Answer:
[tex]\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}[/tex]
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
[tex]Pr(X=0)=\dfrac{1}{16}[/tex]
2. When X=1, then there are 4 possible outcomes GBBB, BGBB, BBGB, BBBG, so
[tex]Pr(X=1)=\dfrac{4}{16}=\dfrac{1}{4}[/tex]
3. When X=2, then there are 6 possible outcomes BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, so
[tex]Pr(X=2)=\dfrac{6}{16}=\dfrac{3}{8}[/tex]
4. When X=3, then there are 4 possible outcomes GGGB, GGBG, GBGG, BGGG, so
[tex]Pr(X=3)=\dfrac{4}{16}=\dfrac{1}{4}[/tex]
5. When X=4, then there is one possible outcome GGGG, so
[tex]Pr(X=4)=\dfrac{1}{16}[/tex]
Now, the probability distribution table is
[tex]\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}[/tex]
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:
