Probability Test

Result

Probability Test - 5

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Two dice are rolled one after the other. Find the probability that the number on the first is smaller than the number on the second.

A

B

C

D

Solution

Total no. of cases = 6² = 36
Favourable cases (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 3) (2, 4) (2, 5) (2, 6), (3, 4) (3, 5) (3, 6), (4, 5) (4, 6) & (5, 6) i.e. 15 in numbers
Required probability = = .
Answer: (4)
Question 2
1 / -0

Two cards are drawn from a pack of 52 playing cards without replacement. What is the probability that both are red cards?

A
0.240

B
0.25

C
0.38

D
0.47

Solution

Probability that both are red cards, without replacement = (26/52) x (25/51) = 0.240
Question 3
1 / -0

A bag contains 6 white balls and 4 black balls. A person picks the balls randomly from the bag (without replacement). If the first 3 picks are white balls, what is the probability that the fourth ball picked by him is also white?

A

B

C

D

Solution

A bag contains 6 white and 4 black balls.
If the first 3 picks are white balls, remaining balls will be 3 white and 4 black.
Probability that the fourth ball drawn is also white = =
Question 4
1 / -0

The first purse contains 4 copper and 3 silver coins; the second purse contains 6 copper and 2 silver coins. A coin is taken out from any one purse. What is the probability that it is a copper coin?

A

B

C

D

Solution

The probability of choosing the first purse = 1/2
The probability of choosing coin in first purse = 4/7
Now, the probability that the copper coin is picked from the first purse = (1/2) x (4/7)
The probability of choosing the second purse = 1/2
The probability of choosing coin in second purse = 6/8
Now, the probability that the copper coin is picked from the first purse = (1/2) x (6/8)
Required probability = (1/2) x (4/7) + (1/2) x (6/8) = 2/7 + 3/8 = 37/56
Question 5
1 / -0

If a date and a month are selected at random from a leap year, what is the probability that the day is not a Tuesday?

A

B

C

D

Solution

The answer is . It does not matter, whether the year is a leap year, or not. But, the day selected can be any day from Sunday to Saturday.
Hence, the probability that the day will not be a Tuesday is .
Question 6
1 / -0

The probability of winning a match is 0.2. If a person plays three matches, what are the chances that he will win at least one of the three matches?

A
0.001

B
0.002

C
0.488

D
0.168

Solution

The probability of winning a match is 0.2.
If the person wins 1 match and looses 2 matches, then the probability = 0.2 × (0.8)² = 0.128
If the person wins 2 matches and looses 1 match, then the probability = (0.2)² × 0.8 = 0.032
If the person wins 3 matches, then the probability = (0.2)³= 0.008
Therefore, the probability that he will win at least one of the three matches = 0.128 + 0.032 + 0.008
= 0.168
Question 7
1 / -0

Two cards are drawn from a pack of 52 playing cards without replacement. What is the probability that both are red cards with numbers less than or equal to 10?

A
0.11

B
0.112

C
0.115

D
0.119

Solution

Probability that both are red cards with numbers less than or equal to 10, without replacement = (18/52) x (17/51) = 0.115
Question 8
1 / -0

An integer is chosen at random out of the numbers ranging from 1 to 50. What is the probability that the integer chosen is a multiple of 2 or 3 or 10?

A

B

C

D

Solution

Integers from 1 to 50, which are not multiples of 2 or 3 or 10, are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47 and 49.
So, required number of integers = 50 - 17 = 33
Probability = 33/50
Question 9
1 / -0

A dice is rolled and a coin is tossed. What is the probability of getting heads and an even number?

A

B

C

D

Solution

Possible outcomes are {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}
Favourable outcomes = {(2, H), (4, H), (6, H)}
Hence, required probability = Number of favourable outcomes/Total number of outcomes
Required Probability = =
Question 10
1 / -0

From a pack of 52 playing cards, five cards fall down – ten, jack, queen, king and ace of diamonds. After that, a card is drawn from the pack at random. What is the probability the card drawn is a face card?

A
5/47

B
9/47

C
3/47

D
7/47

Solution

As five cards – ten, jack, queen, king and ace of diamonds fall down, there are only 47 cards left in the pack, with 9 face cards.
Therefore, P(a face card) = 9/47
So, option 2 is correct.
Question 11
1 / -0

Twenty cards, numbered 1, 2, 3 … 20, are put in a box and mixed thoroughly. A card is then drawn from the box. Find the probability that the number on the selected card is divisible by 3 and 5 both.

A
1/10

B
3/20

C
1/5

D
1/20

Solution

Total number of cases = 20
Number divisible by 3 and 5 both = 15 (one only)
Therefore, Probability of card with a number divisible by 3 and 5 =
So, option 4 is correct.
Question 12
1 / -0

In a dice game, if a player rolls two dice, then his score will be the larger of the two numbers on the dice and if he rolls a doublet, his score will be the same number as the number in doublet. For example, if he rolls 4 and 6, his score will be 6 and if he rolls a doublet, e.g. 3 and 3, then his score will be 3. What is the probability that his score is 4 or more?

