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# Comprehensive Guide Questions with Answers Quant Notes | EduRev

## UPSC : Comprehensive Guide Questions with Answers Quant Notes | EduRev

The document Comprehensive Guide Questions with Answers Quant Notes | EduRev is a part of the UPSC Course General Aptitude for GATE.
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Example 1: An urn contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

Solution: Let A = Getting a white ball in the first draw B = Getting a Black Ball in the second draw. Required Probability = Probability of getting a white ball in the first draw and a black ball in second draw = P(A and B) = P (A ∩B) = P (A) .P(B/A)

P(A) = (10C1/25C1) = (10/25) = (2/5) and P(B/A) = Probability of getting a black ball in the second draw
when a white ball has already been in first draw  = (15C1/24C1) = (15/24) = (5/8)
⇒ Required probability Example 2: A standard die is rolled. Let C be the event that the outcome is an odd number and D be the event that the outcome is not more than 3. Find the probability C knowing that D has happened?
Solution: Let's find the conditional probability of C given that D has occurred.
If we know D has occurred, the possible outcomes are {1,2,3}.
For C to happen the outcome must be in C∩D={1,3}.
P(C|D) = P(C∩D)/P(D) = 2/3

Example3: Hunar wrote two sections of CAT paper ; Verbal and QA in the same order. The probability of her passing both sections is 0.6. The probability of her passing the verbal section is 0.8. What is the probability of her passing the QA section given that she has passed the Verbal section?

Solution: Let P(QA) = passing QA’s section. And P(V) = passing Verbal section. So, P(QA/V) = P(QA∩V)/P(V) = 0.6/0.8⇒0.75
Example 4: Amika has 2 pouches. One pouch contains 3 golden and 4 silver crystals. The second pouch contains 2 golden and 3 silver crystals. One pouch is selected at random and from the selected pouch, one crystal is drawn. Find the probability that the crystal drawn is golden.
Solution: Since one of the two pouches is selected randomly, So,P(E1) =1/2 and P(E2) = 1/2 .
Now P(A/E1) = Probability of drawing a golden crystal when the first pouch has been chosen =3/7 and P(A/E2)=Probability of drawing an golden crystal when the second pouch has been selected = 2/5
[∴ The second pouch contains 2 golden and 4 silver crystals]

Using the law of total probability P(golden crystal) = P(A) = P(E1) .  Example 5: A year is selected at random. What is the probability that it contains 53 Mondays if every fourth year is a leap year?( It is a very famous question and appears frequently in many competitive exams.)
Solution: Let P(L) = probability of selecting a leap year. P(NL) = probability of selecting a non - leap year. P(M) = probability of getting 53 Mondays in a year.

Then by Law of total probability: P(M) = P (M| L) . P(L) + P (M | NL) . P(NL)

= (2/7).(1/4) + (1/7).(3/4) = 5/28

Example 6: There are three cartons, each containing a different number of soda bottles. The first carton has 10 bottles, of which four are flat, the second has six bottles, of which one is flat, and the third carton has eight bottles of which three are flat. What is the probability of a flat bottle being selected when a bottle is chosen at random from one of the three cartons?
Solution: Let A be the event that the selected bottle is flat, X be the event that it is chosen from carton 1, Y be the event that it is chosen from carton 2, Z be the event that it is chosen from carton 3.

Then by the Law of total probability:

P(A) = P(A | X)-P(X) + P(A | Y) P(Y ) + P(A/Z).P(Z).

= (4/ 10).(1/ 3) + (1/ 6) . (1/ 3) + (3/8) .(1/3) = 113/360 .

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## General Aptitude for GATE

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