Question

Bag A contains 7 Red Balls, 'X' Green Balls, and 5 Yellow Balls. The probability to
pick Green Ball at random is 2/5. Another Bag B contains 'X-3' Red Balls, 'X-4'
Yellow Balls and 6 Green Balls. If two balls are picked one after the other from
Bag B at random then what is the probability for the Balls to be Red?​

Answer 1

Answer:

2. 2/21

Explanation:

X/(7+X+5) =2/5

X =8

Bag B = 5 R, 4 Y , 6 G

Probability = 5/15*4/14 = 2/21

Answer 2

Given,

Bag A has 7 red balls, 'X' green balls, and 5 yellow balls.

The chance of randomly selecting Green Ball from bag A is $\frac{2}{5}$.

Bag B also includes 'X-3' Red Balls, 'X-4' Yellow Balls, and 6 Green Balls.

To find,

The probability of the balls being red when they are drawn one after the other from bag B.

Solution,

We may easily answer this mathematical issue by following the steps below.

The following is the method for calculating the probability.

We know that,

The chance of randomly selecting a green ball from Bag A is $\frac{x}{12+x}$.

But, it is given that the probability is $\frac{2}{5}$.

Equating both the values;

⇒ $\frac{x}{12+x}= \frac{2}{5}$

⇒ 5x = 24 + 2x

⇒ 3x = 24

⇒ x = 8

Thus, bag A contains eight green balls.

Now,

The composition of Bag B includes 5 red balls, 4 yellow balls, and 6 green balls.

Thus,

The probability of the balls being red when they are drawn one after the other from bag B will be,

⇒ $\frac{5}{15} * \frac{4}{14}$

⇒ $\frac{2}{21}$

As a consequence, the probability for the balls to be red when picked one after another from bag B is $\frac{2}{21}$.