Math, asked by karankannanme, 1 year ago

There are two boxes I and II . Box I contains 3 red and 6 black balls .Box II contains 5 red and 'n' black balls. One of the two boxes, box I and box II is selected at random and a ball is drawn at random. The ball drawn is found to be red. If the probability that this red all comes out from box II is 3/5, find the value of 'n'.

Answers

Answered by ankurbadani84
8

Answer:

Step-by-step explanation:

Box 1 - Red Ball Probability = 3/9 = 1/3

Box 1 - Black Ball Probability = 6/9 = 2/3

Box 2 - Red Ball Probability = 5/(5+n)

Box 2 - Black Ball Probability = n/(5+n)

Probability that red all comes out from box II is 3/5 when box is selected at random.

Probability of red  from Box 2 / (Probability of red  from Box1 + Probability of red  from Box 2) = 3/5

∴ n/(5+n) / (  n/(5+n) + 1/3 )  = 3/5

∴ 2n = 5 + n

∴ n = 5

Hence there are 5 Black Balls.

Answered by amitnrw
7

Answer:

5 Black Balls

Step-by-step explanation:

There are two boxes I and II . Box I contains 3 red and 6 black balls .Box II contains 5 red and 'n' black balls. One of the two boxes, box I and box II is selected at random and a ball is drawn at random. The ball drawn is found to be red. If the probability that this red all comes out from box II is 3/5, find the value of 'n'.

Box 1 -  3 Red  , 6 Black

Probability of red  from Box 1 =  3/(3+6) = 3/9  = 1/3

Probability of Black from Box 1 = 6/(3+6) = 6/9 = 2/3

Box 2 -  5 Red  , n Black

Probability of red  from Box 2 =  5/(5+n)

Probability of Black from Box 2 = n/(5+n)

Probability of red  from Box 2 / (Probability of red  from Box1 + Box 2) = 3/5

=> n/(5+n) / (  n/(5+n) + 1/3 )  = 3/5

=> 5n/ (5+n) = 3n/(5+n) + 1

=> 2n/(5+n) = 1

=> 2n = 5 + n

=> n = 5

5 Black Balls

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