1. **Problem statement:** We have two boxes U1 and U2 with colored balls. U1 contains 2 red and 3 green balls, U2 contains 3 red and 2 green balls. We draw 3 balls as follows: - Draw one ball from U1 and note its color. - If red, replace it in U1 and draw 2 balls simultaneously from U1. - If green, put it in U2 and draw 2 balls simultaneously from U2. We define events: - $R$: drawing a red ball - $V$: drawing a green ball - $A$: drawing 3 balls of the same color - $B$: drawing at least one green ball 2. **Probability tree completion:** - Probability of drawing red from U1: $P(R) = \frac{2}{5}$ - Probability of drawing green from U1: $P(V) = \frac{3}{5}$ If first ball is red (R): - Since ball is replaced, U1 still has 2 red and 3 green balls. - Probability of drawing 2 red balls simultaneously from U1: $P(RR|R) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$ - Probability of drawing 1 red and 1 green ball simultaneously from U1: $P(RV|R) = \frac{2 \times 3}{\binom{5}{2}} = \frac{6}{10} = \frac{3}{5}$ - Probability of drawing 2 green balls simultaneously from U1: $P(VV|R) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10}$ If first ball is green (V): - The green ball is moved to U2, so U2 now has 3 red and 3 green balls. - Probability of drawing 2 red balls simultaneously from U2: $P(RR|V) = \frac{\binom{3}{2}}{\binom{6}{2}} = \frac{3}{15} = \frac{1}{5}$ - Probability of drawing 1 red and 1 green ball simultaneously from U2: $P(RV|V) = \frac{3 \times 3}{15} = \frac{9}{15} = \frac{3}{5}$ - Probability of drawing 2 green balls simultaneously from U2: $P(VV|V) = \frac{\binom{3}{2}}{15} = \frac{3}{15} = \frac{1}{5}$ 3. **Calculate $P(A)$: probability of 3 balls of the same color** - Case 1: All red balls - First ball red: $\frac{2}{5}$ - Then 2 red balls: $\frac{1}{10}$ - So $P(3R) = \frac{2}{5} \times \frac{1}{10} = \frac{2}{50} = \frac{1}{25}$ - Case 2: All green balls - First ball green: $\frac{3}{5}$ - Then 2 green balls from U2: $\frac{1}{5}$ - So $P(3V) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$ - Total $P(A) = P(3R) + P(3V) = \frac{1}{25} + \frac{3}{25} = \frac{4}{25}$ 4. **Calculate $P(B)$: probability of at least one green ball** - Complement event: no green balls, i.e., all red balls - $P(\text{no green}) = P(3R) = \frac{1}{25}$ - So $P(B) = 1 - \frac{1}{25} = \frac{24}{25}$ 5. **Random variable $X$ = number of red balls drawn in the 3-ball draw** - Possible values: 0,1,2,3 Calculate probabilities: - $P(X=3) = P(3R) = \frac{1}{25}$ - $P(X=0) = P(3V) = \frac{3}{25}$ - $P(X=1)$: first ball red then 2 green or first ball green then 1 red 1 green - From first ball red: $\frac{2}{5} \times \frac{3}{10} = \frac{6}{50} = \frac{3}{25}$ - From first ball green: $\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$ - Total $P(X=1) = \frac{3}{25} + \frac{9}{25} = \frac{12}{25}$ - $P(X=2)$: first ball red then 1 red 1 green (already counted in $X=1$?), or first ball green then 2 red balls - From first ball red: $\frac{2}{5} \times \frac{6}{10} = \frac{12}{50} = \frac{6}{25}$ - From first ball green: $\frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$ - Total $P(X=2) = \frac{6}{25} + \frac{3}{25} = \frac{9}{25}$ Check sum: $\frac{1}{25} + \frac{3}{25} + \frac{12}{25} + \frac{9}{25} = \frac{25}{25} = 1$ 6. **Calculate expectation $E(X)$:** $$E(X) = 0 \times \frac{3}{25} + 1 \times \frac{12}{25} + 2 \times \frac{9}{25} + 3 \times \frac{1}{25} = 0 + \frac{12}{25} + \frac{18}{25} + \frac{3}{25} = \frac{33}{25} = 1.32$$ **Final answers:** - $P(A) = \frac{4}{25}$ - $P(B) = \frac{24}{25}$ - $E(X) = \frac{33}{25} = 1.32$