Question

2.(3 pts) Among 8 PS4s, four are good and four have defects. Unaware of this, a customerbuys 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?

Answer 1

G : Good : non-defective

B : Defective (Bad)

Number of Defective PS4s = 4

Number of Good (non-defective ) PS4s = 4

Number of PS4s customer buys = 5

a.

Probability of exactly 2 defective PS4s among them = P(2B3G)

= Number of ways of selecting 2 defective PS4s from 4 defective PS4s x Number of ways of selecting 3 good PS4s from 4 good PS4s / Number of ways of selecting 5 PS4s from 8 PS4s

Number of ways of selecting 2 defective PS4s from 4 defective PS4s = = 6

Number of ways of selecting 3 good PS4s from 4 good PS4s = =4

Number of ways of selecting 5 PS4s from 8 PS4s = =56

Probability of exactly 2 defective PS4s among them = P(2B3G)

= Number of ways of selecting 2 defective PS4s from 4 defective PS4s x Number of ways of selecting 3 good PS4s from 4 good PS4s / Number of ways of selecting 5 PS4s from 8 PS4s

= (6x4)/56 = 24/56 = 3/7 = 0.428571429

Probability of exactly 2 defective PS4s among them = 3/7 = 0.428571429

b.

Given that ​at least​ 2 purchased PS4s are defective, probability that exactly ​2 are defective

X : Event of at least​ 2 purchased PS4s are defective

A : Event of exactly ​2 are defective

Given that ​at least​ 2 purchased PS4s are defective, probability that exactly ​2 are defective = P(A|X)

By Multiplication theorem of probability

P(X and A) = P(X) P(A|X)

X and A : i.e Event of at least​ 2 purchased PS4s are defective and Event of exactly ​2 are defective;

If exactly 2 are defective then at least 2 purchased will always be defective : therefore A is subset of X

X and A = A

P(A) = P(X) P(A|X)

P(A|X) = P(A)/P(X)

From (a) P(A) = 3/7

X : Event of at least​ 2 purchased PS4s are defective can happen if :

2B3G : 2 Defective 3 Good : From a . P(2B3G) = 3/7

3B2G : 3 defective 2 Good :

P(3B2G) =

Number of ways of selecting 3 defective PS4s from 4 defective PS4s x Number of ways of selecting 2 good PS4s from 4 good PS4s / Number of ways of selecting 5 PS4s from 8 PS4s

Number of ways of selecting 3 defective PS4s from 4 defective PS4s = = 4

Number of ways of selecting 2 good PS4s from 4 good PS4s = =6

Number of ways of selecting 5 PS4s from 8 PS4s = =56

P(3B2G) =

Number of ways of selecting 3 defective PS4s from 4 defective PS4s x Number of ways of selecting 2 good PS4s from 4 good PS4s / Number of ways of selecting 5 PS4s from 8 PS4s

= (4x6)/56 = 3/7

4B1G : 4 defective 1 Good :

P(4B1G) =

Number of ways of selecting 4 defective PS4s from 4 defective PS4s x Number of ways of selecting 1 good PS4s from 4 good PS4s / Number of ways of selecting 5 PS4s from 8 PS4s

Number of ways of selecting 3 defective PS4s from 4 defective PS4s = = 1

Number of ways of selecting 2 good PS4s from 4 good PS4s = =4

Number of ways of selecting 5 PS4s from 8 PS4s = =56

P(4B1G) =

Number of ways of selecting 4 defective PS4s from 4 defective PS4s x Number of ways of selecting 1 good PS4s from 4 good PS4s / Number of ways of selecting 5 PS4s from 8 PS4s

= (1x4)/56 = 4/56 = 1/14

P(X) = P(2B3G) + P(3B2G)+P(4B1G) = 3/7+3/7+1/14 = 13/14

Given that ​at least​ 2 purchased PS4s are defective, probability that exactly ​2 are defective = 6/13 = 0.461538462