contestada

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)

Respuesta :

Answer:

(a) The correct answer is P (CBM) = 0.79.

(b) The probability of selecting an American female who is not red-green color-blind is 0.996.

(c) The probability that neither are red-green color-blind is 0.9263.

(d) The probability that at least one of them is red-green color-blind is 0.0737.

Step-by-step explanation:

The variables CBM and CBW are denoted as the events that an American man or an American woman is colorblind, respectively.

It is provided that 79% of men and 0.4% of women are colorblind, i.e.

P (CBM) = 0.79

P (CBW) = 0.004

(a)

The probability of selecting an American male who is red-green color-blind is, 0.79.

Thus, the correct answer is P (CBM) = 0.79.

(b)

The probability of the complement of an event is the probability of that event not happening.

Then,

P(not CBW) = 1 - P(CBW)

                   = 1 - 0.004

                   = 0.996.

Thus, the probability of selecting an American female who is not red-green color-blind is 0.996.

(c)

The probability the woman is not colorblind is 0.996.

The probability that the man is  not color- blind is,

P(not CBM) = 1 - P(CBM)

                   = 1 - 0.004  

                   = 0.93.

The man and woman are selected independently.

Compute the probability that neither are red-green color-blind as follows:

[tex]P(\text{Neither is Colorblind}) = P(\text{not CBM}) \times P(\text{not CBW})\\ = 0.93 \times 0.996 \\= 0.92628\\\approx 0.9263[/tex]

Thus, the probability that neither are red-green color-blind is 0.9263.

(d)

It is provided that a one man and one woman are selected at random.

The event that “At least one is colorblind” is the complement of part (d) that “Neither is  Colorblind.”

Compute the probability that at least one of them is red-green color-blind as follows:

[tex]P (\text{At least one is Colorblind}) = 1 - P (\text{Neither is Colorblind})\\ = 1 - 0.9263 \\= 0.0737[/tex]

Thus, the probability that at least one of them is red-green color-blind is 0.0737.

Otras preguntas