A

B

C

D

Solution

When a player rolls two dice, there are 6 possibilities for the outcome on each dice.
So, there are 36 possibilities for the outcomes when two dice are rolled.
The possibilities that give a score of 3 or less are:
1 and 1, 1 and 2, 1 and 3, 2 and 1, 2 and 2, 2 and 3, 3 and 1, 3 and 2, 3 and 3.
So, 9 of the 36 possibilities give a score of 3 or less and the probability of getting 3 or less = =
Thus, probability of getting 4 or more = 1 -
Question 13
1 / -0

A natural number is chosen at random from amongst the first 500. Find the probability of it being divisible by 3 or 5.

A

B

C

D

Solution

Natural numbers which are divisible by 3 = Integer part of 500/3 = 166
Natural numbers which are divisible by 5 = Integer part of 500/5 = 100
Natural number which are divisible by 15 = Integer part of 500/15 = 33
Natural numbers which are divisible by either 3 or 5 = P(3) + P(5) - P(3 and 5)
= 166 + 100 - 33 = 233
Required probability = 233/500
Question 14
1 / -0

Sahil draws a card at random from a pack of well-shuffled cards. Find the probability that it is neither a king nor a red card.

A

B

C

D

Solution

Total number of cards = 52
Total number of kings = 4
Total number of red cards = 26
Total number of red king cards = 2
So, total number of cards which are neither king nor red = 52 – 26 – 2 = 24
(As 26 red cards contain 2 red kings and there are 2 black kings also)
Thus, number of favourable cases = 24
Hence, required probability = = =
Question 15
1 / -0

The probability of solving a problem by Steve is , while the probability of solving a problem by Jack is . Find the probability that the problem is solved if they both try together.

A

B

C

D

Solution

Probability of solving the problem by Steve P(S) = 1/7
Probability of solving the problem by Jack P(J) = 1/8
Probability that Steve solves the problem or Jack solves the problem = P(S) + P(J) - P(S) . P(J)
= (1/7) + (1/8) - {(1/7) . (1/8)}
= (1/7) + (1/8) - (1/56)
= 14/56 = 1/4
Question 16
1 / -0

The number of students in class A and class B are in the ratio 3 : 4. In class A, 75% of the students are boys, and in class B, 40% of the students are girls. One student is selected at random from among the total students in both the classes. What is the probability that the selected student is a girl?

A

B

C

D

Solution

25% of the students of class A and 40% of the students of class B are girls.
_{Required probability =}
= = =
Question 17
1 / -0

Jannat has ten black pens, four blue pens and six red pens in her pen stand. If she chooses one pen at random, what is the probability that it will either be a red pen or a black pen?

A

B

C

D

Solution

Probability of choosing a red pen or a black pen =
Question 18
1 / -0

In a bag, there are 100 bulbs, out of which 30 are defective. A bulb is selected from the bag at random. What is the probability that the selected bulb is a good one?

A
0.50

B
0.70

C
0.30

D
0.10

Solution

Number of defective bulbs = 30
Total number of bulbs = 100
Probability that the selected bulb is a good one =
Question 19
1 / -0

A letter is selected at random from the letters of the word PROBABILITY. The probability that it is a vowel is

A

B

C

D

Solution

PROBABILITY - Total 11 characters - 4 vowels, 7 consonants
Hence, total number of cases = 11
Favourable cases = 4
Required probability = 4/11
Question 20
1 / -0

Probability that a student will succeed in GRE entrance test is 0.7 and that he will succeed in GMAT entrance test is 0.5. What is the probability that he will succeed in both the tests, if the probability that he will not be successful in both the tests is 0.2?

A
0.2

B
0.3

C
0.4

D
0.6

Solution

Probability that he succeeds in both the tests = 1 - Probability (does not succeed in any one of the two) = 1 - (0.3 + 0.5 - 0.2) = 0.4
Question 21
1 / -0

The probabilities of A, B and C solving a problem are , and , respectively. If all three try to solve the problem simultaneously, then find the probability that exactly one of them will solve it.

A

B

C

D
1/4

Solution

Let E₁, E₂and E₃ be the events that the problem is solved by A, B and C, respectively. Then,
P(E₁) = , P(E₂) = and P(E₃) =

Exactly one of A, B and C can solve the problem in the following mutually exclusive ways.
(i) A solves and B and C do not solve.
(ii) B solves and A and C do not solve.
(iii) C solves and A and B do not solve.

