Question

Among 8 PS4s, four are good and four have defects. Unaware of this, a customer buys 5 PS4s.

Answer 1

just buy one ps4 it is $299

Answer 2

Answer:

(a) The probability of exactly 2 defective PS4s among them is 0.3125.

(b) The probability that exactly ​ 2 are defective given that  ​at least ​ 2 purchased PS4s are defective is 0.3846.

Step-by-step explanation:

Let X = number of defective PS4s.

It is provided that 4 PS4s of 8 are defective.

The probability of selecting a defective PS4 is:

$P(X)=p=\frac{4}{8}=0.50$

A customer bought n = 5 PS4s.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.50.

The probability function of a Binomial distribution is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2, 3...$

(a)

Compute the probability of exactly 2 defective PS4s among them as follows:

$P(X=2)={5\choose 2}(0.50)^{2}(1-0.50)^{5-2}=10\times0.25\times0.125=0.3125$

Thus, the probability of exactly 2 defective PS4s among them is 0.3125.

(b)

Compute the probability that exactly ​ 2 are defective given that  ​at least ​ 2 purchased PS4s are defective as follows:

$P(X=2|X\geq 2)=\frac{P(X=2\cap X\geq2)}{P(X\geq2)} =\frac{P(X=2)}{P(X\geq 2)}$

The value of P (X = 2) is 0.3125.

The value of P (X ≥ 2) is:

$P(X\geq 2)=1-P(X$

Then the value of P (X = 2 | X ≥ 2) is:

$P(X=2|X\geq 2)=\frac{P(X=2)}{P(X\geq 2)}=\frac{0.3125}{0.8125} =0.3846$

Thus, the probability that exactly ​ 2 are defective given that  ​at least ​ 2 purchased PS4s are defective is 0.3846.