Question

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 a 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?

Answer 1

a) 0.2503 = 25.03% probability that exactly 3 have green seeds.

b) 0.4744 = 47.44% probability that more than 2 have green seeds.

c) 0.1020 = 10.20% probability that more than 30 have green seeds.

d) 0.1414 = 14.14% probability that between 30 and 35 inclusive have green seeds.

e) 0.8508 = 85.05% probability that fewer than 80 have yellow seeds

Explanation:

For questions a and b, the binomial probability distribution is used. For questions c, d and e, the binomial distribution is approximated to the normal.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

A) Out of 10 offspring of heterozygous plants, what is the probability that exactly 3 have green seeds?

10 offspring of heterozygous plants means that
`n = 3`

Probability 0.25 of having green seeds means that
`p = 0.25`

This probability is P(X = 3). So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 3) = C_(10,3).(0.25)^(3).(0.75)^(7) = 0.2503`

0.2503 = 25.03% 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?

This is

`P(X > 2) = 1 - P(X \leq 2)`

In which

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)`

So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(10,0).(0.25)^(0).(0.75)^(10) = 0.0563`

`P(X = 1) = C_(10,1).(0.25)^(1).(0.75)^(9) = 0.1877`

`P(X = 2) = C_(10,2).(0.25)^(2).(0.75)^(8) = 0.2816`

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0563 + 0.1877 + 0.2816 = 0.5256`

`P(X > 2) = 1 - P(X \leq 2) = 1 - 0.5256 = 0.4744`

0.4744 = 47.44% 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?

Now

`n = 100`

`\mu = E(x) = np = 100*0.25 = 25`

`\sigma = √(V(X)) = √(np(1-p)) = √(100*0.25*0.75) = 4.33`

Using continuity correction, this is
`P(X > 30 + 0.5) = P(X > 30.5)`, which is 1 subtracted by the pvalue of Z when X = 30.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (30.5 - 25)/(4.33)`

`Z = 1.27`

`Z = 1.27` has a pvalue of 0.8980

1 - 0.8980 = 0.1020

0.1020 = 10.20% 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?

Using continuity correction, this is
`P(30-0.5 \leq X \leq 35 + 0.5) = P(29.5 \leq X \leq 35.5)`, which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 29.5.

X = 35.5

`Z = (X - \mu)/(\sigma)`

`Z = (35.5 - 25)/(4.33)`

`Z = 2.42`

`Z = 2.42` has a pvalue of 0.9922

X = 29.5

`Z = (X - \mu)/(\sigma)`

`Z = (29.5 - 25)/(4.33)`

`Z = 1.04`

`Z = 1.04` has a pvalue of 0.8508

0.9922 - 0.8508 = 0.1414

0.1414 = 14.14% 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?

0.75 probability of yellow seeds, so
`p = 0.75`. So

`\mu = E(x) = np = 100*0.75 = 75`

`\sigma = √(V(X)) = √(np(1-p)) = √(100*0.75*0.25) = 4.33`

This probability is
`P(X < 80 - 0.5) = P(X < 79.5)`, which is the pvalue of Z when X = 79.5.

`Z = (X - \mu)/(\sigma)`

`Z = (79.5 - 75)/(4.33)`

`Z = 1.04`

`Z = 1.04` has a pvalue of 0.8508

0.8508 = 85.05% probability that fewer than 80 have yellow seeds