end{pmatrix}## is the number of jack's available
- ##\begin{pmatrix} 48 \\ 4 \end{pmatrix}## is the number of ways to pick the remaining 4 cards from 48 non-jack cards in the deck.

Does this look right for (a)?

For part (b), I thought it would be ##\frac{1}{3}##, but I may need to give this more thought...

Thanks in advance.
 
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  • #2
Looks OK to me.
 
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  • #3
Be careful with the wording in part b. "Considering that in the first 3 cards there is a jack"

The results are different depending upon whether you take that to mean "exactly one jack", or whether you take it to mean "one or more jacks". Personally I what have thought it meant one or more jacks, but in any case, it's very important to be clear on the details in this type of question.

In one interpretation the correct answer is 1/3 (can you tell me which interpretation that is?), but in the other interpretation the answer is greater than 1/3. Perhaps you could have a go at tackling each separate case.
 
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  • #4
uart said:
Be careful with the wording in part b. "Considering that in the first 3 cards there is a jack"

The results are different depending upon whether you take that to mean "exactly one jack", or whether you take it to mean "one or more jacks". Personally I what have thought it meant one or more jacks, but in any case, it's very important to be clear on the details in this type of question.

In one interpretation the correct answer is 1/3 (can you tell me which interpretation that is?), but in the other interpretation the answer is greater than 1/3. Perhaps you could have a go at tackling each separate case.
Thanks @uart ! You have raised a fair point. I think the 1/3 would be for the interpretation where there is only 1 jack within the three cards.

Then for the interpretation that the phrase 'considering that in the first 3 cards there is a jack' means there is at least one jack, then I need to consider the cases whereby there are 2 or 3 jacks. So would I need to set up an expression along the following lines?
[tex] \text{P(first card jack| at least one jack)} = \text{P(first card jack| 1 jack)} \cdot \text{P(1 jack| at least one jack)} + \text{P(first card jack| 2 jacks)} \cdot \text{P(2 jack| at least one jack)} + \text{P(first card jack| 3 jack)} \cdot \text{P(3 jack| at least one jack)} [/tex]

So I can get:
- ##\text{P(first card jack| 1 jack)} = 1/3##
- ##\text{P(first card jack| 2 jacks)} = 2/3##
- ##\text{P(first card jack| 3 jack)} = 1##

Then I would just need to calculate the other terms. Would I do something as follows for ##\text{P(2 jack| at least one jack)}##?
[tex] \text{P(2 jack| at least one jack)} = \frac{\begin{pmatrix} 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 48 \\ 1 \end{pmatrix}}{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 48 \\ 2 \end{pmatrix} + \begin{pmatrix} 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 48 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 48 \\ 0 \end{pmatrix}} [/tex]

Then I could change the numerator for the other two terms?
 
  • #5
An alternative approach for part a)

The probability of precisely one Jack is five times the probability that only the first card is a Jack. And that's quite easy to calculate using direct probabilities.

My advice is to do it both ways and check you get the same answer!
 
  • #6
Master1022 said:
Thanks @uart ! You have raised a fair point. I think the 1/3 would be for the interpretation where there is only 1 jack within the three cards.
That's correct.
Then for the interpretation that the phrase 'considering that in the first 3 cards there is a jack' means there is at least one jack, then I need to consider the cases whereby there are 2 or 3 jacks. So would I need to set up an expression along the following lines?
You can do it either by direct counting of permutations, or by conditional probability (Bayes' thm).

If by counting, let n1 be the total number of permutation containing exactly one jack (in the first 3 cards), n2 be the total number of permutations containing exactly 2 jacks, and similarly for n3.

Because they're mutually exclusive, n1+n2+n3 is then the total number of permutations containing one or more jack in the first 3 cards. Now for the numerator, you just have to work out what fraction of n1 has a jack in the first position, and similarly for n2 and n3 (as you've already done above)

Alternately you can use conditional probability.
Define "A" as the (unconditional) event that the first card dealt is a jack, and "B" as the (unconditional) event that there are one of more jacks in the first 3 cards dealt. You need to find P(A|B) = P(A and B)/P(B).

P(A) is very easy. P(B) is best calculated by considering the complementary event, 1-P(B). And P(A and B) is very easy if we recognize that one of those two events is a subset of the other.
 
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Related to Probability of Getting 1 Jack when taking 5 cards from a deck

1. What is the probability of getting 1 Jack when taking 5 cards from a deck?

The probability of getting 1 Jack when taking 5 cards from a deck is approximately 0.4224, or 42.24%. This can be calculated by dividing the number of ways to get 1 Jack (4) by the total number of possible combinations (comb(52,5) = 2,598,960).

2. How does the number of Jacks in the deck affect the probability?

The number of Jacks in the deck does not affect the probability. The probability remains the same regardless of the number of Jacks in the deck, as long as there is at least 1 Jack.

3. What is the probability of getting more than 1 Jack when taking 5 cards from a deck?

The probability of getting more than 1 Jack when taking 5 cards from a deck is approximately 0.0391, or 3.91%. This can be calculated by adding the number of ways to get 2 Jacks (comb(4,2) = 6) and 3 Jacks (comb(4,3) = 4), and dividing by the total number of possible combinations (comb(52,5) = 2,598,960).

4. How does the order of the cards affect the probability?

The order of the cards does not affect the probability. The probability remains the same regardless of the order in which the cards are drawn, as long as the total number of cards drawn is 5.

5. Can the probability be increased by taking more than 5 cards from the deck?

No, the probability of getting 1 Jack when taking 5 cards from a deck is the maximum probability possible. Taking more than 5 cards from the deck will not increase the probability, as the number of possible combinations will also increase.

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