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 . Part: 0 / 20 of 2 Parts Complete Part 1 of 2 Your Answer is incorrect (a) Find the number of possible values for the random variable . There are possible values for the random variable .

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fichoh fichoh · Answer 1 · 2021-03-05T09:08:44+01:00

Answer:

X : ___ 0 ____ 1 _____ 2 _____ 3 _____ 4

P(x): _ 1/16 __ 1/4 ____ 3/8 ____ 1/4 ____ 1/16

Step-by-step explanation:

Given the listed possible occurrence of boys and girls :

B = boys ; G = girls

BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, GBGB, BGGB, GBBG, BBGG, GGBB, BGGG, GBGG, GGBG, GGGB, and GGGG

Let x = distribution of girls

P = required outcome / Total possible outcomes

For :

x = 0 ;(BBBB) = 1

P(x) = 1/16

x = 1 ; (BBBG, BBGB, BGBB, GBBB) = 4

P(x) = 4 / 16 = 1/4

P(x = 2) ; (BGBG, GBGB, BGGB, GBBG, BBGG, GGBB) = 6

P(x) = 6 / 16 = 3/8

P(x = 3) ; (BGGG, GBGG, GGBG, GGGB) = 4

P(x) = 4 / 16 = 1/4

P(x = 4) ; (GGGG) = 1

P(x) = 1 / 16

X : ___ 0 ____ 1 _____ 2 _____ 3 _____ 4

P(x): _ 1/16 __ 1/4 ____ 3/8 ____ 1/4 ____ 1/16