Question

A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible outcomes: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, BGGG, GBGG, GGBG, GGGB, and GGGG. Assume these outcomes are equally likely. Let represent the number of children that are girls. Find the probability distribution of .

Answer 1

Answer:

A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible outcomes: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, BGGG, GBGG, GGBG, GGGB, and GGGG.

Let X represent the number of children that are girls. Then

1. When X=0, there is one possible outcome BBBB. So

2. When X=1, then there are 4 possible outcomes GBBB, BGBB, BBGB, BBBG, so

3. When X=2, then there are 6 possible outcomes BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, so

4. When X=3, then there are 4 possible outcomes GGGB, GGBG, GBGG, BGGG, so

5. When X=4, then there is one possible outcome GGGG, so

Now, the probability distribution table is

Step-by-step explanation:

Answer 2

Answer:

$\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}$

Step-by-step explanation:

A family has four children. If the genders of these children are listed in the order they are born, there are sixteen possible outcomes: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, BGGG, GBGG, GGBG, GGGB, and GGGG.

Let X represent the number of children that are girls. Then

1. When X=0, there is one possible outcome BBBB. So

$Pr(X=0)=\dfrac{1}{16}$

2. When X=1, then there are 4 possible outcomes GBBB, BGBB, BBGB, BBBG, so

$Pr(X=1)=\dfrac{4}{16}=\dfrac{1}{4}$

3. When X=2, then there are 6 possible outcomes BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, so

$Pr(X=2)=\dfrac{6}{16}=\dfrac{3}{8}$

4. When X=3, then there are 4 possible outcomes GGGB, GGBG, GBGG, BGGG, so

$Pr(X=3)=\dfrac{4}{16}=\dfrac{1}{4}$

5. When X=4, then there is one possible outcome GGGG, so

$Pr(X=4)=\dfrac{1}{16}$

Now, the probability distribution table is

$\begin{array}{cccccc}X&0&1&2&3&4\\Pr&\dfrac{1}{16}&\dfrac{1}{4}&\dfrac{3}{8}&\dfrac{1}{4}&\dfrac{1}{16}\end{array}$