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.

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

CASE 2:

2 of the 4 men having blindness.

The probability in this case will be

CASE 3:

3 of the 4 men having blindness.

The required probability is

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

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

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

i hope this is what you were looking for.

see attached graph