Baye's Theorem MCQ Quiz - Objective Question with Answer for Baye's Theorem - Download Free PDF

Calculation

E₁ : Bag B₁ is selected

\(\begin{array}{lll} \mathrm{B}_{1} & \mathrm{~B}_{2} & \mathrm{~B}_{3} \\ 6 \mathrm{~W} 4 \mathrm{~B} & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B} \end{array}\)

E₂ : bag B₂ is selected

E₃ : Bag B₃ is selected

A : Drawn ball is white

We have to find \(P\left(\frac{E_{2}}{A}\right)\)

\(P\left(\frac{E_{2}}{A}\right)=\frac{P\left(E_{2}\right) P\left(\frac{A}{E_{2}}\right)}{P\left(E_{1}\right) P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) P\left(\frac{A}{E_{3}}\right)}\)

= \(\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}}=\frac{4}{15}\)

Hence option 2 is correct

Concept Used:

Conditional Probability: P(A|B) = \(\frac{P(A \cap B)}{P(B)}\)

Bayes' Theorem: P(A|B) = \(\frac{P(B|A) P(A)}{P(B)}\)

Calculation:

Given:

Probability that A speaks truth = \(\frac{4}{5}\)

A coin is tossed. A reports that a head appears.

Let H be the event that a head appears.

Let R be the event that A reports a head.

P(H) = \(\frac{1}{2}\) (probability of getting a head)

P(¬H) = \(\frac{1}{2}\) (probability of not getting a head, i.e., a tail)

P(R|H) = \(\frac{4}{5}\) (probability that A reports a head given that there was a head - A speaks the truth)

P(R|¬H) = \(1 - \frac{4}{5} = \frac{1}{5}\) (probability that A reports a head given that there was a tail - A lies)

P(R) = P(R|H)P(H) + P(R|¬H)P(¬H)

P(R) = \(\frac{4}{5} \times \frac{1}{2} + \frac{1}{5} \times \frac{1}{2} = \frac{4}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}\)

Using Bayes' theorem:

P(H|R) = \(\frac{P(R|H)P(H)}{P(R)}\)

P(H|R) = \(\frac{\frac{4}{5} \times \frac{1}{2}}{\frac{1}{2}}\)

P(H|R) = \(\frac{4}{5}\)

Hence option 2 is correct

Calculation

B: Probability of drawing a black ball.

B₁: Probability associated with drawing from the first set of bags, where each bag contains 3 white and 5 black balls.

B₂: Probability associated with drawing from the second set of bags, where each bag contains 6 white and 4 black balls.

Given:

Probability of choosing a bag from the first set (P(B₁)) is \(\frac{1}{2}\).

Probability of choosing a bag from the second set (P(B₂)) is also \(\frac{1}{2}\) since there are two bags in the first set and four bags in the second set but we simplify by considering overall probabilities.

Probabilities of drawing a black ball from each type of bag:

P(B | B₁) = \(\frac{5}{8}\) for Bag 1.

P(B | B₂) = \(\frac{4}{10}\) = \(\frac{2}{5}\) for Bag 2.

The total probability of drawing a black ball (P(B)) is calculated by adding the probabilities of drawing a black ball from each type of bag, weighted by the probability of choosing each type of bag:

P(B) = P(B₁) · P(B | B₁) + P(B₂) · P(B | B₂)

= \(\frac{1}{2}\) × \(\frac{5}{8}\) + \(\frac{1}{2}\) × \(\frac{2}{5}\)

= \(\frac{5}{16}\) + \(\frac{1}{5}\)

= \(\frac{41}{80}\) 

Final Probability:

Using Bayes' Theorem, we find the probability that a black ball is from the first set of bags:

P(B₁ | B) = \(\frac{P(B|B_1) \cdot P(B_1)}{P(B)}\)

= \(\frac{\frac{5}{8} \times \frac{1}{2}}{\frac{41}{80}}\)

= \(\frac{5}{16} \times \frac{80}{41}\)

= \(\frac{25}{41}\)

Thus, the probability that the black ball is from the first set of bags is \(\frac{25}{41}\).

Hence option 2 is correct

Given:

Refrigerators from 3 companies: C1 (25%), C2 (35%), C3 (40%).

Defective percentages: C1 (3%), C2 (2%), C3 (1%).

