Answer :
Answer:
a) P=0.0000843
b) P=0.0000454
c) P=0.0066113
Step-by-step explanation:
6 cards are drawn at random without replacement from a standard deck.
a) We calculate the probability that all the cards are hearts.
We calculatethe number of possible combinations:
[tex]C^{52}_6=\frac{52!}{6!(52-6)!}=20358520[/tex]
We calculate the number of favorable combinations:
[tex]C_6^{13}=\frac{13!}{6!(13-6)!}=1716[/tex]
Therefore, the probability is
[tex]P=\frac{1716}{20358520}\\\\P=0.0000843[/tex]
b) We calculate the probability that all the cards are face cards.
We calculatethe number of possible combinations:
[tex]C^{52}_6=\frac{52!}{6!(52-6)!}=20358520[/tex]
We calculate the number of favorable combinations:
[tex]C_6^{12}=\frac{12!}{6!(12-6)!}=924[/tex]
Therefore, the probability is
[tex]P=\frac{924}{20358520}\\\\P=0.0000454[/tex]
c) We calculate the probability that all the cards are even.
We calculatethe number of possible combinations:
[tex]C^{52}_6=\frac{52!}{6!(52-6)!}=20358520[/tex]
We calculate the number of favorable combinations:
[tex]C_6^{24}=\frac{24!}{6!(24-6)!}=134596[/tex]
Therefore, the probability is
[tex]P=\frac{134596}{20358520}\\\\P=0.0066113[/tex]
Using the hypergeometric distribution, it is found that there is a:
a) 0.0000843 = 0.00843% probability that all the cards are hearts.
b) 0.0000454 = 0.00454% probability that all the cards are faces.
c) 0.0019039 = 0.19039% probability that all the cards are even.
The cards are chosen without replacement, which is why the hypergeometic distribution is used to solve this question.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- A standard deck has 52 cards, hence [tex]N = 52[/tex].
- 6 are chosen, hence [tex]n = 6[/tex].
Item a:
- 13 of the cards are hearts, hence [tex]k = 13[/tex].
The probability is P(X = 6), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 6) = h(6,52,6,13) = \frac{C_{13,6}C_{39,0}}{C_{52,6}} = 0.0000843[/tex]
0.0000843 = 0.00843% probability that all the cards are hearts.
Item b:
- 12 of the cards are faces, hence [tex]k = 12[/tex].
The probability is P(X = 6), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 6) = h(6,52,6,12) = \frac{C_{12,6}C_{40,0}}{C_{52,6}} = 0.0000454[/tex]
0.0000454 = 0.00454% probability that all the cards are faces.
Item c:
- 20 of the cards are even, hence [tex]k = 20[/tex].
The probability is P(X = 6), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 6) = h(6,52,6,20) = \frac{C_{20,6}C_{32,0}}{C_{52,6}} = 0.0019039[/tex]
0.0019039 = 0.19039% probability that all the cards are even.
A similar problem is given at https://brainly.com/question/24826394