There are three bags A, B and C. In each bag there are three types of colored balls Yellow, Green and Black. In bag A, no. of yellow colored balls are y and no. of green colored balls are g. Number of green colored balls are 4 more than the number of yellow colored balls. When one ball is picked at random then the probability of getting black color ball is 5 13 . The value of y is 18 2 11 % less then g. In bag B, number of yellow colored balls is 22 2 9 % more than that of bag A. If two balls are picked at random from bag B then the probability of getting both green color ball is 4 37 . Total number of balls in bag B is 75. In bag C, the ratio of number of green colored balls and number of black colored balls is 7 : 5. Total number of green and black colored balls is 36. If one ball is picked at random then the probability of getting one yellow ball is 7 13 .
If one ball is drawn from each of bags A and B, what is the probability that both balls are of the same color?

Question

Accepted Answer

**Solution:**

1. **Bag A:**
   - Given $g = y + 4$.
   - *y* is $18 \frac{2}{11}\%$ less than *g*.
     This means:
     $$ y = g 	imes \left(1 - \frac{20}{110}ight) = g 	imes \frac{90}{110} = \frac{9}{11} g $$
   - From $g = y + 4$, substitute $y = \frac{9}{11} g$:
     $$ g = \frac{9}{11} g + 4 \implies g - \frac{9}{11} g = 4 \implies \frac{2}{11} g = 4 \implies g = 22 $$
   - Then, $y = \frac{9}{11} 	imes 22 = 18$.
   - Total balls in bag A = $y + g + b$ where $b$ is black balls.
   - Probability of black ball = $\frac{b}{y + g + b} = \frac{5}{13}$.
   - Substitute $y=18$, $g=22$:
     $$ \frac{b}{18 + 22 + b} = \frac{5}{13} \implies \frac{b}{40 + b} = \frac{5}{13} $$
   - Cross multiply:
     $$ 13b = 5(40 + b) = 200 + 5b \implies 13b - 5b = 200 \implies 8b = 200 \implies b = 25 $$
   - Total balls in bag A = $18 + 22 + 25 = 65$.

2. **Bag B:**
   - Yellow balls in B are $22 \frac{2}{9}\%$ more than yellow balls in A.
   - Percentage increase = $22 \frac{2}{9}\% = \frac{200}{9}\%$.
   - Yellow balls in B:
     $$ 18 	imes \left(1 + \frac{200}{900}ight) = 18 	imes \frac{11}{9} = 22 $$
   - Total balls in B = 75.
   - Probability of both balls green when two balls are drawn is $\frac{4}{37}$.
   - Let green balls in B be $G$.
   - Probability of both green:
     $$ \frac{G}{75} 	imes \frac{G-1}{74} = \frac{4}{37} $$
   - Multiply both sides by $75 	imes 74$:
     $$ G(G-1) = \frac{4}{37} 	imes 75 	imes 74 $$
   - Calculate right side:
     $$ \frac{4}{37} 	imes 75 	imes 74 = 4 	imes 2 	imes 75 = 600 $$
     (Since $74/37 = 2$)
   - So,
     $$ G^2 - G - 600 = 0 $$
   - Solve quadratic:
     $$ G = \frac{1 \pm \sqrt{1 + 2400}}{2} = \frac{1 \pm 49}{2} $$
   - Positive root:
     $$ G = \frac{50}{2} = 25 $$
   - Green balls in B = 25.
   - Total balls in B = 75, yellow balls = 22, green balls = 25, so black balls = $75 - 22 - 25 = 28$.

3. **Bag C:**
   - Ratio of green to black balls = $7:5$.
   - Total green + black = 36.
   - Let green = $7x$, black = $5x$, so:
     $$ 7x + 5x = 36 \implies 12x = 36 \implies x = 3 $$
   - Green = $21$, black = $15$.
   - Probability of yellow ball = $\frac{7}{13.6} = \frac{7}{13.6}$.
   - Total balls in C = green + black + yellow = $21 + 15 + y_c$.
   - Probability yellow = $\frac{y_c}{21 + 15 + y_c} = \frac{7}{13.6}$.
   - Let total balls = $36 + y_c$.
   - So,
     $$ \frac{y_c}{36 + y_c} = \frac{7}{13.6} $$
   - Cross multiply:
     $$ 13.6 y_c = 7 (36 + y_c) = 252 + 7 y_c $$
   - Rearranged:
     $$ 13.6 y_c - 7 y_c = 252 \implies 6.6 y_c = 252 \implies y_c = \frac{252}{6.6} = 38.18 $$
   - Total balls in C = $36 + 38.18 = 74.18$ (approx).

4. **Probability that balls drawn from A and B are of the same color:**
   - Probability both yellow:
     $$ P_A(yellow) 	imes P_B(yellow) = \frac{18}{65} 	imes \frac{22}{75} = \frac{396}{4875} $$
   - Probability both green:
     $$ \frac{22}{65} 	imes \frac{25}{75} = \frac{550}{4875} $$
   - Probability both black:
     $$ \frac{25}{65} 	imes \frac{28}{75} = \frac{700}{4875} $$
   - Total probability:
     $$ \frac{396 + 550 + 700}{4875} = \frac{1646}{4875} $$
   - Simplify numerator and denominator:
     $$ \frac{1646}{4875} = \frac{21 	imes 47}{65 	imes 75} $$

**Answer:** $\boxed{\frac{21 	imes 47}{65 	imes 75}}$

Answer

Answer

Answer