Concept Used:

Bayes' Theorem: \(P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}\)

where \(P(A|B)\) is the probability of A given B, \(P(B|A)\) is the probability of B given A, \(P(A)\) is the probability of A, and \(P(B)\) is the probability of B.

Calculation:

Let D be the event that a refrigerator is defective.

\(P(C1) = 0.25\)

\(P(C2) = 0.35\)

\(P(C3) = 0.40\)

\(P(D|C1) = 0.03\)

\(P(D|C2) = 0.02\)

\(P(D|C3) = 0.01\)

\(P(D) = P(D|C1)P(C1) + P(D|C2)P(C2) + P(D|C3)P(C3)\)

\(P(D) = (0.03 \times 0.25) + (0.02 \times 0.35) + (0.01 \times 0.40)\)

\(P(D) = 0.0075 + 0.0070 + 0.0040 = 0.0185\)

Now, using Bayes' Theorem:

\(P(C2|D) = \frac{P(D|C2) \times P(C2)}{P(D)}\)

\(P(C2|D) = \frac{0.02 \times 0.35}{0.0185}\)

\(P(C2|D) = \frac{0.0070}{0.0185}\)

\(P(C2|D) = \frac{70}{185} = \frac{14}{37}\)

Hence option 3 is correct

Formula Used:

Bayes' Theorem:

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

Calculation:

Let:

A1 = Event that the person goes to college by car

A2 = Event that the person goes to college by bus

A3 = Event that the person goes to college by train

B = Event that the person reaches the college in time

We need to find P(A1|B)

\(P(A_1) = \frac{1}{5}\)

\(P(A_2) = \frac{2}{5}\)

\(P(A_3) = \frac{3}{5}\)

\(P(B|A_1) = 1 - \frac{2}{7} = \frac{5}{7}\)

\(P(B|A_2) = 1 - \frac{4}{7} = \frac{3}{7}\)

\(P(B|A_3) = 1 - \frac{1}{7} = \frac{6}{7}\)

Using the law of total probability:

\(P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3)\)

\(P(B) = \frac{5}{7} \times \frac{1}{5} + \frac{3}{7} \times \frac{2}{5} + \frac{6}{7} \times \frac{3}{5}\)

\(P(B) = \frac{5}{35} + \frac{6}{35} + \frac{18}{35} = \frac{29}{35}\)

Using Bayes' Theorem:

\(P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)}\)

\(P(A_1|B) = \frac{\frac{5}{7} \times \frac{1}{5}}{\frac{29}{35}}\)

\(P(A_1|B) = \frac{\frac{5}{35}}{\frac{29}{35}}\)

\(P(A_1|B) = \frac{5}{29}\)

Hence option 3 is correct

Concept:

Law of Total Probability:

• If E₁, E₂, E₃, ⋯ is a partition of the sample space S, then for any event A we have

\(\Rightarrow P\;\left( A \right) = \mathop \sum \limits_{i\; = \;1} P\left( {{E_i}} \right)\;P\left( {\frac{A}{{{E_i}}}} \right)\)

Calculation:

Given:

Bag one contain 3 white and 2 black balls and bag two contains 5 white and 3 black balls

Let E₁, E₂ and A be the three events as defined below:

E₁ = Selecting bag one

E₂ = Selecting bag two

A = Drawing a white ball

Now, Bag is chosen randomly;

∴ P (E₁) = P (E₂) = 1/2

Now,

Probability of getting white ball from bag 1 = P (A/ E₁) = 3/5

Probability of getting white ball from bag 2 = P (A/ E₂) = 5/8

Using the law of total probability, we get

Required probability P (A) = P (E₁) P (A/ E₁) + P (E₂) P (A/ E₂) = (1/2) × (3/5) + (1/2) × (5/8)

Concept:

Let E₁, E₂,…, E_n be a set of events associated with a sample space S, where all the events E₁, E₂,…, E_n have nonzero probability of occurrence and they form a partition of S. Let A be any event associated with S, then by Bayes theorem,

P(E_i/A) = \({P(E_i)P(A|E_i)\over\sum_{k=1}^nP(E_k)P(A|E_k)}\)

Explanation:

Let b: black, r: red

\(\begin{matrix} A & B & C \\\ 7r, 5b & 5r, 7b & 6r, 6b \end{matrix}\)

Selecting any urn is P(A) = P(B) = P(C) = 1/3

Probability that black ball is from urn A is P(A/b) = \(5\over 12\)

