Probability MCQ Quiz - Objective Question with Answer for Probability - Download Free PDF

Last updated on Apr 17, 2025

Latest Probability MCQ Objective Questions

Probability Question 1:

If two events A and B such that P(A ∪ B) = \(\frac{7}{8}\) and P(A ∩ B) = \(\frac{1}{4}\) and P(A) = \(\frac{5}{8}\), then P\((\overline{A} ∪ \overline{B})\) = ?

  1. \(​​\frac{3}{8}\)
  2. \(​​\frac{1}{8}\)
  3. \(​​\frac{3}{4}\)
  4. \(​​\frac{1}{4}\)
  5. \(​​\frac{3}{9}\)

Answer (Detailed Solution Below)

Option 3 : \(​​\frac{3}{4}\)

Probability Question 1 Detailed Solution

Explanation:

Given, P(A ∪ B) = \(\frac{7}{8}\) and P(A ∩ B) = \(\frac{1}{4}\) and P(A) = \(\frac{5}{8}\)

To find: P\((\overline{A} ∪ \overline{B})\) = ?

Since, \(P(A\cap B)+P(\overline{A\cap B})=1\)

⇒ \(P(\overline{A\cap B})=1-P(A\cap B)\)

⇒ \(P(\overline{A\cap B})=1-\frac{1}{4}\)

⇒ \(P(\overline{A\cap B})=\frac{3}{4}\)

We know that, \(P(\overline{A}\cup \overline{B})=P(\overline{A\cap B})\)

Hence, \( P(\overline{A}\cup \overline{B})=\frac{3}{4}\)

Probability Question 2:

Probability that a leap year selected at random containing 53 Sundays, is equal to

  1. \(\frac{53}{365}\)
  2. \(\frac{1}{7}\)
  3. \(\frac{2}{7}\)
  4. \(\frac{1}{365}\)
  5. \(\frac{1}{10}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{2}{7}\)

Probability Question 2 Detailed Solution

Concept:

Number of days in leap year = 366

Calculation:

A week has 7 days and total days are 366.

⇒ Number of Sundays in a leap year

= 366/7

= 52 Sundays + 2 days

⇒ Total outcomes with 2 days

= (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

= 7

⇒ Number of outcomes without Sundays = 5

⇒ Probability of leap year with 53 Sundays = 2/7

∴ Probability of leap year with 53 Sundays is 2/7.

The correct answer is Option 3.

Probability Question 3:

If A and B are two events such that P(A) ≠ 0 and P(A) ≠ 1, then \(\rm P \left( {\frac{{\bar A}}{{\bar B}}} \right)\)

  1. \(1 - P\left( {\frac{A}{B}} \right)\)
  2. \(1 - P\left( {\frac{{\bar A}}{B}} \right)\)
  3. \(\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)
  4. \(\frac{{P\left( {\bar A} \right)}}{{P\left( {\bar B} \right)}}\)
  5. \(1 - P\left( {\frac{B}{A}} \right)\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)

Probability Question 3 Detailed Solution

\(\rm P \left( {\frac{{\bar A}}{{\bar B}}} \right) = \frac{{P\left( {\bar A \cap \bar B} \right)}}{{P\left( {\bar B} \right)}}\)

\( {{P\left( {\bar A \cap \bar B} \right)}}\) = \(P({\overline {A \cup B}})\)

\(= \frac{{P\left( {\overline {A \cup B} } \right)}}{{P\left( {\bar B} \right)}} = \frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)

Probability Question 4:

If A and B are two events such that P(A) ≠ 0 and P(A) ≠ 1, then \(\rm P \left( {\frac{{\bar A}}{{\bar B}}} \right)\)

  1. \(1 - P\left( {\frac{A}{B}} \right)\)
  2. \(1 - P\left( {\frac{{\bar A}}{B}} \right)\)
  3. \(\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)
  4. \(\frac{{P\left( {\bar A} \right)}}{{P\left( {\bar B} \right)}}\)
  5. None of these 

Answer (Detailed Solution Below)

Option 3 : \(\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)

Probability Question 4 Detailed Solution

\(\rm P \left( {\frac{{\bar A}}{{\bar B}}} \right) = \frac{{P\left( {\bar A \cap \bar B} \right)}}{{P\left( {\bar B} \right)}}\)

\( {{P\left( {\bar A \cap \bar B} \right)}}\) = \(P({\overline {A \cup B}})\)

\(= \frac{{P\left( {\overline {A \cup B} } \right)}}{{P\left( {\bar B} \right)}} = \frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)

Probability Question 5:

For two events A and B, which of the following relations is true?

