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

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

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

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 $=\frac{w}{n}$

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 $=\frac{w-1}{n-1}$

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

$\Rightarrow \frac{w}{n} \times \frac{w-1}{n-1}=\frac{14}{33}$

$\Rightarrow \frac{w(w-1)}{n(n-1)} =\frac{14}{33}$

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

$\frac{w(w-1)}{n(n-1)} =\frac{14\alpha}{33\alpha}$

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

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.

$n(n-1)=11\times (3\alpha)$

$\Rightarrow n(n-1)=12\times11$ [as $\alpha=4$]

$\Rightarrow n=12$

So, the number of total balls =12

From equation (iv)

$w(w-1)=7\times (2\alpha)$

$\Rightarrow w(w-1)=8\times7$

$\Rightarrow w=8$

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

$=\frac{b}{n} \times \frac{b-1}{n-1}$

$= \frac{4}{15} \times \frac{4-1}{15-1}$

$= \frac{4}{15} \times \frac{3}{14}$

$= \frac{2}{35}$

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