The sample space listing the eight simple events that are possible when a couple has three children is {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}. After identifying the sample space for a couple having four children, find the probability of getting three girls and one boy (in any order ).

Using probability and sample space concepts, it is found that there is a 0.25 = 25% probability of getting three girls and one boy (in any order ).

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For 4 children, the sample space is given by:

B - B - B - B

B - B - B - G

B - B - G - B

B - B - G - G

B - G - B - B

B - G - B - G

B - G - G - B

B - G - G - G

G - B - B - B

G - B - B - G

G - B - G - B

G - B - G - G

G - G - B - B

G - G - B - G

G - G - G - B

G - G - G - G

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Thus:

0.25 = 25% probability of getting three girls and one boy (in any order ).

A similar problem is given at link

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Step-by-step explanation:

BBBG

BBGB

BGBB

GBBB

GGGB

GGBG

GBGG

BGGG

BBGG

GGBB

GBGB

BGBG

GBBG

BGGB

3G and 1B

4/16 = 1/4

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1