Question

A certain group of women has a 0.61% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she doesn't have red/green blindness? What is the probability that the woman selected does not have red/green color blindness?

Answer 1

Answer: 99.39%

Explanation:

Given : A certain group of women has a 0.61% rate of red/green color blindness.

We can write 0.61%= 0.0061 [by dividing it by hundred]

Let E be the event of a woman has red/gree.n blindness

Then, the probability that a woman has red/green blindness :

P(E)= 0.0061

Now, there are only two possible events , one is "E" and the other is "E not" .

`P(E)+P(E\ not)=1\\\\\Rightarrow\ P(E\ not)=1-P(E)\\\\\Rightarrow\ P(E\ not)=1-0.0061=0.9939`

Hence, the probability that the woman selected does not have red/green color blindness = 0.9939=99.39%

Answer 2

Hey there!

Not sure if it's my eyes, but I'm pretty sure you asked the same question twice. It looks like they're both asking what the probability of choosing a woman without red/green color blindness from a specified group. Do let me know if they really are two different questions, though, and I'll be happy to help you out.

If there is a 0.61% rate of choosing a woman with red/green colorblindness, then there is a 99.39% chance of choosing a woman without red/green colorblindness. You could also express this as decimals (which you do by simply dividing your percentages by 100), which would be 0.0061 and 0.9939, respectively.

Your answer will be 99.39% or 0.9939 (without a percent sign!).

Hope this helped you out! :-)