Bayes's Theorem MCQ Quiz - Objective Question with Answer for Bayes's Theorem - 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:

20% of the pens are red,4% pens are red and defective.

Formula used:

Bayes Theorem:- Let E1, E2,... En be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E1 or E2 or... or En such that P(A) ≠ 0. Then
Bayes Formula:

P(Ei | A) = \(\frac{P(E_i) \times P(A|E_i)}{\sum_{i = 1}^ {n}P(E_i) \times P(A| E_i)}\), i = 1, 2, ... n
Calculation: 

Let A be the event that the pens are red

B be the event that the pens are defective

P(A) = \(\frac{20}{100}\) = \(\frac{1}{5}\)

Probability that the pen is red and defective is:

P(A ∩ B) = \(\frac{4}{100}\) = \(\frac{1}{25}\)

Probability that the pen is defective, given that it is red, 

P(\(\frac{B}{A}\)) = \(\frac{P(A \cap B)}{P(A)}\) = \(\frac{1}{25} \times \frac{5}{1}\)

⇒ \(\frac{1}{5}\)

⇒ 0.2

The Correct Answer is 0.2

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

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

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

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:

Let A be the event of choosing box A, and B be the event of choosing box B, and R is the event of drawing a red ball.

Baye's theorem:

\(P\left( {\frac{B}{R}} \right) = \frac{{P\left( B \right)\;P\left( {\frac{R}{B}} \right)}}{{P\left( A \right) \times \left( {\frac{R}{A}} \right)\;+\;P\left( B \right) \times \left( {\frac{R}{B}} \right)}}\)

where

P(A) = Probability of event A.

P(B) = Probability of event B.

\(P\left( {\frac{R}{B}} \right)\) = Probability of event R when event B already occur

\(P\left( {\frac{R}{A}} \right)\) = Probability of event R when event A already occur

\(P\left( {\frac{B}{R}} \right)\) = Probability of event B when event R already occur

Calculation:

Given:

P(A) = 2P(B), \(P\left( {\frac{R}{B}} \right) = \frac{4}{7}\), \(P\left( {\frac{R}{A}} \right) = \frac{3}{5}\)

Now, the probability of drawing a ball from box B, provided it is a red ball

\(P\left( {\frac{B}{R}} \right) = \frac{{P\left( B \right)\;P\left( {\frac{R}{B}} \right)}}{{P\left( A \right) \times P\left( {\frac{R}{A}} \right)\;+\;P\left( B \right) \times P\left( {\frac{R}{B}} \right)}}\)

\(P\left( {\frac{B}{R}} \right)= \frac{{P\left( B \right) \times \left( {\frac{4}{7}} \right)}}{{2P\left( B \right) \times \left( {\frac{3}{5}} \right)\;+\;P\left( B \right) \times \left( {\frac{4}{7}} \right)}}\)

\(P\left( {\frac{B}{R}} \right)= \frac{{\left( {\frac{4}{7}} \right)}}{{2 \times \left( {\frac{3}{5}} \right)\;+\;\left( {\frac{4}{7}} \right)}}\)

\(\therefore P\left( {\frac{B}{R}} \right)= \frac{{10}}{{31}}\)

Given that P(A) = 2P(B) = 3P(C) and A, B, C are mutually exclusive

⇒ P(A) + P(B) + P(C) = 1

Let P(A) = 2P(B) = 3P(C) = k

⇒ P(A) = k, P(B) = \(\frac{{\rm{k}}}{2}\) P(C) = \(\frac{{\rm{k}}}{3}\)

∴ k + \(\frac{k}{2} + \frac{k}{3}\) = 1 ⇒ k = \(\frac{6}{{11}}\)

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

Concept:

According to Bayes’ theorem, if multiple events A_i form an exhaustive set with another event B.

\(P\left( {{A_i}/B} \right) = \frac{{{\rm{P}}\left( {{\rm{B}}/{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)}}{{P\left( B \right)}}\)

Where, \(P\left( B \right) = \mathop \sum \limits_{i = 1}^n {\rm{P}}\left( {{\rm{B}}/{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)\)

Calculation:

Let P(A) be the probability of choosing a ball from box A

P(A) = 1/2

Let P(B) be the probability of choosing a ball from box B

P(B) = 1/2

P(R) be the probability of getting a red ball.

P(R/A) be the probability of getting red given that we are drawing a ball from box A

\(P\left( {R/A} \right) = \frac{3}{5}\)

P(R/B) be the probability of getting red given that we are drawing a ball from box B

\(P\left( {R/B} \right) = \frac{5}{9}\)

From the total probability

P(R) = P(R/A) P(A) + P(R/B) P(B)

\(= \left( {\frac{3}{5} \times \frac{1}{2}} \right) + \left( {\frac{5}{9} \times \frac{1}{2}} \right) = \frac{{26}}{{45}}\)

Let P(B/R) be the probability of chosen red ball given that it was from box B,

\(P\left( {B/R} \right) = \frac{{{\rm{P}}\left( {\frac{{\rm{R}}}{{\rm{B}}}} \right){\rm{P}}\left( {\rm{B}} \right)}}{{P\left( R \right)}} = \frac{{\frac{5}{9} \times \frac{1}{2}}}{{\frac{{26}}{{45}}}} = \frac{{25}}{{52}}\)

(i) True only when events A and B are independent.

(ii) True only when A and are exclusive.

(iii). \(P(A/B)=P(A∩B)/P(B)\)

(iv). \(P(A∪B)≤P(A)+P(B)-P(A∩B)\)

Concept:

M for men, W for women, E for employed & U for unemployed

Referring to the diagram below,

∴ Total probability when the selected person is unemployed,

P(U) = P(M). P(U/M) + P(W) ⋅ P(U/W)       ...(1)

Total probability when the selected person is emlpoyed,

 ⇒ P(E) = 1 – P(U)      ...(2)

Calculation:

Given:

P(M) = 0.5, P(W) = 0.5, P(U/M) = 0.2 & P(U/W) = 0.5

Using equation (1),

⇒ P(U) = (0.5× 0.2) + (0.5× 0.5)

⇒ P(U) = 0.35

Using equation (1),

⇒ P(E) = 1 - 0.35

⇒ P(E) = 0.65