Answer to Question #303471 in Statistics and Probability for angel

Question

. From a box containing 4 black balls and 3 green balls, 4 balls are drawn in succession. Let Y be a random variable representing the number of green balls that occur.

Answer 1

Let's denote B - black ball, G - green ball.

Sample space S is all possible outcomes.

$$S=\{BBBB, BBBG, BBGB, BGBB, GBBB,$$

$$BBGG, BGBG, BGGB, GBGB, GBBG, GGBB,$$

$$BGGG, GBGG, GGBG, GGGB\}$$

The possible values of the random variable $Y$ are $0, 1, 2, 3.$

We will assume that the probability of getting heads and tails is the same:

$$p = q =1/2$$ $$\def\arraystretch{1.5} \begin{array}{c:c} Possible \ Outcomes & Z \\ \hline BBBB & 0 \\ \hdashline BBBG & 1 \\ \hdashline BBGB & 1 \\ \hdashline BGBB & 1 \\ \hdashline GBBB & 1 \\ \hdashline BBGG & 2 \\ \hdashline BGBG & 2 \\ \hdashline BGGB & 2 \\ \hdashline GBGB & 2 \\ \hdashline GBBG & 2 \\ \hdashline GGBB & 2 \\ \hdashline BGGG & 1 \\ \hdashline GBGG & 1 \\ \hdashline GGBG & 1 \\ \hdashline GGGB & 1 \\ \hdashline \end{array}$$

Construct the probability distribution of the random variable

$$P(BBBB)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{2}{5})(\dfrac{1}{4})=\dfrac{1}{35}$$

$$P(BBBG)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{2}{5})(\dfrac{3}{4})=\dfrac{3}{35}$$

$$P(BBGB)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{3}{5})(\dfrac{2}{4})=\dfrac{3}{35}$$

$$P(BGBB)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{3}{5})(\dfrac{2}{4})=\dfrac{3}{35}$$

$$P(GBBB)=\dfrac{3}{7}(\dfrac{4}{6})(\dfrac{3}{5})(\dfrac{2}{4})=\dfrac{3}{35}$$

$$P(BBGG)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{3}{5})(\dfrac{2}{4})=\dfrac{3}{35}$$

$$P(BGBG)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{3}{5})(\dfrac{2}{4})=\dfrac{3}{35}$$

$$P(BGGB)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{2}{5})(\dfrac{3}{4})=\dfrac{3}{35}$$

$$P(GBGB)=\dfrac{3}{7}(\dfrac{4}{6})(\dfrac{2}{5})(\dfrac{3}{4})=\dfrac{3}{35}$$

$$P(GBBG)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{3}{5})(\dfrac{2}{4})=\dfrac{3}{35}$$

$$P(GGBB)=\dfrac{3}{7}(\dfrac{2}{6})(\dfrac{4}{5})(\dfrac{3}{4})=\dfrac{3}{35}$$

$$P(BGGG)=\dfrac{4}{7}(\dfrac{3}{6})(\dfrac{2}{5})(\dfrac{1}{4})=\dfrac{1}{35}$$

$$P(GBGG)=\dfrac{3}{7}(\dfrac{4}{6})(\dfrac{2}{5})(\dfrac{1}{4})=\dfrac{1}{35}$$

$$P(GGBG)=\dfrac{3}{7}(\dfrac{2}{6})(\dfrac{4}{5})(\dfrac{1}{4})=\dfrac{1}{35}$$

$$P(GGGB)=\dfrac{3}{7}(\dfrac{2}{6})(\dfrac{1}{5})(\dfrac{4}{4})=\dfrac{1}{35}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} g & 0 & 1 & 2 & 3 \\ \hline p(g) & 1/35 & 12/35 & 18/35 & 4/35 \end{array}$$