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

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

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

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

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

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)

option 1 is correct.

2) P(A̅ ∪ B̅) = 1 – P(A ∪ B)

∴ option 2 is incorrect.

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

w.k.t

Probability of two tails coming from the fair coin =

\(\frac{1}{{\underbrace 2_{\begin{array}{*{20}{c}} {choosing\;}\\ {fair} \end{array}}}} \times \frac{1}{{\underbrace 2_{\begin{array}{*{20}{c}} {first\;}\\ {tails} \end{array}}}} \times \frac{1}{{\underbrace 2_{\begin{array}{*{20}{c}} {second\;}\\ {tails} \end{array}}}} = P\left( F \right)\)

Also,

Probability of two tail coming from unfair coin =

\(\frac{1}{{\underbrace 2_{\begin{array}{*{20}{c}} {choosing\;}\\ {fair} \end{array}}}} \times \underbrace 1_{\begin{array}{*{20}{c}} {first\;}\\ {tails} \end{array}} \times \underbrace 1_{\begin{array}{*{20}{c}} {second\;}\\ {tails} \end{array}} = P\left( U \right)\)

Hence probability that both tails came from fair coin =

\(\begin{array}{l} \frac{{P\left( F \right)}}{{P\left( F \right) + P\left( U \right)}}\\ = \frac{{\frac{1}{8}}}{{\frac{1}{8} + \frac{1}{2}}} = \frac{1}{5} \end{array}\)

Given:

A, B, and C are three mutually exclusive and exhaustive events such that if P(B) = \(\frac{3}{2}\)P(A) and P(C) = \(\frac{1}{2}\)P(B).

Concept:

If A, B, and C are three mutually exclusive and exhaustive events then,

P (A U B U C) = P(A) + P(B) + P(C) = 1

Solution:

According to the question,

A, B, and C are three mutually exclusive and exhaustive events such that if P(B) = \(\frac{3}{2}\)P(A) and P(C) = \(\frac{1}{2}\)P(B).

P (A U B U C) = P(A) + \(\frac{3}{2}P(A)\) + \(\frac{1}{2}P(B)\)

P (A U B U C) = P(A) + \(\frac{3}{2}P(A)\) + \(\frac{1}{2}\times \frac{3}{2}P(A)\)

P (A U B U C) = P(A) + \(\frac{3}{2}P(A)\) + \(\frac{3}{4}P(A)\)

P (A U B U C) =  \(\frac{13}{4}P(A)\)

Also,

P (A U B U C) = 1

\(\frac{13}{4}P(A)=1 \\ P(A)=\frac{4}{13}\)

Hence, option 4 is correct.

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:

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

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

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

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

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)

Concept:

Bayes' Theorem: It is a mathematical formula for determining conditional probability.

\(P(A|B) = \frac {P(B|A)\ P(A)}{P(B)}\)

Calculation:

Given:

P(raining) = \(7 \over 10\) ⇒ P(not-raining) = \(3 \over 10\); P(loss/rain) = \(8 \over 10\) ⇒ P(no-loss/rain) = \(2 \over 10\);

P(loss/no-rain) = \(1 \over 10\) ⇒ P(no-loss/no-rain) = \(9 \over 10\);

Probability of no-rain for no loss on a given day is calculated as

P(no-rain / no-loss) = \(\frac {P(no-loss/no-rain)\ \times \ P(not-raining)}{P(raining)\ \times\ P(no-loss/raining) \ +\ P(not-rainnig)\ \times \ P(no-loss/no ~raining) }\)

P(no-rain / no-loss) = \(\frac {\frac {9}{10}\ \times \ \frac {3}{10}}{\frac {7}{10}\ \times\ \frac {2}{10} \ +\ \ \frac {9}{10}\ \times \ \frac {3}{10} }\)

P(no-rain / no-loss) = \(\frac {27}{41}\)

Concept:

P(R ∪ S) = P(R) + P(S) - P(R \(\cap\) S)

If R and S are mutually disjoint P(R \(\cap\) S) = 0

Calculation:

Given:

P(R) = 0.4, P(S) = p and P(R ∪ S) = 0.6

Let us assume R and S are mutually disjoint, then

P(R ∪ S) = P(R) + P(S) - P(R \(\cap\) S)

0.6 = 0.4 + p - 0

p = 0.2

Explanation:

Total number of questions = 18

Choices for each question = 4

Here it is mentioned that one option is incorrect this means for each question there will be only three choices to guess the correct answer.

So we have a total of 18 questions which are will be considered as independent events.

The probability of choosing one correct answer from 3 choices is \(\frac1 3\).

\(For\;18\;questions = \frac{1}{3} \times 18 = 6\)

∴The expected number of correct questions = 6