According to a theory in genetics, if tall and colorful plants are crossed with short and colorless plants, four types of plants will result: tall and colorful, tall and colorless, short and colorful, and short and colorless, with corresponding probabilities of nine sixteenths , three sixteenths , three sixteenths , and one sixteenth . In a multinomial experiment, if 10 plants are selected, find the probability that 3 will be tall and colorful, 3 will be tall and colorless, 3 will be short and colorful, and 1 will be short and colorless.

The probability that 3 will be tall and colorful, 3 will be tall and colorless, 3 will be short and colorful, and 1 will be short and colorless out of 10 plants is 0.0081.

Step-by-step explanation:

As the values of different probabilities are given as

X1=tall and colorful, P(X1)=9/16

X2=tall and colorless, P(X2)=3/16

X3=short and colorful, P(X3)=3/16

X4=short and colorless, P(X4)=1/16

Now the total number of plants is 10 and the X1=3, X2=3,X3=3,X4=1

so

[tex]P(X1=n_1,X2=n_2,X3=n_3,X4=n_4)=\dfrac{n!}{n_1!n_2!n_3!n_4!}p_1^{n_1}p_2^{n_2}p_3^{n_3}p_4^{n_4}\\=\dfrac{10!}{3!3!3!1!}\left(\dfrac{9}{16}\right)^{3}\left(\dfrac{3}{16}\right)^{3}\left(\dfrac{3}{16}\right)^{3}\left(\dfrac{1}{16}\right)^{1}\\=\dfrac{279006525}{2^{33}\cdot \:4}\\=0.0081[/tex]

So the probability that 3 will be tall and colorful, 3 will be tall and colorless, 3 will be short and colorful, and 1 will be short and colorless out of 10 plants is 0.0081.

MrRoyal MrRoyal · Answer 2 · 2022-01-18T14:19:22+01:00

Probabilities are used to determine the chances of events.

The required probability is 0.008121

To determine the required probability, we make use of the following representations

a : tall and colorful
b : tall and colorless
c : short and colorful
d : short and colorless

So, we have:

[tex]P(a)=\frac{9}{16}[/tex].
[tex]P(b)=\frac{3}{16}[/tex].
[tex]P(c)=\frac{3}{16}[/tex].
[tex]P(d)=\frac{1}{16}[/tex].
Sample size: n = 10

The probability that 3 will be tall and colorful, 3 will be tall and colorless, 3 will be short and colorful, and 1 will be short and colorless is then calculated as:

[tex]P(A = a,B=b,C =c,D=d) = \frac{n!}{a!b!c!d!} \times p_1^a p_2^bp_3^cp_4^d}[/tex]

This gives

[tex]P(a = 3,b=3,c =3,d=1) = \frac{10!}{3!3!3!1!} \times (\frac{9}{16})^3 \times (\frac{3}{16})^3 \times (\frac{3}{16})^3 \times (\frac{1}{16})^1[/tex]

Evaluate the factorials

[tex]P(a = 3,b=3,c =3,d=1) = \frac{3628800}{6 \times 6 \times 6 \times 1} \times (\frac{9}{16})^3 \times (\frac{3}{16})^3 \times (\frac{3}{16})^3 \times (\frac{1}{16})^1[/tex]

[tex]P(a = 3,b=3,c =3,d=1) = 16800 \times (\frac{9}{16})^3 \times (\frac{3}{16})^3 \times (\frac{3}{16})^3 \times (\frac{1}{16})^1[/tex]

Evaluate the exponents

[tex]P(a = 3,b=3,c =3,d=1) = 16800 \times 0.177979 \times 0.006592 \times 0.006592 \times 0.0625[/tex]

[tex]P(a = 3,b=3,c =3,d=1) = 0.008121[/tex]

Hence, the probability is 0.008121