Question

A bag contains some white and black balls . The probability of picking two white balls one after other without replacement from that bag is 14/33. Then what will be the probability of picking two black balls from that bag if bag can hold maximum 15 balls only?

Answer 1

2/35

Explanation:

Let w and b be the numbers of white and black balls in the bag respectively.

So, the total numbers of the balls in the bag is

`n=w+b\;\cdots(i)`

As the bag can hold maximum 15 balls only, so

`n\leq15 \;\cdots(ii)`

Probability of picking two white balls one after other without replacement

=Probability of the first ball to be white and the probability of second ball to be white

=(Probability of picking first white balls) x( Probability of picking 2nd white ball)

Here, the probability of picking the first white ball
`=(w)/(n)`

After picking the first ball, the remaining

white ball in the bag
`= w-1`

and the remaining total balls in the bag
`=n-1`

So, the probability of picking the second white ball
`=(w-1)/(n-1)`

Given that, the probability of picking two white balls one after other without replacement is 14/33.

`\Rightarrow (w)/(n) * (w-1)/(n-1)=(14)/(33)`

`\Rightarrow (w(w-1))/(n(n-1)) =(14)/(33)`

Here,
`w` and
`n` are counting numbers (integers) and 14 and 33 are co-primes.

Let,
`\alpha` be the common factor of the numbers
`w(w-1)` (numerator) and
`n(n-1)` (denominator), so

`(w(w-1))/(n(n-1)) =(14\alpha)/(33\alpha)`

`\Rightarrow w(w-1)=14\alpha\cdots(iii)`

And
`n(n-1)=33\alpha.`

As from eq. (ii),
`n\leq 15`, so, the possible value of
`\alpha` for which multiplication od two consecutive positive integers (n and n-1) is
`33\alpha` is 4.

`n(n-1)=11* (3\alpha)`

`\Rightarrow n(n-1)=12*11` [as
`\alpha=4`]

`\Rightarrow n=12`

So, the number of total balls =12

From equation (iv)

`w(w-1)=7* (2\alpha)`

`\Rightarrow w(w-1)=8*7`

`\Rightarrow w=8`

So, the number of white balls =8

From equations (i), the number of black balls =12-8=4

In the similar way, the required probability of picking two black balls one after other in the same way (i.e without replacement) is

`=(b)/(n) * (b-1)/(n-1)`

`= (4)/(15) * (4-1)/(15-1)`

`= (4)/(15) * (3)/(14)`

`= (2)/(35)`

probability of picking two black balls one after other without replacement is 2/35.