  1. \(P(\bar A \cup \bar B) = 1 - P(A)P\left(\frac B A\right)\)
  2. P(A̅ ∪ B̅) = 1 - P(A ∪ B)
  3. \(P(\bar A \cup \bar B) = P(\overline {A\cup B})\)
  4. P(A̅ ∪ B̅) = P(A̅) + P(B̅)
  5. None of these 

Answer (Detailed Solution Below)

Option 1 : \(P(\bar A \cup \bar B) = 1 - P(A)P\left(\frac B A\right)\)

Probability Question 5 Detailed Solution

Let us go by options, one by one

1. \(P(\frac{B}{A}) = \frac{P(A ~∩~ B)}{P(A)}\)

\(P(\bar A \cup \bar B) = 1 - P(A)P\left(\frac B A\right)\)

\(P(\bar A \cup \bar B) = 1 - P(A)\frac{P(A∩ B)}{P(A)}\)

= 1 - P(A ∩ B)

F1 Neha 23.12.20 Pallavi D1

option 1 is correct.

2) P(A̅ B̅) = 1 – P(A B)

F1 Neha Madhu 23.12.20 D1

option 2 is incorrect.

3) \(P\left( {\bar A \cup \bar B} \right) = P\left( {\overline {A \cup B} } \right)\) 

F1 Neha Madhu 23.12.20 D2

Top Probability MCQ Objective Questions

A bag contains 3 white, 2 blue and 5 red balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is not red?

  1. 3/10
  2. 1/5
  3. 1/2
  4. 4/5

Answer (Detailed Solution Below)

Option 3 : 1/2

Probability Question 6 Detailed Solution

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Calculation:

A bag contains 3 white, 2 blue and 5 red balls.

Total number of balls = 3 + 2 + 5 = 10

Number of balls that are not red = 10 - 5 = 5

Probability of balls drawn is not red = (number of balls which are not red)/(total number of balls) = 5/10 = 1/2

An urn contains 5 red ball and 5 black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a red ball in the second draw is

  1. \(\frac{1}{2}\)
  2. \(\frac{4}{9}\)
  3. \(\frac{5}{9}\)
  4. \(\frac{6}{9}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{2}\)

Probability Question 7 Detailed Solution

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Concept:

In probability theory, the probability measure of an event is made if another event has already occurred is referred to as conditional probability.

For calculating conditional probability, the probability of the preceding event and the probability of succeeding event is multiplied.

Conditional probability is given by

\(P\left( {{E_1}/{E_2}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_2}} \right)}}\)

\(P\left( {{E_2}/{E_1}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}}\)

Where E1 and E2 are the events.

Calculation:

Given:

Urn contains 5 red balls, 5 black balls.

One ball is picked at random.

Case (i): The first ball is red ball

Probability to get a red ball in the second draw is

\({P_1} = \frac{5}{{10}} \times \frac{4}{9} = \frac{2}{9}\)

Case (ii): The first ball is black ball

Probability to get a red ball in the second draw is

\({P_2} = \frac{5}{{10}} \times \frac{5}{9} = \frac{5}{{18}}\)

Required probability (P) \(= {P_1} + {P_2} = \frac{2}{9} + \frac{5}{{18}} = \frac{1}{2}\)

A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve a problem, selected at random from the book?

  1. 0.16
  2. 0.69
  3. 0.97
  4. 0.20

Answer (Detailed Solution Below)

Option 3 : 0.97

Probability Question 8 Detailed Solution

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Concept:

when two independent A and B events occur.

then the probability of occurring of at least one event is given by:

P = 1 - (P(\({\rm{\bar A}}\)) × P(\({\rm{\bar B}}\)))

Calculation:

Given:

A can solve 90% of problems and B can solve 70% of problems.

Therefore, A and B are independent of each other.

