Some iPhones produced within a time period have faulty battery issues. Thirty percent of such iPhones will need battery service while under warranty. A family owns 5 iPhones produced within the time frame of potential faulty production.
(a) What is the probability that exactly two iPhones will end up being serviced under warranty?
(b) What is the probability that at least two iPhones will end up being serviced under warranty?
(c) What is the expected number of iPhones in the family that require service?
(d) What is the standard deviation of the number of iPhones in the family that requires service?

Answer:

a) 0.3087 = 30.87% probability that exactly two iPhones will end up being serviced under warranty.

b) 0.4718 = 47.18% probability that at least two iPhones will end up being serviced under warranty

c) The expected number of iPhones in the family that require service is 1.5.

d) The standard deviation of the number of iPhones in the family that requires service is 1.0247.

Step-by-step explanation:

For each iPhone, there are only two possible outcomes. Either they have battery issues under warranty, or they do not. iPhones are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Thirty percent of such iPhones will need battery service while under warranty.

This means that [tex]p = 0.3[/tex]

A family owns 5 iPhones produced within the time frame of potential faulty production.

This means that [tex]n = 5[/tex]

(a) What is the probability that exactly two iPhones will end up being serviced under warranty?

This is P(X = 2).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{3} = 0.3087[/tex]

0.3087 = 30.87% probability that exactly two iPhones will end up being serviced under warranty.

(b) What is the probability that at least two iPhones will end up being serviced under warranty?

This is:

[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{3} = 0.3087[/tex]

[tex]P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{2} = 0.1323[/tex]

[tex]P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{1} = 0.0284[/tex]

[tex]P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{0} = 0.0024[/tex]

[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.3087 + 0.1323 + 0.0284 + 0.0024 = 0.4718[/tex]

0.4718 = 47.18% probability that at least two iPhones will end up being serviced under warranty.

(c) What is the expected number of iPhones in the family that require service?

[tex]E(X) = np = 5*0.3 = 1.5[/tex]

The expected number of iPhones in the family that require service is 1.5.

(d) What is the standard deviation of the number of iPhones in the family that requires service?

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{5*0.3*0.7} = 1.0247[/tex]

The standard deviation of the number of iPhones in the family that requires service is 1.0247.

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