Question
Question: A bag contains \(40\) pen drives out of which x are non-defective. If one pen drive is drawn at rand...
A bag contains 40 pen drives out of which x are non-defective. If one pen drive is drawn at random, the probability of drawing a non-defective pen drive is y. Now place this pen drive and 20 more non-defective pen drives in this bag. Now, if a pen drive is drawn from the bag the probability of drawing the non-defective pen drive is 4y. Find x.
Solution
Probability defines the chance of a certain event to occur. The sum of all the events that can occur in a given situation is equal to1. We will find the probability of drawing non defective pen drives in both the cases and equate them with the expression given to us. Hence, we will get two relations between x and y. Solving both, we can find the value of x.
Complete step by step answer:
Given, total number of pen drives is n(s)=40
Number of pen drives that are non-defective in the bag =x
As per the question, the probability of drawing a non-defective pen drive is y=Total no. of pen drivesNumber of non defective pen drives
⇒y=40x−−−−(1)
Now, 20 more non-defective pen drives are added.
Now, the total number of pen drives in the bag becomes, n(s)=40+20
⇒n(s)=60
The number of non-defective pen drives in the bag =x+20
Now, we are given that the probability of drawing a non-defective pen drive becomes4y.
Therefore, according to the question, 4y=Total number of pen drives in the bagNumber of non - defective pen drives in the bag
⇒4y=60x+20−−−(2)
So, we have two relations in x and y.
Now, substituting the value of y from (1) in (2).
⇒4(40x)=(60x+20)
Cancelling the common factors in numerator and denominator, we get,
⇒10x=60x+20
Now, cross multiplying the terms of the equation,
⇒60(x)=10(x+20)
Dividing both sides of the equation by 10,
⇒6x=x+20
Subtracting xfrom both sides of equation,
⇒6x−x=20
⇒5x=20
Dividing both sides of the equation by 5, we get,
∴x=4
So, the value of x is 4.
Therefore, the number of non-defective pen drives is 4.
Note: We should always remember the basic formula of probability of an event as (Total number of outcomesNumber of favourable outcomes). We must have a clear understanding of simplification rules and transposition in order to solve the equation formed in two variables. Also, if the event has p probability to occur then the probability that the event doesn’t occur is (1−p). If any event has probability equal to1, then such the event is said to be an obvious event. On the other hand, if an event has 0 probability to occur, then such an event is said to be a null event.