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 . What will be the probability of
picking two black balls from that bag if bag can hold maximum 15 balls only?

Answer 1

Answer:

$\frac{1}{11}$

Step-by-step explanation:

Edited Question : the probability of picking two white balls one after without replacement from that bag is $\frac{14}{33}$

1) We have, Probability of picking two white balls.

Let the number of white balls in the bag be 'w' and that of black balls be 'b'.

Then,$w+b=n(say)$

Now,

According to the question, we get:

$\frac{^wC_2}{^nC_2}=\frac{14}{33}\\ \\=>\frac{w(w-1)}{n(n-1)}=\frac{14}{33}=\frac{14*4}{33*4}\\ \\=\frac{8*7}{12*11}\\ \\=>\frac{w(w-1)}{n(n-1)}=\frac{8*7}{12*11}\\ \\=>\boxed{w=8,n=12}$

2)Now, Coming to what is asked in the question :

Hence, b = 12-8=4

Probability of picking two black balls from that bag if bag can hold maximum of 15 balls only is :

$\frac{^bC_2}{^nC_2} =\frac{^4C_2}{^{12}C_2}=\frac{4*3}{12*11}=\frac{1}{11}$

Hence, required probability is 1/11

Answer 2

the answer to your question is 1 divided by 11 .

$\huge \: thx$