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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.

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

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

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

That is,

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

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:

$$\begin{gathered} P\left(\stackrel{炉}{A}\right)=\left(0.95\right)脳\left(0.95\right)脳\left(0.95\right) \\ =0.857 \end{gathered}$$

The probability of $\stackrel{炉}{A}$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

$$\begin{gathered} P\left(A\right)+P\left(\stackrel{炉}{A}\right)=1 \\ P\left(A\right)=1-P\left(\stackrel{炉}{A}\right) \\ P\left(A\right)=1-\left(\left(0.95\right)脳\left(0.95\right)脳\left(0.95\right)\right) \\ P\left(A\right)=0.143 \end{gathered}$$

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.