A magician shuffles a standard deck of playing cards and allows an audience member to pull out a card, look at it, and replace it in the deck. Two additional people do the same. Find the probability that of the 3 cards drawn, at least 1 is a face card. (Round your answer to one decimal place.)

Respuesta :

Answer: 0.54

Step-by-step explanation:

We know that there are 12 face cards out of the 52 cards of a standard deck. So the probability of drawing a face card will be:  

 [tex]\text{ P(face)}=\frac{12}{52}=\frac{3}{13}[/tex]

Logically the probability of not drawing a face card will be:  

  [tex]\text{ P(not face)}=1-\text{ P(face)}=\frac{10}{13}[/tex]

Since the drawn cards by audience member are put back in the deck, each draw is independent from the previous draw.  

Therefore, the probability that of the 3 cards drawn, no card is  is a face card is given by :

[tex]\text{P(not face card)}=\frac{10}{13}\times\frac{10}{13}\times\frac{10}{13}=\frac{1000}{2197}[/tex]

Now, the probability that of the 3 cards drawn,  at least 1 is a face card is given by :-

[tex]\text{P(atleast one face card)}=1-\text{P(not face card)}=1-\frac{1000}{2197}=\frac{1197}{2197}=0.54483\approx0.54[/tex]

Using the binomial distribution, it is found that there is a 0.545 = 54.5% probability that of the 3 cards drawn, at least 1 is a face card.

For each card drawn, there are only two possible outcomes, either it is a face card, or it is not. The probability of a card being a face card is independent of any other card, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 3 cards are drawn, thus [tex]n = 3[/tex].
  • In a deck of 52 cards, 12 are faces, thus [tex]p = \frac{12}{52} = 0.2308[/tex].

The probability of at least one face is given by:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{3,0}.(0.2308)^{0}.(0.7692)^{3} = 0.455[/tex]

Then:

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.455 = 0.545[/tex]

0.545 = 54.5% probability that of the 3 cards drawn, at least 1 is a face card.

A similar problem is given at https://brainly.com/question/24863377