Question

About 7% of men in the United States have some form of red-green color blindness. Suppose we

Answer 1

Answer:

The probability will be  $\frac{9538}{37345}$

Step-by-step explanation:

Let the total population of United States is 100.

As per the given scenario, among them 7 men have the red-green color blindness.

Among the 4 selected males having at least 1 man having color blindness there could be total 4 possible cases.

CASE 1:

1 of the 4 men having blindness.

The probability will be $\frac{7C1\times93C3}{100C4} = \frac{7\times\frac{93\times92\times91}{6} }{\frac{100\times99\times98\times97}{24} } = \frac{28\times93\times92\times91}{100\times99\times98\times97}$

CASE 2:

2 of the 4 men having blindness.

The probability in this case will be $\frac{7C2\times93C2}{100C4} = \frac{7\times6\times\frac{93\times92}{2} }{\frac{100\times99\times98\times97}{24} } = \frac{56\times93\times92}{100\times99\times98\times97}$

CASE 3:

3 of the 4 men having blindness.

The required probability is $\frac{7C3\times93C1}{100C4} = \frac{7\times6\times\frac{93\times5}{6} }{\frac{100\times99\times98\times97}{24} } = \frac{35\times93\times24}{100\times99\times98\times97}$

CASE 4:

All of the 4 men that will be chosen, have the blindness.

In this case all of the men will be chosen from the 7% of the total population.

Hence, the probability is $\frac{7C4}{100C4} = \frac{7\times6\times5\times4 }{100\times99\times98\times97 } = \frac{35\times24}{100\times99\times98\times97}$

As any of the above 4 cases could be possible, in order to get the desired answer we need to add them.

hence, the answer is $\frac{9538}{37345}$