Question

A bag contains 3 white balls, 4 green balls and 5 red balls. Three balls are drawn from bag which out replacement, find probability that balls are all of different colours
(A) $\dfrac{3}{{11}}$
(B) $\dfrac{3}{{110}}$
(C) $\dfrac{4}{{11}}$
(D) $\dfrac{4}{{110}}$

Answer 1

Hint:Probability- Probability is numerical description of how likely an event is to occur.The probability of an event is a number between 0 and 1, where, 0 indicates impossibility of the event and 1 indicates certainty.

For example
When we flip a coin: There are two possible outcomes—heads or tails.
50% chances of getting heads and 50% of getting tails.
But also we can find out this by using probability formula
Probability $ = \dfrac{{favourable\,outcomes}}{{Total\,outcomes}}$
Here in our case favourable outcome for heads as well as tails is 1.
(As a coin can have only one heads and only one tails)

Total number of outcomes
= no. of heads + no of tails
So, total outcomes \[ = 1 + 1 = 2\]

Now, We will find probability of getting heads on coin flip.
For that we will use above formula
And from formula we have
Probability $ = \dfrac{{favourable\,outcomes}}{{Total\,outcomes}}$
Therefore, $P = \dfrac{1}{2}$
Which is 50% chances for both.
In above question we will use combination formula to find out outcomes.
And i.e.
Combinations formula
$^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Where
n $ = $ total number of items
r $ = $ number of items to be selected from sample
Symbol Represents
Here \[n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times \left( {n - 3} \right) \times ........{\text{ }} \times 1\]
And \[r! = r \times \left( {r - 1} \right) \times \left( {r - 2} \right) \times \left( {r - 3} \right) \times ........ \times 1\]
Lets understand this formula by taking an example
$_n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$_5{C_2} = \dfrac{{5!}}{{2!(5 - 2)!}}$
$_5{C_2} = \dfrac{{5!}}{{2!(3)!}}$
$_5{C_2} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{(2 \times 1) \times (3 \times 2 \times 1)}} = 10$


Complete step by step answer:

Number of green balls $ = 4$
Number of white balls $ = 3$
Number of red balls $ = 5$
So, we have total number of balls $ = 4 + 3 + 5 = 12$
As balls are drawn without replacement so,
1. We can draw white coloured balls in $^3{C_1}$ ways and $^3{C_1} = \dfrac{{3!}}{{(3 - 1)!1!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1}} = 3$ …..(1)
2. Ways of drawing green balls are $^4{C_1}$ and $^4{C_1} = \dfrac{{4!}}{{(4 - 1)!1!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3! \times 1!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 1}} = 4$ …..(2)
3. Ways of drawing red balls are $^5{C_1} = \dfrac{{5!}}{{(5 - 1)! \times 1!}}$ or, $^5{C_1} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 1}} = 5$ …..(3)
4. Ways of drawing 3 balls out of 12 balls ${ = ^{12}}{C_3}$ and $^{12}{C_3} = \dfrac{{12!}}{{(2 - 3)! \times 3!}}$
$^{12}{C_3} = \dfrac{{12 \times 11 \times 12 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (3 \times 2 \times 1)}}$
$^{12}{C_3} = 220$ …..(equation (4))
Result obtained above is total outcomes.
Probability of getting 3 balls of different colours,
P $ = $ ways of drawing red balls $ \times $ ways of drawing green balls $ \times $ ways of drawing white balls $/$ total outcomes
From equation (1), (2), (3) and (4)
$P = \dfrac{{5 \times 4 \times 3}}{{220}}$
$P = \dfrac{3}{{11}}$
So option A is the correct option $P = \dfrac{3}{{11}}$

Note:
Probability ranges between 0 and 1.
First we need to make a set(S) for total number of outcomes before solving the problem.
Like
¡. When a coin is tossed, \[S = \left\{ {H,T} \right\}\]
¡¡. When two coins are tossed, \[S = \left\{ {HH,HT,TH,TT} \right\}\]
where H $ = $ Head and T $ = $ Tail