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Kryger [21]
3 years ago
6

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

icate 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
Mathematics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

(a)

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

(b)

i.

1\ girl = \{GBB, BBG, BGB\}

P(1\ girl) = 0.375

ii.

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

P(Atleast\ 2 \ girls) = 0.5

iii.

No\ girl = \{BBB\}

P(No\ girl) = 0.125

Step-by-step explanation:

Given

Children = 3

B = Boys

G = Girls

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

The possible elements are:

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

And the number of elements are:

n(S) = 8

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

1\ girl = \{GBB, BBG, BGB\}

And the number of the list is;

n(1\ girl) = 3

The probability is calculated as;

P(1\ girl) = \frac{n(1\ girl)}{n(S)}

P(1\ girl) = \frac{3}{8}

P(1\ girl) = 0.375

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

And the number of the list is;

n(Atleast\ 2 \ girls) = 4

The probability is calculated as;

P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}

P(Atleast\ 2 \ girls) = \frac{4}{8}

P(Atleast\ 2 \ girls) = 0.5

Solving (biii) No girl

From the list of possible elements, we have:

No\ girl = \{BBB\}

And the number of the list is;

n(No\ girl) = 1

The probability is calculated as;

P(No\ girl) = \frac{n(No\ girl)}{n(S)}

P(No\ girl) = \frac{1}{8}

P(No\ girl) = 0.125

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