Probability that only one will solve the problem
= P(A) and P(not B) and P(not C) + P(not A) and P(B) and P(not C) + P(not A) and P(not B) and P(C)
=
Question 22
1 / -0

Two C.I.D. inspectors, A and B, are given a case to solve. The chance that A solves the case is 1/5 and that for B is 1/6. Find the probability that

(a) the case is solved
(b) B solves the case but A does not
(c) the case is not solved

A
, ,

B
, ,

C
, ,

D
, ,

Solution

P(A) = Probability that A solves the case =
P(A') =
P(B) = Probability that A solves the case =
P(B') =
Probability that the case is solved = 1 - (Probability that the case is not solved)
= 1 - [P(A') P(B')] = 1 - [(4/5) (5/6)] = 1 - =
Probability that B solves the case but A does not = P(A') P(B) = (4/5) (1/6) = 2/15
Probability that the case is not solved = [P(A') P(B')] = [(4/5)(5/6)] = 2/3

Question 23

1 / -0

Three dice are thrown simultaneously. Match the probability of events in Column A with the values in Column B.

Column A	Column B
(i) 6 appearing on all the dice	A. 1/36
(ii) 4 appearing on at least one die	B. 91/216
(iii) Sum of the numbers appearing on the dice to be greater than 3	C. 1/216
(iv) Sum of the numbers appearing on the dice to be 5	D. 215/216

(i) - C, (ii) - B, (iii) - A, (iv) - D

(i) - D, (ii) - A, (iii) - C, (iv) - B

(i) - D, (ii) - B, (iii) - A, (iv) - C

(i) - C, (ii) - B, (iii) - D, (iv) - A

Solution

Total number of possibilities = 6 × 6 × 6 = 216

(i) 6 appearing on all the dice:
There is only one possibility for dice bearing all sixes (6, 6, 6) out of 216 possible outcomes.
Required probability = 1/216

(ii) 4 appearing on at least one die:
This event comprises all the events when 4 appeared on the die, whether once, or twice, or thrice.
Number of outcomes when no 4 appeared = 5 × 5 × 5 = 125
Number of outcomes when 4 appeared on the die at least once = Total number of possibilities - Number of outcomes when no 4 appeared = 216 - 125 = 91
Required probability = 91/216

(iii) Sum of the numbers appearing on the dice to be greater than 3:
Whatever be the possibilities, the least numbers appearing on the dice are (1, 1, 1) and sum = 1 + 1 + 1 = 3; that is, there is only one possibility in which sum is 3.
So, number of outcomes when the sum is greater than 3 = 216 - 1 = 215
Required probability = 215/216

(iv) Sum of the numbers appearing on the dice to be 5:
The possible outcomes of the event are: (1, 1, 3); (1, 2, 2); (1, 3, 1); (2, 1, 2); (2, 2, 1) and (3, 1, 1).
Required probability = 6/216 = 1/36

Question 24
1 / -0

The given figure represents the dimensions of different fields of a sports complex. A sportsman while playing in one of the given fields has lost his car key.

Find the probability that the key was lost in the

(i) hockey field
(ii) tennis court
(iii) cricket field

A
(i) , (ii) , (iii)

B
(i) , (ii) , (iii)

C
(i) , (ii) , (iii)

D
(i) , (ii) , (iii)

Solution

Area of the whole sports complex = 3 × 4 = 12 km²Now,
(i) Area of the hockey field = 3 × 1.5 = 4.5 km²
So, probability that the key was lost in the hockey field =
(ii) Area of the tennis court = 2.5 × 1 = 2.5 km²
So, probability that the key was lost in the tennis court =
(iii) Area of the cricket field = 2.5 × 2 = 5 km²
So, probability that the key was lost in the cricket field =

Question 25

1 / -0

Fill in the blanks with the help of the given table.

(i) The probability of an event that is certain to happen is P . Such an event is called a/an Q event.
(ii) The sum of the probabilities of all the elementary events of an experiment is R .
(iii) The event which gives the probability of non occurrence of that event is called its S event.

	P	Q	R	S
A	1	complementary	1	certain
B	0	impossible	0	certain
C	0	complementary	0	impossible
D	1	certain	1	complementary

A

B

C

D

Solution

(i) The probability of an event that is certain to happen is 1. Such an event is called a certain event.
(ii) The sum of the probabilities of all the elementary events of an experiment is 1.
(iii) The event which gives the probability of non occurrence of that event is called its complementary event.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Probability Test - 5

Result

Probability Test - 5

Score

-

Rank

-

SHARING IS CARING

Question 1

Solution

Question 2

0.240

0.25

0.38

0.47

Solution

Question 3

Solution

Question 4

Solution

Question 5

Solution

Question 6

0.001

0.002

0.488

0.168

Solution

Question 7

0.11

0.112

0.115

0.119

Solution

Question 8

Solution

Question 9

Solution

Question 10

5/47

9/47

3/47

7/47

Solution

Question 11

1/10

3/20

1/5

1/20

Solution

Question 12

Solution

Question 13

Solution

Question 14

Solution

Question 15

Solution

Question 16

Solution

Question 17

Solution

Question 18

0.50

0.70

0.30

0.10

Solution

Question 19

Solution

Question 20

0.2

0.3

0.4

0.6

Solution

Question 21

1/4

Solution

Question 22