Question

There are 9 white and 1 black ball in an urn. 3 balls are chosen randomly without replacement. What is the probability of all three being white?

Answer 1

There are 9 White Balls and 1 Black in an Urn. Therefore totally we have 10 Balls in an Urn.

In Without Replacement case the Size of "n" will decreases from one trial to another trial.

We know that

22 PE) =

Where m = Favourable Number of events

n = Total Number of events

1st Draw:

The way of drawing a ball from 10 balls is

$n = \binom{10}{1} =10$

The way of drawing a White ball from 9 White balls is

$m = \binom{9}{1} =9$

Therefore P( 1st Ball is White ) is

m P(EL) = n

9 → P(E1) = 10

2nd Draw:

Since it's the case of Wihtout Replacement; So, now the total number of balls will be 9 and total number of white balls is 8.

The way of drawing a ball from 9 balls is

$n = \binom{9}{1} =9$

The way of drawing a White ball from 8 White balls is

$m = \binom{8}{1} =8$

Therefore P( 2nd Ball is White ) is

$P(E_{2}) = \frac{m}{n}$

$\Rightarrow P(E_{1}) = \frac{8}{9}$

3rd Draw:

Since it's the case of Wihtout Replacement; So, now the total number of balls will be 8 and total number of white balls is 7.

The way of drawing a ball from 8 balls is

$n = \binom{8}{1} =8$

The way of drawing a White ball from 7 White balls is

$m = \binom{7}{1} =7$

Therefore P( 3rd Ball is White ) is

$P(E_{3}) = \frac{m}{n}$

$\Rightarrow P(E_{3}) = \frac{7}{8}$

Therefore the required probability is

$\Rightarrow P(E) = \frac{9}{10}*\frac{8}{9}*\frac{7}{8}$

$\Rightarrow P(E) = \frac{7}{10}$

$\Rightarrow P(E) = 0.7$