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Question: A bag contains \[18\] balls out of which x balls are red (i) If one ball is drawn at random from t...

A bag contains 1818 balls out of which x balls are red
(i) If one ball is drawn at random from the bag, then what is the probability that it is a red ball?
(ii) If 22 more red balls are put in the bag then the probability of drawing a red ball will be 98\dfrac{9}{8} time the probability of the red ball coming in part, find the value of x.

Explanation

Solution

Hint : To solve the first part of the question, i.e., probability of getting a red ball, here the total outcomes, i.e., the total number of balls in the bag is 18,18, and the favourable outcome is x. So, on applying the probability formula we will get our required answer. And to find the second part of the question, i.e., the value of x, here the total outcomes will be 20,20, and the favourable outcomes will be x + 2,x{\text{ }} + {\text{ }}2, and also we will be applying the given condition, and hence we will get our required answer.

Complete step-by-step answer :
We have been given a bag which contains 1818 balls out of which x balls are red.
(i) It is given that one ball is drawn at random from the bag, then we need to find the probability that that ball is red.
So, the total outcomes of number of balls in the bag = 18 = {\text{ }}18
And, the number of favourable outcomes of red balls in the bag = x = {\text{ }}x
We know that, Probability =favourable outcomestotal outcomes = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}
On applying the value in the above formula, we get
Probability of getting red ball =x18 = \dfrac{x}{{18}}
Thus, probability of red ball is x18.\dfrac{x}{{18}}.

(ii) It is given that 22 more red balls are put in the bag then the probability of drawing a red ball will be 98\dfrac{9}{8} time that of probability of red ball coming in part, we need to find the value of x.
We have been given two more red balls to put in the bag.
Now, the total outcomes of number of balls = (2+18) = 20 = {\text{ }}\left( {2 + 18} \right){\text{ }} = {\text{ }}20
So, now the number of favourable outcomes of red balls = x + 2 = {\text{ }}x{\text{ }} + {\text{ }}2
The probability of getting red ball =x+220 = \dfrac{{x + 2}}{{20}}
In the question it is given that the probability of drawing a red ball will be 98\dfrac{9}{8} time that of the probability of the red ball coming in part.
So, x+220= 98(x18)\dfrac{{x + 2}}{{20}} = {\text{ }}\dfrac{9}{8}(\dfrac{x}{{18}})
x+220=x16\dfrac{{x + 2}}{{20}} = \dfrac{x}{{16}}

{16x + 32 = 20x} \\\ {4x = 32} \\\ {x = 8} \end{array}$$ Thus, value of x is $$8.$$ **Note** : Students should take care that, in the second part we have been given a condition that is “If $$2$$ more red balls are put in the bag then probability of drawing a red ball will be $\dfrac{9}{8}$ time that of probability of red ball coming in part” so, it should be noted that this condition is actually an extended part of first part.