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

Concept:

For a uniform distribution defined over the interval \([a, b]\).The formulas for the mean and variance are:

• Mean: \(\mu = \frac{a + b}{2}\)
• Variance: \(\sigma^2 = \frac{(b - a)^2}{12}\)

Given:

a = -4 & b =1

Calculation:

Mean:

\(\mu = \frac{-4 + 1}{2} = \frac{-3}{2} = -1.5\)

Variance:

\(\sigma^2 = \frac{(1 - (-4))^2}{12} = \frac{(5)^2}{12} = \frac{25}{12}\)

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

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.

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

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

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

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

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

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

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 

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

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

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

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

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

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

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

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.

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

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

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

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

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} = ?

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

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

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)