Mathematics, 25.01.2020 01:31 angeleyes4u610p6np54

Exercise 3.23 introduces a husband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes

(a) what is the probability that their first child will have green eyes and the second will not?

(b) what is the probability that exactly one of their two children will have green eyes?

(c) if they have six children, what is the probability that exactly two will have green eyes?

(d) if they have six children, what is the probability that at least one will have green eyes?

Answers: 2

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Answer from: purplefish53

a) There is a 10.9375% probability that their first child will have green eyes and the second will not.

b) There is a 21.875% probability that exactly one of their two children will have green eyes.

c) There is a 13.74% probability that exactly two will have green eyes.

d) There is a 55.12% probability that at least one will have green eyes.

Step-by-step explanation:

In this problem, the binomial probability distribution is going to be important for itens b,c and d.

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

We have these following probabilities:

0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes.

(a) What is the probability that their first child will have green eyes and the second will not?

There is a 0.125 probability a child will have green eyes and an 1-0.125 = 0.875 probability a child will not have green eyes.

0.125*0.875 = 0.109375

There is a 10.9375% probability that their first child will have green eyes and the second will not.

(b) What is the probability that exactly one of their two children will have green eyes?

Here we use the binomial probability distribution, with n = 2, p = 0.125 .

We want P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{2,1}*(0.125)^{1}*(0.875)^{1} = 0.21875$

There is a 21.875% probability that exactly one of their two children will have green eyes.

Again the binomial probability distribution, with n = 6, p = 0.125

We want P(X = 2)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{6,2}*(0.125)^{2}*(0.875)^{4} = 0.1374$

There is a 13.74% probability that exactly two will have green eyes.

(d) If they have six children, what is the probability that at least one will have green eyes?

n = 6, p = 0.125

Either none has green eyes, or at least one has. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = C_{6,0}*(0.125)^{0}*(0.875)^{6} = 0.4488$

Finally

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4488 = 0.5512$

There is a 55.12% probability that at least one will have green eyes.

Answer from: Quest

113.1

step-by-step explanation:

v=4

3πr3=4

3·π·33≈113.09734

Answer from: Quest

yes.. it's proven to be 36.

Can someone me i think the answer is d. but not sure.

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