Question

A bag contains $3$ white and $5$ black balls. If one ball is drawn, then the probability that is black is
$1)$ $\dfrac{3}{8}$
$2)$ $\dfrac{5}{8}$
$3)$ $\dfrac{6}{8}$
$4)$ $\dfrac{10}{20}$

Answer 1

To get the probability that the drawn ball is black, we will calculate total ways of drawing a ball and favourable ways of drawing a ball by using formula ${}^{n}{{C}_{r}}$ , where $n$ represents total number of balls and $r$ represents number of drawn balls. Then, the ratio of favourable ways of drawing a ball and the total ways of drawing a ball will give the required probability of drawing a black ball.

Complete step-by-step solution:
Since, in the question, a bag contains $3$ white and $5$ black balls. So, the bag has a total $8$ balls and now we need to draw a ball. Therefore, the no. of ways of drawing a ball from the bag will be as:
$\Rightarrow {}^{n}{{C}_{r}}$
Here, we will apply the necessary value of $n$ and $r$ that are $8$ and $1$ respectively as:
$\Rightarrow {}^{8}{{C}_{1}}$
We can write the above formula as:
$\Rightarrow \dfrac{8!}{1!.\left( 8-1 \right)!}$
After solving denominator, we will have the above step below as:
$\Rightarrow \dfrac{8!}{1!.7!}$
After solving above step, we have the value of total ways of drawing a ball as:
$\Rightarrow 8$
Now, we will calculate the favourable ways of drawing a black ball and we have some data as:
Total black balls $=5$
So, the number of ways of drawing a black ball from the bag is:
$\Rightarrow {}^{5}{{C}_{1}}$
Here, we can write it in expand way as:
$\Rightarrow \dfrac{5!}{1!.\left( 5-1 \right)!}$
After Simplifying denominator term, we will have the denominator as:
$\Rightarrow \dfrac{5!}{1!.4!}$
We will have the favourable ways after solving the above step as:
$\Rightarrow 5$
Now, we will use the method of ration to get the required probability as:
$\Rightarrow \dfrac{Favourable\text{ }Ways}{Total\text{ }Ways}$
Here, we will put the necessary values as:
$\Rightarrow \dfrac{5}{8}$
Hence, this is the required probability of drawing a black ball from the bag.

Note: Here is the expression of $n!$ that will be multiple of natural numbers up to $n$ as:
$\Rightarrow n!=1\times 2\times 3\times ...\times \left( n-1 \right)\times n$
Similarly, we will have the values for:
$8!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8$ and $7!=1\times 2\times 3\times 4\times 5\times 6\times 7$
Now, we can use this expansion to get the solution of the total number of possible ways.
Similarly, we are able to do the calculation in a favourable number of ways.