Question

A bag contains 40 white balls some of which are red some of balls are yellow and the rest are black the probability of drawing a red ball and yellow ball at random from the bag are 1 by 4 and 2 by 5 respectively find the probability of drawing a black ball at random from the bag 2 X + 1 Red balls and X + 2 yellow balls are added to the bag while X - 3 black balls are removed from the bag the probability of drawing a yellow ball at random from the bag is now 3 by 7 find an expression in terms of x for the total number of balls in the bag now, find the number of yellow balls in the bag now

Answer 1

Step-by-step explanation:

Total no. of balls = 40

Since, Probability = No. of Favourable Outcomes / Total no. of Outcomes

For Red Balls

$\frac{1}{4} = \frac{red \: balls}{40}$

$red \: balls \: = \frac{40}{4}$

$red \: balls \: = 10$

For Yellow Balls

$\frac{2}{5} = \frac{yellow \: balls}{40}$

$yellow \: balls \: = \frac{80}{5}$

$yellow \: balls \: = 16$

Therefore, No. of Black balls = 40 - (16 + 10)

= 40 - 26

= 14

After the ball are added

Now no. of balls in the bag = 40 + 2x + 1 + x + 2 - x +3

= 40 + 2x + 6

= 46 + 2x

No. of yellow balls now = 16 + x + 2

= 18 + x

Probability = 3/7

3 / 7 = 18 + x / 46 + 2x

3(46 + 2x) = 7(18 +x)

138 + 6x = 126 + 7x

7x - 6x = 138 - 126

x = 12

Now, No. of black balls = 14 + 12 - 3

= 23

No. of total balls = 46 + 2(12)

= 46 + 24

= 70

Now, probability of black balls = 23/70

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