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

a. What is the probability of exactly 2 defective PS4s among them?

b. Given that at least 2 purchased PS4s are defective, what is the probability that exactly 2 are defective?

(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:

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:

(a)

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

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:

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

The value of P (X ≥ 2) is:

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

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

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1