P(A) = 0.90 and P(B) = 0.70

Therefore, P(at least one of them will solve a problem)  = 1 - (P(\({\rm{\bar A}}\)) × P(\({\rm{\bar B}}\)))

∴ P = 1 - [(1 - 0.9) × (1 - 0.7)] ⇒ 1 - 0.03

P = 0.97 

P and Q are considering to apply for a job. The probability that P applies for the job is \(\frac{1}{4}\), the probability that P applies for the job given that Q applies for the job is \(\frac{1}{2}\), and the probability that Q applies for the job given that P applies for the job is \(\frac{1}{3}\). Then the probability that P does not apply for the job given that Q does not apply for the job is

  1. \(\frac{4}{5}\)
  2. \(\frac{5}{6}\)
  3. \(\frac{7}{8}\)
  4. \(\frac{{11}}{{12}}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{4}{5}\)

Probability Question 9 Detailed Solution

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Data:  

\(p\left( P \right) = \frac{1}{4}\)

\(P\left( {\frac{P}{Q}} \right) = \;\frac{1}{2}\) , \(P\left( {\frac{Q}{P}} \right) = \;\frac{1}{3}\)

Formula

\(P\left( {\frac{A}{B}} \right) = \;\frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\)

Calculation:

\(P\left( {\frac{Q}{P}} \right) = \;\frac{{P\;\left( {P \cap Q} \right)}}{{P\left( P \right)}},\;\;\) 

\(\frac{1}{3} = \;\frac{{P\left( {P \cap Q} \right)}}{{\frac{1}{4}}}\)  , 

\(P\left( {P \cap Q} \right) = \;\frac{1}{{12}}\)

Also, \(P\left( {\frac{P}{Q}} \right) = \frac{{P\;\left( {P \cap Q} \right)}}{{P\left( Q \right)}},\;\;\)

\(\frac{1}{2} = \frac{{\frac{1}{{12}}}}{{P\left( Q \right)}}\) ,

\(P\left( Q \right) = \frac{1}{6}\) 

Required probability, \(P\left( {\frac{{P'}}{{Q'}}} \right) = \frac{{P\left( {P' \cap Q'} \right)}}{{P\left( {Q'} \right)}}\)

\(= \frac{{P{{\left( {P \cup Q} \right)}'}}}{{1 - P\left( Q \right)}}\; = \frac{{1 - P\left( {P \cup Q} \right)}}{{1 - P\left( Q \right)}}\)

\(= \frac{{1 - \left( {P\left( P \right) + P\left( Q \right) - P\;\left( {P \cap Q} \right)} \right)}}{{1 - P\left( Q \right)}}\;\)

\(= \frac{{1 - \left( {\frac{1}{4} + \frac{1}{6} - \frac{1}{{12}}} \right)}}{{1 - \frac{1}{6}}}\)

\(= \frac{{\frac{8}{{12}}}}{{\frac{5}{6}}} = \frac{4}{5}\)

In a simultaneous throw of two coins, the probability of getting at least one head is

  1. 1/2
  2. 1/3
  3. 2/3
  4. 3/4

Answer (Detailed Solution Below)

Option 4 : 3/4

Probability Question 10 Detailed Solution

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Explanation:

When a coin is tossed, there are only two possible outcomes, either heads or tails.

We know that the sample space S = {HH, HT, TH, TT}

The Event E that at least one of them is head = {HH, HT, TH}

Probability P(E) = \(\frac{n(e)}{n(s)}\) = \(\frac34\).

The probability of getting at least one head is \(\frac34\).

If for a moderately symmetrical distribution mean deviation is 12, then the value of standard deviation is

  1. 15
  2. 12
  3. 24
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : 15

Probability Question 11 Detailed Solution

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Concept:

  • The standard deviation (SD) measures the amount of variability, or dispersion, from the individual data values to the mean, while the standard error of the mean (SEM) or mean deviation measures how far the sample mean (average) of the data is likely to be from the true population mean.
  • The SEM is always smaller than the SD.
  • In a symmetrical distribution, mean deviation equals 4/5 of standard deviation.

Calculation:

Since the distribution is symmetrical,

⇒ Mean deviation = 4/5 of Standard deviation

⇒ 12 = (4/5) × Standard deviation

⇒ Standard deviation = 15

Hence, the value of the standard deviation is 15.

