Question

The most common form of color blindness is an inability to distinguish red from green. However, this particular form of color blindness is much more common in men than in women (this is because the genes corresponding to the red and green receptors are located on the X-chromosome). Approximately 79% of American men and 0.4% of American women are red-green color-blind.1 Let CBM and CBW denote the events that a man or a woman is color-blind, respectively.
(a) If an Americal male is selected at random, what is the probability that he is red-green color-blind? P(CBM) =
(b) If an American female is selected at random, what is the probability that she is NOT red-green color-blind? P (not CBW) =
(c) If one man and one woman are selected at random, what is the probability that neither are red-green color-blind? P=(neither is color-blind) =
(d) If one man and one woman are selected at random, what is the probability that at least one of them is red-green color-blind? P=(at least one is color-blind)

Answer 1

Answer:

(a) P(CBM) = 0.07

(b) P(not CBW) = 0.996

(c ) P(neither is color-blind) = 0.926

(d) P=(at least one is color-blind) = 0.074

Step-by-step explanation:

The correct data is  that Approximately 7% of American men and 0.4% of American women are red-green color-blind.

(a) Probability that he is red-green color-blind:

$P(CBM) = 0.07$

(b) Probability that she is NOT red-green color-blind:

$P(not\ CBW) =1- P(CBW)\\P(not\ CBW) = 1 -0.004\\P(not\ CBW) =0.996$

(c) Probability that neither are red-green color-blind

$P(neither) = P(not\ CBW)*P(not\ CBM) \\P(neither) = 0.996 *(1-0.07)\\P(neither)=0.926$

(d) Probability that at least one of them is red-green color-blind

$P(at\ least\ one) = 1- P(neither) \\P(at\ least\ one) = 1-0.926\\P(at\ least\ one) = 0.074$