Suppose a couple planned to have three children. Let X be the number of girls the couple has.

a. List all possible arrangements of girls and boys. For example, if the couple had 2 girls and then a boy, then type GGB.
b. List the sample space for X.
c. What is the probability that X=2?
d. Find the probability that the couple have three boys.

Respuesta :

Answer:

a) {GGG, GGB, GBG, BGG, BBG, BGB, GBB, BBB}

b) {0,1,2,3}

c)

[tex]P(X=2) = \dfrac{3}{8}[/tex]

d)

[tex]P(\text{3 boys}) = \dfrac{1}{8}[/tex]

Step-by-step explanation:

We are given the following in the question:

Suppose a couple planned to have three children. Let X be the number of girls the couple has.

a) possible arrangements of girls and boys

Sample space:

{GGG, GGB, GBG, BGG, BBG, BGB, GBB, BBB}

b) sample space for X

X is the number of girls couple has. Thus, X can take the values 0, 1, 2 and 3 that is 0 girls, 1 girl, 2 girls and three girls from three children.

Sample space: {0,1,2,3}

c) probability that X=2

P(X=2)

That is we have to compute the probability that couple has exactly two girls.

[tex]\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}[/tex]

Favorable outcome: {GGB, GBG, BGG}

[tex]P(X=2) =\dfrac{3}{8}[/tex]

d) probability that the couple have three boys.

Favorable outcome: {BBB}

[tex]P(BBB) = \dfrac{1}{8}[/tex]

Using probability concepts, it is found that:

a) The possible arrangements are: {GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB}

b) The sample space is: S = {0, 1, 2, 3}.

c) [tex]\frac{3}{8}[/tex] probability that X=2.

d) [tex]\frac{1}{8}[/tex] probability that the couple have three boys.

  • A probability is the number of desired outcomes divided by the number of total outcomes.
  • The sample space is the set that contains all possible outcomes.

Item a:

Taking into account the different orders, then:

GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB.

Then, the possible arrangements is: {GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB}.

Item b:

The sample space is the number of girls, which is: S = {0, 1, 2, 3}.

Item c:

Out of 8 outcomes, 3 have two girls(GGB, GBG, BGG), thus:

[tex]p = \frac{3}{8}[/tex]

[tex]\frac{3}{8}[/tex] probability that X=2.

Item d:

Out of 8 outcomes, 1 has three boys, which is BBB, thus:

[tex]\frac{1}{8}[/tex] probability that the couple have three boys.

A similar problem is given at https://brainly.com/question/15006554