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The probability of selecting a defective iPhone is 0.20.

Fifteen iPhones are selected at random.

The engineer needs at least one defective iPhone.

The probability of an event happening鈥渁t least once鈥� is the probability of the eventhappening once or more than once.

It is also equal to one minus the probability of the event not happening at all.

$P\left(\text{A}\text{happening}\text{at}\text{least}\text{once}\right)=1-P\left(\text{not}\text{A}\right)$

Let A be the event of selecting a defective iPhone and be the event of selecting any good iPhone.

The probability of selecting a defective iPhone is given by:

$$\begin{gathered} P\left(A\right)=\frac{20}{100} \\ =0.2 \end{gathered}$$

Thus,P(A) is equal to 0.2.

The probability of selecting a good iPhone is given by:

$$\begin{gathered} P\left(\stackrel{炉}{A}\right)=1-P\left(A\right) \\ =1-0.2 \\ =0.8 \end{gathered}$$

Thus,$P\left(\stackrel{炉}{A}\right)$ is 0.8.

Fifteen iPhones are selected.

The probability of getting all good iPhones or not getting a defective iPhone is given by:

$$\begin{gathered} P\left(\text{all}\text{good}\right)=\left(0.8\right)脳\left(0.8\right)脳...脳\left(0.8\right) \\ ={\left(0.8\right)}^{15} \\ =0.0352 \end{gathered}$$

The probability of getting at least one defective iPhone is one minus the probability of getting all good iPhones:

$$\begin{gathered} P\left(\text{at}\text{least}\text{1}\text{defective}\right)=1-P\left(\text{all}\text{good}\right) \\ =1-0.0352 \\ =0.965 \end{gathered}$$

Therefore, the probability of getting at least one defective iPhone is equal to 0.965.

The probability of getting at least one defective iPhone is very high; 0.965.

Also, the engineer needs at least one defective iPhone to identify the problem(s).

The probability value of selecting at least one defective iPhone is fairly high. So, the engineer can be sure of getting a defect to work upon.