Question

A bag contains $5$ white, $7$ black and $4$ red balls. Three balls are drawn from the bag at random. The probability that all the three balls are white is
A) $\dfrac{3}{{16}}$
B) $\dfrac{3}{5}$
C) $\dfrac{1}{{60}}$
D) $\dfrac{1}{{56}}$

Answer 1

Hint: The joining or merging of different parts or qualities in which the component elements are individually distinct . First we find the total number of balls in all colours and then we find the combination of all balls and again we find the combination over white balls . Then find the probability that the three balls are white . Formula of combination is $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$

Complete step by step answer:
Given that in a bag there are $5$ white , $7$ black and $4$ red balls .
Therefore the total number of balls are $5 + 7 + 4$
$ = 16$
Now we calculate the combination of collecting three balls over $16$ balls is $^{16}{C_3}$ .
${ ^{16}}{C_3} = \dfrac{{16!}}{{(16 - 3)! \times 3!}}$
Separate the factorial term which may be divided by itself and get
$ = \dfrac{{16 \times 15 \times 14 \times 13!}}{{13! \times 3!}}$
$ = \dfrac{{16 \times 15 \times 14}}{{3 \times 2 \times 1}}$
$ = 8 \times 5 \times 14$
Multiplying and get
$ = 560$
And we calculate the combination of collecting three white balls over $5$ white balls is $^5{C_3}$
${ ^5}{C_3} = \dfrac{{5!}}{{(5 - 3)! \times 3!}}$
Separate the factorial term which may be divided by itself and get
$ = \dfrac{{5 \times 4 \times 3!}}{{2! \times 3!}}$
$ = \dfrac{{5 \times 4}}{{2 \times 1}}$
$ = 5 \times 2$
Multiplying and get
$ = 10$
Now we find the probability of all three balls are white balls is $\dfrac{{n(S)}}{{n(A)}}$ , where \[n(S)\] is the favourable number and $n(A)$ is the total number .
Here $n(S) = 10$ and $n(A) = 560$ , we put this and we get
$ P(A) = \dfrac{{10}}{{560}}$
Divide both the numerator and denominator by $10$
$ \Rightarrow P(A) = \dfrac{1}{{56}}$
Therefore option (D) is correct.

Note:
The formula of probability we used in the solution that is \[P(A) = \dfrac{{n(S)}}{{n(A)}}\] , where \[n(S)\] is the favourable number and $n(A)$ is the total number . Factorial is the product of factors from $1$ to the given term .
i.e., $n! = n \times (n - 1) \times (n - 2) \times ....... \times 3 \times 2 \times 1$