Question

A bag contains $6$ red, $5$ white and $4$ black balls. Two balls are drawn. The probability that none of them is red, is
$1)$ $\dfrac{12}{35}$
$2)$ $\dfrac{6}{35}$
$3)$ $\dfrac{4}{35}$
$4)$ None of these

Answer 1

Hint: For getting the probability that both the drawn balls are not red, we will do the procedure for getting the probability that at least one ball is red. Since, we can get the total number of possible ways of drawing two balls with use of formula ${}^{n}{{C}_{r}}$ , where $n$ is for total value and $r$ number of drawn of balls. Since, we got the total number of possible ways of drawing two balls, we will do calculation for drawing at least one red ball that has two conditions. First condition is that both the balls may be red and second condition is one may be red and other would be white or black. After completion of getting the number of favourable ways, we will find the probability of getting at least one ball red after drawing. Then, we will subtract this probability by $1$ and will get the required probability.

Complete step by step answer:
Since, from the question, we have a total $15$ balls in the bag as the total of all balls. So the total number of possible ways of drawing two balls is:
$\Rightarrow {}^{15}{{C}_{2}}$
And this will give the expression below as:
$\Rightarrow \dfrac{15!}{2!.\left( 15-2 \right)!}$
Here, we can write the numerator and denominator as:
$\Rightarrow \dfrac{15\times 14\times 13!}{1\times 2\times 13!}$
Now, we will cancel out equal like terms.
$\Rightarrow \dfrac{15\times 14}{1\times 2}$
After doing the necessary calculation, we will get the value of total number of possible ways as:
$\Rightarrow 15\times 7$
$\Rightarrow 105$
Now, we will do the calculation for favourable number of ways that have two conditions as:
First condition is the both drawn ball is red. Since, total number of red balls are six, the number of ways will be as:
$\Rightarrow {}^{6}{{C}_{2}}$
We can further write the above step below as:
$\Rightarrow \dfrac{6!}{2!.\left( 6-2 \right)!}$
Here, we can write the above step below as:
$\Rightarrow \dfrac{6\times 5\times 4!}{1\times 2\times 4!}$
Now, we will cancel out equal like terms.
$\Rightarrow \dfrac{6\times 5}{1\times 2}$
After doing the necessary calculation, we will get the value of total number of possible ways as:
$\Rightarrow 3\times 5$
We will solve the above step and will get the value for first condition as:
$\Rightarrow 15$
Now, second condition has two possibilities as; one is that one drawn ball is red and other one is white and second is that one ball is red and other one is black. So, the possible ways for this condition will be as:
$\Rightarrow {}^{6}{{C}_{1}}\times {}^{5}{{C}_{1}}+{}^{6}{{C}_{1}}\times {}^{4}{{C}_{1}}$
Here, we will calculate the values for possible ways and will get:
$\Rightarrow 6\times 5+6\times 4$
Now, we will complete the multiplication and will have:
$\Rightarrow 30+24$
Here, we will get the sum from the above step as:
$\Rightarrow 54$
Now, we will calculate the favourable number of ways by summing both the conditions as:
$\Rightarrow 15+54$
After solving the above step, we will have the value as:
$\Rightarrow 69$
Since, we got the total number of possible ways and favourable number of ways. So, we will get the probabilities of drawing at least one red ball as:
$\Rightarrow \dfrac{69}{105}$
Here, $3$ is the common factor in the numerator and denominator. So, we will cancel out it and will get the probability of drawing at least one red ball as:
$\Rightarrow \dfrac{23}{35}$
Now, we can calculate required probability by subtracting above obtained probability from one as:
$\Rightarrow 1-\dfrac{23}{35}$
Here, we will multiply and divide by $35$ in first term of above step as:
$\Rightarrow 1\times \dfrac{35}{35}-\dfrac{23}{35}$
The above step will be as:
$\Rightarrow \dfrac{35}{35}-\dfrac{23}{35}$
Here, we will solve it as:
$\Rightarrow \dfrac{35-23}{35}$
$\Rightarrow \dfrac{12}{35}$
Hence, the required probability is $\dfrac{12}{35}$ .

So, the correct answer is option 1).

Note:
Here, we will find the required probability in another. Since, we got the total number of possible ways in the solution that is $105$. Now, we will calculate the favourable number of ways. Since, both drawn balls are not red. So, we will exclude the number of red balls from the total number of balls. So, the total remaining number of balls without red is $15$ minus $6$ that is equal to $9$ . Now, we will draw the ball from remaining balls that is favourable number of ways of drawing two balls also as:
$\Rightarrow {}^{9}{{C}_{2}}$
And this will give the expression below as:
$\Rightarrow \dfrac{9!}{2!.\left( 9-2 \right)!}$
After solving the above step, we will find the value as:
$\Rightarrow 36$
So, the required probability is:
$\Rightarrow \dfrac{36}{105}$
Since, $3$ is a common factor in numerator and denominator. So, we will have:
$\Rightarrow \dfrac{12}{35}$
Hence, the solution is correct.