Question

Metal detector A boy uses a homemade metal detector to look for valuable metal objects on a beach. The machine isn’t perfect—it beeps for only 98% of the metal objects over which it passes, and it beeps for 4% of the nonmetallic objects over which it passes. Suppose that 25% of the objects that the machine passes over are metal. Choose an object from this beach at random. If the machine beeps when it passes over this object, find the probability that the boy has found a metal object

Short answer

Probability that the boy has found a metal object is approx. $0.8909.$

Step-by-step solution

Step 1:Given information

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.

Step 2:Calculation

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,

$P(\mathrm{A}\text{and}\mathrm{B})=P(\mathrm{A}\cap \mathrm{B})=P(\mathrm{A})\times P(\mathrm{B}\mid \mathrm{A})=P(\mathrm{B})\times P(\mathrm{A}\mid \mathrm{B})$

According to the addition rule for mutually exclusive events,

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

Definition for conditional probability:

$P(\mathrm{B}\mid \mathrm{A})=\frac{P(\mathrm{A}\cap \mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

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,

$P(\mathrm{B}\mid \mathrm{M})=0.98$

Probability for getting non - metal when machine beeps,

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

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,

$P(\mathrm{B}\text{and}\mathrm{M})=P\left(\mathrm{M}\right)\times P(\mathrm{B}\mid \mathrm{M})=0.25\times 0.98=0.245$

Probability for machine beeps and getting non - metal,

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

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:

$P(\mathrm{M}\mid \mathrm{B})=\frac{P(\mathrm{B}\text{and}\mathrm{M})}{P(\mathrm{B})}=\frac{0.245}{0.275}=\frac{49}{55}\approx 0.8909$

Thus,

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