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Metal detector data when it passes over the objects (metal or non-metal):

Machine beeps for $98\%$ of the metal objects over which it passes.

Machine beeps for $4\%$ of the nonmetallic objects over which it passes.

Machine passes over $25\%$of the metal objects.

According to complement rule,

$P\left({\mathrm{A}}^{\mathrm{c}}\right)=P(not\mathrm{A})=1-P(\mathrm{A})$

According to the general multiplication rule,

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}鈭�\mathrm{B})=P(\mathrm{A})脳P(\mathrm{B}鈭�\mathrm{A})=P(\mathrm{B})脳P(\mathrm{A}鈭�\mathrm{B})$

According to the addition rule for mutually exclusive events,

$P(\mathrm{A}鈭�\mathrm{B})=P(\mathrm{A}\text{or}\mathrm{B})=P(\mathrm{A})+P(\mathrm{B})$

Definition for conditional probability:

$P(\mathrm{B}鈭�\mathrm{A})=\frac{P(\mathrm{A}鈭�\mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

Let

M: Metal

$M^c:$Non - metal

B: Machine beeps

$B^c:$Machine does not beep

NoW,

The corresponding probabilities:

Probability for getting metal,

$P\left(\mathrm{M}\right)=0.25$

Probability for getting metal when machine beeps,

$P(\mathrm{B}鈭�\mathrm{M})=0.98$

Probability for getting non - metal when machine beeps,

$P\left(\mathrm{B}鈭�{\mathrm{M}}^{\mathrm{c}}\right)=0.04$

Now,

Apply the complement rule:

Probability for getting non - metal,

$P\left({\mathrm{M}}^{\mathrm{c}}\right)=1-P\left(\mathrm{M}\right)=1-0.25=0.75$

Then

Apply general multiplication rule:

Probability for machine beeps and getting metal,

$P(\mathrm{B}\text{and}\mathrm{M})=P\left(\mathrm{M}\right)脳P(\mathrm{B}鈭�\mathrm{M})=0.25脳0.98=0.245$

Probability for machine beeps and getting non - metal,

$P\left(B\text{and}{\mathrm{M}}^{\mathrm{c}}\right)=P\left({\mathrm{M}}^{\mathrm{c}}\right)脳P\left(\mathrm{B}鈭�{\mathrm{M}}^{\mathrm{c}}\right)=0.75脳0.04=0.030$

We know that

Machine passes over either metal or non - metal.

Since both events are not possible at same time,

Apply addition rule for mutually exclusive events:

Probability for machine beeps,

$P\left(B\right)=P\left(B\text{and}\mathrm{M}\right)+P\left(B\text{and}{\mathrm{M}}^{\mathrm{c}}\right)$

$=0.245+0.030$

$=0.275$

Using conditional probability definition:

$P(\mathrm{M}鈭�\mathrm{B})=\frac{P(\mathrm{B}\text{and}\mathrm{M})}{P(\mathrm{B})}=\frac{0.245}{0.275}=\frac{49}{55}鈮�0.8909$

Thus,

If the machine beeps while passing over randomly selected object, the probability of getting a metal object is approx.$0.8909.$