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About 7% of men in the United States have some form of red-green color blindness. Suppose we
randomly select 4 U.S. adult males. What's the probability that at least one of them is red-green color blinare
carry out a simulation to answer this question. Follow the four-step process.

Respuesta :

Answer:

The probability will be  [tex]\frac{9538}{37345}[/tex]

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 [tex]\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}[/tex]

CASE 2:

2 of the 4 men having blindness.

The probability in this case will be [tex]\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}[/tex]

CASE 3:

3 of the 4 men having blindness.

The required probability is [tex]\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}[/tex]

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 [tex]\frac{7C4}{100C4} = \frac{7\times6\times5\times4 }{100\times99\times98\times97 } = \frac{35\times24}{100\times99\times98\times97}[/tex]

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 [tex]\frac{9538}{37345}[/tex]