Updated Thu, 15 Apr 2021 09:52 PM IST

Source: safalta

1. Trimsy speaks the truth 2 out of 9 times. On selecting a card randomly from a pack of cards, she reports that it is either king or ace. Find the probability that it is actually king or ace.

a) 45/91

b) 55/91

c) 4/117

b) 14/117

2. Around a round table, 13 persons are sitting on 13 chairs. Harry is one among them. A person is chosen randomly from them (other than Harry). What is the probability that there will be exactly 2 persons between Harry and the chosen person?

a) 100

b) 200

c) 300

d) 400

a) 45/91

b) 55/91

c) 4/117

b) 14/117

Probability of Truth = 2/9 or probability of lie = 7/9

Since there are 4 ace and 4 king in each pack of 52 cards so,

Probability of getting an king or ace = 8/52

Probability of getting a king or ace when Trimsy reports that it is a king or ace is (2/9) × (8/52)

⇒ P(getting a king or ace when Trimsy reports that it is a king or ace) = 4/1172. Around a round table, 13 persons are sitting on 13 chairs. Harry is one among them. A person is chosen randomly from them (other than Harry). What is the probability that there will be exactly 2 persons between Harry and the chosen person?

a) 1/3

b) 1/4

c) 1/5

d) 1/6

To have exactly 2 persons between Harry and chosen person, the chosen person must be either 3rd to left of Harry or 3rd to right of harry

A person is chosen randomly from them (other than Harry)

∴ The person can be chosen from 12 possible persons

∴ Required probability = 2/12 = 1/6

3. A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color.

a) 5 balls

b) 9 balls

c) 6 balls

d) 7 balls

Suppose he picks 5 balls of all different colours then when he picks up the sixth one, it must match any on of the previously drawn ball colour. thus he must pick 6 balls

4. Six cards King, Queen, Jack, Ace, the five and the nine of spades from a set of cards are well shuffled. One card is picked up at random. Then it is kept aside and other card is drawn. What is the probability that both the cards are Jacks of spade?

a) 1

b) 0

c) 1/2

d) 1/3

There is only one Jack card in the set of 13 spades.

∴ There is no other way that there can be other jack in spade.

∴ Required probability is zero.

5. A small spherical substance was accidently dropped into a cylindrical vessel of height 42 cm and radius 30 cm filled with water. A small cylindrical glass of height 7 cm and radius 5 cm was used to take some water from it. What is the probability that the small spherical substance gets transferred to the glass?

a) 1/64

b) 1/128

c) 1/512

d) 1/216

⇒ Probability of the small spherical substance getting transferred to the glass = Volume of glass/Volume of vessel

⇒ (Radius of glass/Radius of vessel)^2 × (Height of the glass/Height of vessel) [∵ Volume of the cylinder = πr^2 h]

⇒ (5/30)^2 × (7/42) = 1/36 × 1/6

∴ Probability of the small spherical substance getting transferred to the glass = 1/216

6. There are 49 cards in a box numbered from 1 to 49. Every card is numbered with only 1 number. Probability of picking up a card, the number printed on which is a multiple of 5 but not that of 10 or 15 is.

a) 2/49

b) 3/49

c) 4/49

d) 5/49

According to the given information,

Card numbers which are multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45

Card numbers which are multiples of 10 or 15 are 10, 15, 20, 30, 40, 45

The latter ones are to be ruled out from the former ones.

⇒ So, the only card numbers we are left with are 5, 25, 35.

∴ Probability of picking up a card, the number on which is a multiple of 5 but not that of 10 and 15 = 3/49

7. There are 10 people seated in two rows (5 in 1 row) and there are two types of food items. Each row can be served any of the two food items but it must be different from the other row. In how many ways the food can be served?

a) 7247600

b) 7227600

c) 7267600

d) 7257600

We know that,

Number of ways 10 people can be seated in a row = 10C5 = 252

Number of ways 5 people can be seated in one row = 5!

Number of ways 5 people can be seated in the second row = 5!

Number of ways the people of the rows can be arranged = 5! × 5!

Number of ways 2 food items can be arranged among 2 rows = 2! = 2

Total number of ways the food can be served to these people = 252 × 5! × 5! × 2 = 7257600

∴ the people can be served in 7257600 ways.

8. If a license plate has to be made where the first 4 places are numbers from 0 to 9 and the last two letters are made from the alphabets A to F, how many different license plates are possible where the repetition of any letter or number is not allowed?

a) 151200

b) 162800

c) 172800

d) 152400

We know that,

The 10 numbers can be placed in the first 4 places

Out of 10 digits, we need to select one

For the first place:

Out of 10 digits, we need to select one

Hence, combinations for third place = 10C1 = (10 * 9!) / 9! = 10

For the second place:

Out of 10 digits, we need to select one

Hence, combinations for third place = 9C1 = (9 * 8!) / 8! = 9

For the third place:

Out of 10 digits, we need to select one

Hence, combinations for third place = 8C1 = (8 * 7!) / 7! = 8

For the fourth place:

Hence, combinations for third place = 7C1 = (7 * 6!) / 6! = 7

The number of ways the 10 digits can be arranged in 4 places = 10 × 9 × 8 × 7 = 5040

The number of ways the last two letters can be arranged from A to F = 6 × 5 = 30

The total number of ways the license plates can be made = 5040 × 30 = 151200

∴ 151200 license plates can be made if no repetition is allowed

9. Three pen companies A, B and C launched 6, 5 and 6 different models respectively. Find the ways in which they can be displayed in a case with 17 slots such that the models of no two companies are mixed together.

a) (5!)(6!)

b) 180

c) (6!)^3/3!

d) (6!)^3

Here A has 6 models, B has 5 models and C has 6 models

So, A can be kept in 6! Ways

B can be kept in 5! Ways

C can be kept in 6! Ways

And considering them together A, B and C as 3 different slots.

They can be kept in 3! Ways

So, total ways in which they can be kept is 3! 6! 5! 6!

3! = 6 and 6(5!) = 6!

So, total ways are (6!)^3

10. In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl isa) 100

b) 200

c) 300

d) 400

Let there be g girls and b boys.

Number of games between two girls = gC2

Number of games between two boys = bC2

∴ g(g - 1)/2 = 45

∴ g = 10

Also, b(b - 1) / 2 = 190

∴ b = 20

∴ Total number of games = (g + b)C2 = 30C2 = 435

∴ Number of games in which one player is a boy and other is a girl = 435 – 45 - 190 = 200
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