Answer :
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:
[tex]P(X)=p=\frac{4}{8}=0.50[/tex]
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:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2, 3...[/tex]
(a)
Compute the probability of exactly 2 defective PS4s among them as follows:
[tex]P(X=2)={5\choose 2}(0.50)^{2}(1-0.50)^{5-2}=10\times0.25\times0.125=0.3125[/tex]
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:
[tex]P(X=2|X\geq 2)=\frac{P(X=2\cap X\geq2)}{P(X\geq2)} =\frac{P(X=2)}{P(X\geq 2)}[/tex]
The value of P (X = 2) is 0.3125.
The value of P (X ≥ 2) is:
[tex]P(X\geq 2)=1-P(X<2)\\=1-P(X=0)-P(X=1)\\=1-0.03125-0.15625\\=0.8125[/tex]
Then the value of P (X = 2 | X ≥ 2) is:
[tex]P(X=2|X\geq 2)=\frac{P(X=2)}{P(X\geq 2)}=\frac{0.3125}{0.8125} =0.3846[/tex]
Thus, the probability that exactly 2 are defective given that at least 2 purchased PS4s are defective is 0.3846.
