Answer:
2/35
Step-by-step 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)](https://tex.z-dn.net/?f=n%3Dw%2Bb%5C%3B%5Ccdots%28i%29)
As the bag can hold maximum 15 balls only, so
![n\leq15 \;\cdots(ii)](https://tex.z-dn.net/?f=n%5Cleq15%20%5C%3B%5Ccdots%28ii%29)
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 ![=\frac{w}{n}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bw%7D%7Bn%7D)
After picking the first ball, the remaining
white ball in the bag ![= w-1](https://tex.z-dn.net/?f=%3D%20w-1)
and the remaining total balls in the bag ![=n-1](https://tex.z-dn.net/?f=%3Dn-1)
So, the probability of picking the second white ball ![=\frac{w-1}{n-1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bw-1%7D%7Bn-1%7D)
Given that, the probability of picking two white balls one after other without replacement is 14/33.
![\Rightarrow \frac{w}{n} \times \frac{w-1}{n-1}=\frac{14}{33}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7Bw%7D%7Bn%7D%20%5Ctimes%20%5Cfrac%7Bw-1%7D%7Bn-1%7D%3D%5Cfrac%7B14%7D%7B33%7D)
![\Rightarrow \frac{w(w-1)}{n(n-1)} =\frac{14}{33}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7Bw%28w-1%29%7D%7Bn%28n-1%29%7D%20%3D%5Cfrac%7B14%7D%7B33%7D)
Here,
and
are counting numbers (integers) and 14 and 33 are co-primes.
Let,
be the common factor of the numbers
(numerator) and
(denominator), so
![\frac{w(w-1)}{n(n-1)} =\frac{14\alpha}{33\alpha}](https://tex.z-dn.net/?f=%5Cfrac%7Bw%28w-1%29%7D%7Bn%28n-1%29%7D%20%3D%5Cfrac%7B14%5Calpha%7D%7B33%5Calpha%7D)
![\Rightarrow w(w-1)=14\alpha\cdots(iii)](https://tex.z-dn.net/?f=%5CRightarrow%20w%28w-1%29%3D14%5Calpha%5Ccdots%28iii%29)
And ![n(n-1)=33\alpha.](https://tex.z-dn.net/?f=n%28n-1%29%3D33%5Calpha.)
As from eq. (ii),
, so, the possible value of
for which multiplication od two consecutive positive integers (n and n-1) is
is 4.
![n(n-1)=11\times (3\alpha)](https://tex.z-dn.net/?f=n%28n-1%29%3D11%5Ctimes%20%283%5Calpha%29)
[as
]
![\Rightarrow n=12](https://tex.z-dn.net/?f=%5CRightarrow%20n%3D12)
So, the number of total balls =12
From equation (iv)
![w(w-1)=7\times (2\alpha)](https://tex.z-dn.net/?f=w%28w-1%29%3D7%5Ctimes%20%282%5Calpha%29)
![\Rightarrow w(w-1)=8\times7](https://tex.z-dn.net/?f=%5CRightarrow%20w%28w-1%29%3D8%5Ctimes7)
![\Rightarrow w=8](https://tex.z-dn.net/?f=%5CRightarrow%20w%3D8)
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
![=\frac{b}{n} \times \frac{b-1}{n-1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bb%7D%7Bn%7D%20%5Ctimes%20%5Cfrac%7Bb-1%7D%7Bn-1%7D)
![= \frac{4}{15} \times \frac{4-1}{15-1}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4%7D%7B15%7D%20%5Ctimes%20%5Cfrac%7B4-1%7D%7B15-1%7D)
![= \frac{4}{15} \times \frac{3}{14}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4%7D%7B15%7D%20%5Ctimes%20%5Cfrac%7B3%7D%7B14%7D)
![= \frac{2}{35}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B2%7D%7B35%7D)
probability of picking two black balls one after other without replacement is 2/35.