Question

6. Of all the televisions that are sent to a certain workshop for repair, 80% are no longer under warranty

a. What is the expected number of televisions under warranty for a group of 20 televisions? (5 pts)

b. Among those 20 televisions, what is the probability that at least 75% are not covered under the warranty? (8pts)

c. Suppose there are currently 12 televisions in the workshop of which 5 are under warranty. 8 televisions are taken at random to be worked

i. What is the probability that at least 4 of those that are under warranty will be chosen? (5pts)

ii. What is the probability that all televisions of the same type will be chosen? (7pts)

Answer 1

Given, 80% of the televisions are no longer under warranty

Therefore, probability that a television is not under warranty = 80% = 0.8

Probability that a television is under warranty = 1 – 0.8 = 0.2

Answer a:
Let X be a random variable representing the number of televisions under warranty in a group of 20 televisions.

Therefore, X follows Binomial Distribution with parameters, n – number of independent observations = 20 and p – probability of success = 0.2

X follows Bin.(20, 0.2)

The P.M.F. of X, f(x) = (nCx) x (p^x) x (1 - p)^(n - x)     x = 0, 1, 2,……..19, 20

Expected number of televisions under warranty for a group of 20 televisions = np = 20 x 0.2 = 4

Answer b:
According to the situation, at least 75% of the 20 televisions are not covered under the warranty, that is, at least ((75/100) x 20) = 15 televisions are not covered.

Let Y be a random variable representing the number of televisions that are not under warranty in a group of 20 televisions.

Therefore, Y follows Binomial Distribution with parameters, n – number of independent observations = 20 and p – probability of success (not covered under warranty) = 0.8

X follows Bin.(20, 0.8)

The P.M.F. of X, f(y) = (nCx) x (p^y) x (1 - p)^(n - y)     y = 0, 1, 2,……..19, 20

According to given,

The required probability –

Prob.[Y ≥ 15] = P[Y = 15] + P[Y = 16] + P[Y = 17] + P[Y = 18] + P[Y = 19] + P[Y = 20]

                     = [(20C15) x (0.8^15) x (0.2^5)] + [(20C16) x (0.8^16) x (0.2^4)] + [(20C17) x (0.8^17) x (0.2^3)] + [(20C18) x (0.8^18) x (0.2^2)] + [(20C19) x (0.8^19) x (0.2^1)] + [(20C20) x (0.8^20) x 1]

                   = 0.80421

Answer c:
There are 12 televisions of which are 5 are under warranty, which means 7 are not under warranty.

8 televisions are taken at random to be worked.

Number of ways of choosing 8 televisions = 12C8 = 495

i. Let A be a random variable representing the number of televisions that are under warranty in the choice of 8 televisions.

According to given, the required probability –

Prob.[A ≥ 4] = P[A = 4] + P[A = 5] (Since there are a maximum of 5 televisions that are under warranty in the 12 televisions)

                      = {[(5C4) x (7C4)] / 12C8} + {[(5C5) x (7C3)] / 12C8} = 210/495 = 0.42424

ii. There are only two types of televisions – under warranty and not under warranty.

There are 12 televisions out of which 7 are not under warranty and 5 are under warranty. If one is choosing 8 televisions, it is not possible to choose 8 televisions of the same type, there can be at least 1 television of different type.

So, Probability that all televisions of the same type will be chosen = 0/495 = 0