Acertain 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?

99.39%

Step-by-step 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" .

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

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

she gotta workk that a 2

step-by-step explanation: