Permutation, Combination and Probability (UPSC) Part-I

Total Questions: 51

1. A student has 60% chance of passing in English and 54% chance of passing in both English and Mathematics. What is the percentage probability that he will fail in Mathematics? [1995]

Correct Answer: (d) 10
Solution:

P(E) = Probability of passing in English = 0.6
P(E∩M) = Probability of passing in Maths and English = 0.54
P(M) = Probability of passing in Maths
Since, P(M) and P(E), both are independent events.
So, P(E∩M) = P(E) × P(M)

P(M) = P(E∩M)/P(E) = 0.54/0.6 = 0.9

∴ Probability of failing in Maths = 1 - 0.9 = 0.1 = 10%

2. A table has three drawers. It is known that one of the drawers contains two silver coins, another contains two gold coins and the third one contains a silver coin and gold coin. One of the drawers is opened at random and a coin is drawn. It is found to be a silver coin. What is the probability that the other coin in the drawer is a gold coin? [1995]

Correct Answer: (c) 0.50
Solution:

For finding the silver coin, only drawer 1 and 3 remains in consideration, because the open drawer in any case cannot be the drawer that have only gold coins. Now the probability of next coin being a gold coin = 1/2 = 0.5

3. X and Y are two variables whose values at Y time are related to each other as shown in Fig. (i). X is known to vary periodically with reference to time as shown in Fig. (ii) [1995]

Which of the following curves depicts correctly the dependence of Y on time?

Correct Answer: (c)
Solution:

Let the radius of the circle be unity
Equation of the circle, x² + y² = 1

y = √(1 - x²)                                                     ....(i)

and, x = sin t                                                   ....(ii)

From (i) and (ii), y = √(1 - sin²t) = cos t

Now, option (c) is the graph of y = cos t.

4. Two packs of cards are thoroughly mixed and shuffled and two cards are drawn at random, one after the other. What is the probability that both of them are jacks? [1996]

Correct Answer: (a) 1/13
Solution:

Total number of cards = 104 = 2 × 52
and total number of jacks = 8 = 2 × 4

∴ Probability for the jack in first draw = 8/104

and probability for the jack in second draw = 7/103

Since both the events are independent events.
Hence the probability that both of them are jacks.

= (8/104) × (7/103) = 7/1339

5. In the given figure, if QRS is an equilateral triangle and TQS is an isosceles triangle and x = 47°, then the value of y (in degrees) will be [1997]

Correct Answer: (a) 13°
Solution:

As ΔTQS is an isosceles triangle.
∴ ∠TSQ = ∠TQS = 47°
Now, in equilateral triangle ΔQRS, ∠RQS = ∠RSQ = ∠QRS = 60°
Now, ∠RQS = ∠RQT + ∠TQS = 60°, ∠RQT = ∠Y = 60° - 47° = 13°

6. When three coins are tossed together, the probability that all coins have the same face up, is [1997]

Correct Answer: (c) 1/8
Solution:

Probability of Head or Tail on the upper side for a coin = 1/2

∴ Probability of same side on the upper side for the three coins = 1/2 × 1/2 × 1/2 = (1/2)³ = 1/8

7. In a factory quality assurance test is conducted on various samples for a specific characteristic value of the product. The values and the number of samples are as given in the following table: [1999]

Characteristic value, XNo. of Samples
103
117
1210
1315
1428
1533
1624
1711
1810
196
203

Consider the following statements based on the table:

  1. The probability that X ≤ 15 is 0.64
  2. The probability that 13 < X ≤ 17 is greater than 0.64
  3. The probability that X = 15 is less than 0.22

Which of the above statements is/are not true?

Correct Answer: (c) 2 & 3
Solution:

Number of samples = 150

So, probability (P) = Number of samples for (X) / Total number of samples (150)

When we consider the given statements

(1) P(X ≤ 15) = (3 + 7 + 10 + 15 + 28 + 33) / 150 = 96/150 = 0.64

(2) P(13 < X ≤ 17) = (28 + 33 + 24 + 11) / 150 = 96/150 = 0.64

(3) P(X = 15) = 33/150 = 0.22

8. A bag contains 20 balls, 8 balls are green, 7 are white and 5 are red. What is minimum number of balls that must be picked up from the bag blind-folded (without replacing any of it) to be assured of picking at least one ball of each colour? [2002]

Correct Answer: (d) 16
Solution:Since, 8 Green balls + 7 White balls = 15 balls
7 White balls + 5 Red balls = 12 balls
and 8 Green balls + 5 Red balls = 13 balls
Now, if we pick 15 balls, they may be white, green or red but if we pick 16 balls, then its certain that there will be at least one ball of each colour.

9. A complete cycle of a traffic light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds and red for 30 seconds. At a randomly chosen time, the probability that the light will not be green, is [2002]

Correct Answer: (d) 7/12
Solution:Probability that the light is not green
= time for which light is not green / time taken for the entire cycle
= (5+30)/60 = 35/60 = 7/12

10. Three flags, each of different colour, are available for a military exercise. Using these flags, different codes can be generated by waving [2003]

(i) single flag of different colours or
(ii) any two flags in a different sequence of colour
                            Or
(iii) three flags in a different sequence of colours.
The maximum number of codes that can be generated, is

Correct Answer: (c) 15
Solution:
  1. (i) Number of ways of arranging three colours taken 1 at a time = ³P₁ = (3×2!)/2! = 3

(ii) Number of ways of arranging three colours taken 2 at a time = ³P₂ = 3!/1! = 6

(iii) Number of ways of arranging three colours taken 3 at a time = ³P₃ = 6

Hence, Maximum no. of codes = No. of ways of arranging these flags = 3 + 6 + 6 = 15