Question

Assume that the causes of a disease are 29% of the time associated with diet and 71% of the time associated with genetics, and that the causes of the disease are independent from each other. What is the expected number of cases of the disease due to genetics before the first case appears due to diet?

Step 1

**Introduction to geometric distribution:**

Consider a random experiment involving *n* repeated independent Bernoulli trials until a success is obtained.

Moreover, the probability of getting a success is *p*, and it remains a constant for all the *n *independent trials. Denote the probability of failure as *q*. As success and failure are mutually exclusive, *q *= 1 – *p*.

Let the random variable *X* denote the number of trails before the first success. Thus, *X* can take any of the values 0,1,2,….*n.*

Then, the probability distribution of *X* is a Geometric distribution with parameter (*p*) and the probability mass function (pmf) of *X*, that is, of a geometric random variable, is given as:

Step 2

**Obtain the expected number of cases of the disease due to genetics until the first case of the disease appears due to diet:**

It is given that, the causes of a disease are 29% of the time associated with diet.

That is, the probability that a disease causes due to diet is *p *= 0.29.

The probability a disease not causes due to diet is *q *= 1 – *p *= 0.71.

Further it is given that, the causes of a disease are 71% of the time associated with genetics.

That is, the probability that a disease causes due to genetics is *p *= 0.71.

The probability a disease not causes due to genetics is *q *= 1 – *p *= 0.29.

From this, it is clear that if a disease not causes due to diet that means the disease causes due to genetics and vice versa.

Here, it is given that the causes of disease are independent from each other. As a result, the causes of disease may be considered as independent trials.

Consider the event that a disease is caused due to diet as a &ld...

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