Respuesta :
Using the hypergeometric distribution, it is found that there is a
- 0.0588 = 5.88% probability of drawing two clubs from a standard deck of 52 cards.
- 0.0498 = 4.98% probability of drawing two faces from a standard deck of 52 cards.
- 0.0045 = 0.45% probability of drawing two aces from a standard deck of 52 cards.
The cards are chosen without replacement, hence, the hypergeometric 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]
- Two cards are chosen, hence [tex]n = 2[/tex].
To draw 2 clubs, it is P(X = 2) when [tex]k = 13[/tex], as there are 13 clubs in a deck. 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 = 2) = h(2,52,2,13) = \frac{C_{13,2}C_{39,0}}{C_{52,0}} = 0.0588[/tex]
0.0588 = 5.88% probability of drawing two clubs from a standard deck of 52 cards.
To draw 2 faces, it is P(X = 2) when [tex]k = 12[/tex], as there are 12 faces in a deck. 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 = 2) = h(2,52,2,12) = \frac{C_{12,2}C_{40,0}}{C_{52,0}} = 0.0498[/tex]
0.0498 = 4.98% probability of drawing two faces from a standard deck of 52 cards.
To draw 2 aces, it is P(X = 2) when [tex]k = 4[/tex], as there are 4 aces in a deck. 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 = 2) = h(2,52,2,4) = \frac{C_{4,2}C_{48,0}}{C_{52,0}} = 0.0045[/tex]
0.0045 = 0.45% probability of drawing two aces from a standard deck of 52 cards.
A similar problem is given at https://brainly.com/question/24826394
