Question

Probability of At Least One Let A = the event of getting at least 1 defective iPhone when 3 iPhones are randomly selected with replacement from a batch. If 5% of the iPhones in a batch are defective and the other 95% are all good, which of the following are correct?

a. = (0.95)(0.95)(0.95) = 0.857

b. P(A) = 1 - (0.95)(0.95)(0.95) = 0.143

c. P(A) = (0.05)(0.05)(0.05) = 0.000125

Short answer

a. is correct.

b. is correct.

c. is incorrect.

Step-by-step solution

Given information

Three iPhones are selected at random. A is the event of selecting at least one defective iPhone.

The selection is made with a replacement from the batch.

The chances of selecting a defective iPhone are 5%, the rest 95% being good.

Define probability 

If A is the probability of selecting at least n out of N units of a given kind, then it has the following probability:

$P\left(\text{at}\text{least}\text{n}\text{are}\text{of a}\text{kind}\right)=P\left(nareofakind\right)+P\left(n+1areofakind\right)+....+P\left(Nareofakind\right)$

Complementary events

A and $\overline{A}$are said to be complementary when the outcomes of one event do not support the other occurrence of the other event.

That is,

$P\left(A\right)+P\left(\overline{A}\right)=1$

Compute the probability event A and its complement

A is the event of selecting at least one defective iPhone.

The probability of selecting a defective iPhone is:

$$\begin{gathered} P\left(defective\right)=\frac{5}{100} \\ =0.05 \end{gathered}$$

The probability of selecting a good iPhone is:

$$\begin{gathered} P\left(good\right)=\frac{95}{100} \\ =0.95 \end{gathered}$$

a.

The probability of selecting no defective iPhone is given as:

$$\begin{gathered} P\left(\overline{A}\right)=\left(0.95\right)\times \left(0.95\right)\times \left(0.95\right) \\ =0.857 \end{gathered}$$

The probability of $\overline{A}$is the probability that none are defective, or all are good.

Thus, the given probability represents selecting a good iPhone in each of the three selections.

Therefore, option a. is correct

b.

It is known that

$$\begin{gathered} P\left(A\right)+P\left(\overline{A}\right)=1 \\ P\left(A\right)=1-P\left(\overline{A}\right) \\ P\left(A\right)=1-\left(\left(0.95\right)\times \left(0.95\right)\times \left(0.95\right)\right) \\ P\left(A\right)=0.143 \end{gathered}$$

Therefore, option b. is correct.

c.

$P\left(A\right)$is the probability of choosing one or two or all three defective iPhones, which is computed as 0.143.

The given probability only depicts the case of choosing all three defective iPhones but not the cases of:

• Choosing one defective iPhone
• Choosing two defective iPhones

Therefore, option c. is incorrect.