Pea plants contain two genes for seed color, each of which may be Y (for yellow seeds) or G (for green seeds). Plants that contain one of each type of gene are called heterozygous. According to the Mendelian theory of genetics, if two heterozygous plants are crossed, each of their offspring will have probability 0.75 of having yellow seeds and probability 0.25 of having green seeds.

Question

a. Out of 10 offspring of heterozygous plants, what is the probability that exactly 3 have green seeds?
b. Out of 10 offspring of heterozygous plants, what is the probability that more than 2 have green seeds?
c. Out of 100 offspring of heterozygous plants, what is the probability that more than 30 have green seeds?
d. Out of 100 offspring of heterozygous plants, what is the probability that between 30 and 35 inclusive have green seeds?
e. Out of 100 offspring of heterozygous plants, what is the probability that fewer than 80 have yellow seeds?

Answer 1

(a) Let X be the number of plants out of 10 that have green seeds. Then X~ Bin(10, 0.25).





(b)



(c) Let Y be the number of plants out of 100 that have green seeds.

Then Y~ Bin(100, 0.25) so Y is approximately normal with mean = 100(0.25)=25 and standard deviation



To find P(????>30), use the continuity correction and find the z-score for 30.5.

The z-score of 30.5 is (30.5 ? 25) / 4.3301 = 1.27.

The area to the right of z= 1.27 is 1 ? 0.8980 = 0.1020.



(d) To find P(30 ? ????? 35), use the continuity correction and find the z-scores for 29.5 and 35.5.

The z-score of 29.5 is (29.5 ? 25) / 4.3301 = 1.04.

The z-score of 35.5 is (35.5 ? 25) / 4.3301 = 2.42.

The area to between z= 1.04 and z= 2.42 is 0.9922 ? 0.8508 = 0.1414.



(e) Fewer than 80 have yellow seeds if more than 20 have green seeds.

To find P(Y>20), use the continuity correction and find the z-score for 20.5.

The z-score of 20.5 is (20.5 ? 25) / 4.3301 = ?1.04.

The area to the right of z= ?1.04 is 1 ? 0.1492 = 0.8508.