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.

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?

Solution 3SE

Step1 of 6:

Let us consider a pie plant it contains two genes for seeds colour.

Y = yellow seeds and

G = green seeds.

Here each type of gene is called as heterozygous.

Mendelian theory of genetics is stats that:

If two heterozygous plants are crossed, Then

P(Each of their offspring of having yellow seeds) = 0.75 and

P(Each of their offspring of having green seeds) = 0.25.

Let X follows binomial distribution with parameters “n and p ”.

Then the probability mass function of binomial distribution is given by:

, x = 0,1,2,...,n.

Where,

n = sample size

x = random variable

p = probability of success

q = 1 - p (probability of failure)

Here our goal is:

a).We need to find the probability that exactly 3 have green seeds, when Out of 10 offspring of heterozygous plants.

b).We need to find the probability that more than 2 have green seeds, when Out of 10 offspring of heterozygous plants.

c).We need to find the probability that more than 30 have green seeds, when Out of 100 offspring of heterozygous plants.

d).We need to find the probability that between 30 and 35 inclusive have green seeds, when Out of 100 offspring of heterozygous plants.

e).We need to find the probability that fewer than 80 have yellow seeds, when Out of 100 offspring of heterozygous plants.

Step2 of 6:

a).

We have n = 10 and P(Each of their offspring of having green seeds) = 0.25 that is p = 0.25 and

x = 3

Now,

XB(n, p)

XB(10, 0.25)

Consider the pmf:

, x = 0,1,2,...,n.

Substitute n, p and x value in above equation then

0.250

Hence, P(Exactly 3 have green seeds) = 0.250.

Step3 of 6:

b).

We have n = 10 and P(Each of their offspring of having green seeds) = 0.25 that is p = 0.25 and

x > 2.

Now,

XB(n, p)

XB(10, 0.25)

Consider the pmf:

, x = 0,1,2,...,n.

Substitute n, p and x value in above equation then

Consider, can be obtained from Excel by using function “=Binomdist(X,n,p,FALSE)”

...X | Function | P(X=x) |

0 | Binorm(0,10,0.25,FALSE)” | 0.05631351471 |

1 | Binorm(1,10,0.25,FALSE)” |