Question

A box contains a certain number of balls.Some of these balls are marked A,some are marked B and remainings are marked C. When a ball is drawn at ranfom from the box p(A)=1/3 and p (B)=1/4. If there are 40 balls in the box which are marked C,find the number of balls in the box.​

Answer 1

Answer:

96 balls.

Step-by-step explanation:

Let the number of balls be n.

Then P(A) = number of A/n = 1/3

=> number of A = n/3

P(B) = 1/4 = number of B/n

=> number of B = n/4

P(A) + P(B) + P(C) = 1, since, well, we must have a ball out of these three.

So, P(C) = 1 - P(A) - P(B) = 1 - 1/3 - 1/4 = 5/12

Now, P(C) = 5/12 = number of C/n = 40/n

So, n = 40 * 12/5 = 96.

Answer 2

Suppose the total number is A and B marked balls are 'x' and 'y'.

So, total balls in the box=

= x + y + 40.

Now, given,

P(A) = 1/3

x/(x+y+40) = 1/3

3x = x + y + 40

2x = y + 40.......eq. 1 ●

Samely,

P(B) = 1/4

y/(x+y+40) = 1/4

4y = x + y + 40

3y = x + 40.....eq 2 ●

On solving equations 1 and 2, we derive =>

x = 32 ●

y = 24 ●

Thus, Total balls in the box =

= x + y + 40

= 32 + 24 + 40

Total balls = 96......●

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Hope it was helpful.