Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG indicate that the first two children born are boys and the third child is a girl, let GBG indicate that the first and third children born are girls and the second is a boy, and so forth. (a) Use set-roster notation to describe the elements in the sample space whose outcomes are all possible genders of the three children. (b) Write each of the following events as a set, in set-roster notation, and find its probability. (i) the event that exactly one child is a girl set probability (ii) the event that at least two children are girls set probability (iii) the event that no child is a girl set probability

Answer:

(a)

[tex]S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}[/tex]

(b)

i.

[tex]1\ girl = \{GBB, BBG, BGB\}[/tex]

[tex]P(1\ girl) = 0.375[/tex]

ii.

[tex]Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}[/tex]

[tex]P(Atleast\ 2 \ girls) = 0.5[/tex]

iii.

[tex]No\ girl = \{BBB\}[/tex]

[tex]P(No\ girl) = 0.125[/tex]

Step-by-step explanation:

Given

[tex]Children = 3[/tex]

[tex]B = Boys[/tex]

[tex]G = Girls[/tex]

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

[tex]S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}[/tex]

And the number of elements are:

[tex]n(S) = 8[/tex]

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

[tex]1\ girl = \{GBB, BBG, BGB\}[/tex]

And the number of the list is;

[tex]n(1\ girl) = 3[/tex]

The probability is calculated as;

[tex]P(1\ girl) = \frac{n(1\ girl)}{n(S)}[/tex]

[tex]P(1\ girl) = \frac{3}{8}[/tex]

[tex]P(1\ girl) = 0.375[/tex]

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

[tex]Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}[/tex]

And the number of the list is;

[tex]n(Atleast\ 2 \ girls) = 4[/tex]

The probability is calculated as;

[tex]P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}[/tex]

[tex]P(Atleast\ 2 \ girls) = \frac{4}{8}[/tex]

[tex]P(Atleast\ 2 \ girls) = 0.5[/tex]

Solving (biii) No girl

From the list of possible elements, we have:

[tex]No\ girl = \{BBB\}[/tex]

And the number of the list is;

[tex]n(No\ girl) = 1[/tex]

The probability is calculated as;

[tex]P(No\ girl) = \frac{n(No\ girl)}{n(S)}[/tex]

[tex]P(No\ girl) = \frac{1}{8}[/tex]

[tex]P(No\ girl) = 0.125[/tex]

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