If A and B are two events such that P(A) ≠ 0 and P(A) ≠ 1, then \(\rm P \left( {\frac{{\bar A}}{{\bar B}}} \right)\)

  1. \(1 - P\left( {\frac{A}{B}} \right)\)
  2. \(1 - P\left( {\frac{{\bar A}}{B}} \right)\)
  3. \(\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)
  4. \(\frac{{P\left( {\bar A} \right)}}{{P\left( {\bar B} \right)}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)

Probability Question 12 Detailed Solution

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\(\rm P \left( {\frac{{\bar A}}{{\bar B}}} \right) = \frac{{P\left( {\bar A \cap \bar B} \right)}}{{P\left( {\bar B} \right)}}\)

\( {{P\left( {\bar A \cap \bar B} \right)}}\) = \(P({\overline {A \cup B}})\)

\(= \frac{{P\left( {\overline {A \cup B} } \right)}}{{P\left( {\bar B} \right)}} = \frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}\)

The probability that a person  stopping at a gas station will ask to have his tyres checked is 0.12, the probability that he will ask to have his oil checked is 0.29 and the probability that he will ask to have them both checked is 0.07. The probability that a person who has his tyres checked will also have oil checked is

  1. 0.34
  2. 0.58
  3. 0.24
  4. 0.41

Answer (Detailed Solution Below)

Option 2 : 0.58

Probability Question 13 Detailed Solution

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Concept:

Conditional probability:

It gives the probability of happening of any event if the other has already occurred.

\({\rm{P}}\left( {\frac{{{{\rm{E}}_1}}}{{{{\rm{E}}_2}}}} \right) = {\rm{Probability\;of\;getting\;the\;event\;}}{{\rm{E}}_1}{\rm{\;when\;}}{{\rm{E}}_2}{\rm{is\;already\;occured}}.\)

\(P\left( {\frac{{{E_1}}}{{{E_2}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_2}} \right)}}\)

Calculation:

Given:

P (E1) = Probability of stopping at the gas station and ask for tyre checked = 0.12

P (E2) = Probability of stopping at the gas station and ask for oil checked = 0.29

P (E1∩ E2) = Probability of both checked = 0.07

\({\rm{P}}\left( {\frac{{{{\rm{E}}_2}}}{{{{\rm{E}}_1}}}} \right) = {\rm{Probability\;of\;person\;who\;has\;his\;tyre\;checked\;will\;also\;have\;oil\;checked}}\)   

∵ \(P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}}\)

∴ \(P\left( {\frac{{{E_2}}}{{{E_1}}}} \right) = \frac{{0.07}}{{0.12}} = 0.58\)

The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________

  1. 0
  2. 0.25
  3. 0.33
  4. 0.50

Answer (Detailed Solution Below)

Option 2 : 0.25

Probability Question 14 Detailed Solution

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Concept:

In such questions, show various regions represented by equations on the graph.

Calculation:

Given that

0 ≤ x ≤ 10

0 ≤ y ≤ 20

p {x + y ≥ 20} = ?

F1 V.S M.P 26.09.19 D 1

Required probability = Area of right angled triangle ABC/Area of rectangular region OABD

\(p=\frac{\frac{1}{2}\times 10\times 10}{10\times 20}=\frac{1}{4}=0.25\)

The chances of a defective screw in three boxes A, B, C are \(\frac{1}{5},{\rm{\;}}\frac{1}{6}\) and \(\frac{1}{7}\) respectively. A box is selected at random and a screw drawn from it at random is found to be defective. Find the probability that it came from box A.

  1. \(\frac{{40}}{{107}}\)
  2. \(\frac{{41}}{{107}}\)
  3. \(\frac{{42}}{{107}}\)
  4. \(\frac{{43}}{{107}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{42}}{{107}}\)

Probability Question 15 Detailed Solution

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Let E1, E2 and E3 denote the events of selecting box A, B, C respectively and A be the event that a screw selected at random is defective.

Then,

P(E1) = P(E2) = P(E3) = 1/3,

\({\rm{P}}\left( {{\rm{A}}/{{\rm{E}}_1}} \right) = \frac{1}{5}\)

\({\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{E}}_2}}}} \right) = \frac{1}{6} \Rightarrow {\rm{P}}\left( {{\rm{A}}/{{\rm{E}}_3}} \right) = \frac{1}{7}\)

Then, by Baye’s theorem, required probability

= P(E1/A)

\(= \frac{{\frac{1}{3}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{5} + \frac{1}{3}.\frac{1}{6} + \frac{1}{3}.\frac{1}{7}}} = \frac{{42}}{{107}}\)
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