A bag contains black and white balls, such that the probability of picking a black ball is 4/7. If the probability of picking two black balls without replacing the first is 4/13, how many black balls are there in the bag?
Let the bag contain ‘x’ black and ‘y’ white balls Total no. of balls = x + y Probability of picking a black ball = 4/7 ⇒ x/(x + y) = 4/7 ⇒ 7x = 4x + 4y ⇒ 3x = 4y ⇒ y = 3x/4 ⇒ Total no. of balls = x + 3x/4 = 7x/4 Now, probability of picking two black balls without replacement = 4/13 ⇒ [x/(x + y)] × [(x – 1)/(x + y – 1)] = 4/13 ⇒ 4/7 × (x – 1)/(7x/4 – 1) = 4/13 ⇒ 4(x – 1)/(7x – 4) = 7/13 ⇒ 52(x – 1) = 7(7x – 4) ⇒ 52x – 52 = 49x – 28 ⇒ 52x – 49x = 52 – 28 ⇒ 3x = 24 ⇒ x = 24/3 = 8 Therefore, there are 8 black balls in the bag.
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