Question

All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High frequency EM is thought to be a cause of cancer; the lower frequencies associated with household current are generally assumed to be harmless. The following table summarizes the probability distribution for cancer sufferers and their wiring configuration in the Denver area.
Leukemia Lymphoma Other Cancers
High Frequency wiring 0.242 0.047 0.079
Low frequency wiring 0.391 0.098 ???
(a) What is the missing probability (labelled ???) in the above table?

(b) What is the probability of having high frequency wiring among cancer suffers in the Denver area?

(c) Is the event "Having Leukemia" independent of the event "Having high frequency frequency wiring"? Explain.

Answer 1

Answer:

$x = 0.143$

$P(High\ |\ Cancer) = 0.215$

Not independent

Step-by-step explanation:

Given

See attachment for proper table

Solving (a): The missing probability

First, we add up the given probabilities

$Sum = 0.242+0.047+0.079+0.391+0.098$

$Sum = 0.857$

The total probability must add up to 1.

If the missing probability is x, then:

$x + 0.857 = 1$

Collect like terms

$x = -0.857 + 1$

$x = 0.143$

Solving (b): P(High | Cancer)

This is calculated as:

$P(High\ |\ Cancer) = \frac{n(High\ n\ Cancer)}{n(Cancer)}$

So, we have:

$P(High\ |\ Cancer) = \frac{0.079}{0.242+0.047+0.079}$

$P(High\ |\ Cancer) = \frac{0.079}{0.368}$

$P(High\ |\ Cancer) = 0.215$

Solving (c): P(Leukemia) independent of P(High Wiring)

From the attached table

$P(Leukemia\ n\ High\ Wiring) = 0.242$

$P(Leukemia) = 0.242 + 0.391 =0.633$

$P(High\ Wiring) = 0.242+0.047+0.079=0.368$

If both events are independent, then:

$P(Leukemia\ n\ High\ Wiring) = P(Leukemia) * P(High\ Wiring)$

$0.242 = 0.633 * 0.368$

$0.242 \ne 0.232$

Since the above is an inequality, then the events are not independent