Probability that black ball is from urn B is P(B/b) = \(7\over 12\) and

Probability that black ball is from urn C is P(C/b) = \(6\over 12\)

Total probability  = \(\frac{1}{3}\cdot \frac{5}{12}+\frac{1}{3}\cdot \frac{7}{12}+\frac{1}{3}\cdot \frac{6}{12}\)

Then required probability

P(b|A) \(=\frac{\frac{1}{3}\cdot \frac{5}{12}}{\frac{1}{3}\cdot \left[ \frac{5}{12}+\frac{7}{12}+ \frac{6}{12}\right]}=\frac{5}{18}\)

Option (2) is true.

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)

Calculation:

Let E₁ be the event that both A and B solve the problem.

∴ P(E₁) = \(\frac{1}{4}\times \frac{1}{3}\) = \(\frac{1}{12}\)

Let E₂ be the event that both A and B got the incorrect solution of the problem

∴ P(E₂) = \(\frac{2}{3}\times \frac{3}{4}\) = \(\frac{1}{2}\)

Let E be the event of getting the same answers.

⇒ P(E/E₁) = 1

Given probability of common error is \(\frac{1}{20}\)

P(E/E₂) = \(\frac{1}{20}\)

We need to find P(E₁/E) = \(\frac{P(E_1\cap E)}{P(E)}\)

⇒  \(\frac{P(E_1\cap E)}{P(E)}\) = \(\frac{P(E_1)\cdot P(E/E_1)}{P(E_1)\cdot P(E/E_1)+P(E_2)\cdot P(E/E_2)}\)

= \(\frac{\frac{1}{12}\cdot 1}{\frac{1}{12}\cdot 1+\frac{1}{2}\times\frac{1}{20}}\) 

= \(\frac{10}{13}\)

∴ The probability is 10/13.

The correct answer is option 4.

Concept:

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 \(\rm P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

Calculation:

Let E₁: He knows the answer

E2: He does not know the answer

X: He gets the correct answer.

Therefore, P (E1) = 90% = \(\frac {9}{10}\)

P (E2) = \(1- \frac {9}{10} = \frac {1}{10}\) 

P (X | E1)  = 1

P (X | E2)  = \(\frac 1 4\)

As we know that according to Bayes' theorem:

 \(\rm P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

\(\therefore \rm P{\rm{(}}{E_2}\;{\rm{|}}X) = \frac{{\frac{1}{10} \cdot \frac{1}{{4}}}}{{\left[ {\frac{9}{10} \cdot 1 + \;\frac{1}{10} \cdot \frac{1}{{4}}} \right]}} = \frac{{1}}{{37}}\;\)

Concept:

Let A₁, A₂, …., A_n be n mutually exclusive and exhaustive events of the sample space S and A is event which can occur with any of the events then

• \(P\left( \frac{{{A}_{i}}}{A} \right)=\frac{P\left( {{A}_{i}} \right)P\left( \frac{A}{{{A}_{i}}} \right)}{\mathop{\sum }_{i=1}^{n}P\left( {{A}_{i}} \right)P\left( \frac{A}{{{A}_{i}}} \right)}\)

Calculation:

Let A be the event that the man reports that number four is obtained.

Let E₁ be the event that four is obtained and E₂ be its complementary event.

Then, P (E₁) = Probability that four occurs = 1/6

P (E₂) = Probability that four does not occurs = 1 – P (E₁) = 1 − 1/6 = 5/6

Also, P (A|E₁) = Probability that man reports four and it is actually a four = 2/3

P (A|E₂) = Probability that man reports four and it is not a four = 1/3

By using Bayes’ theorem,

Probability that number obtained is actually a four.

\(P\left( {E1{\rm{|}}A} \right) = \frac{{P\left( {E1} \right)P(A|E1)}}{{P\left( {E1} \right)P\left( {A{\rm{|}}E1} \right) + P\left( {E2} \right)P(A|E2)}}\)

Concept:

Bayes' Theorem:

Let E₁, E₂, ….., E_n 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 E_n such that P(A) ≠ 0. Then

\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}}\)

i = 1, 2, ..., n

Calculation:

Let E₁ be the event of choosing bag A, E₂ be the event of choosing the bag B and X be the event of drawing a red ball.

⇒ P (E₁) = P (E₂) = 1/2

⇒ P (X | E₁) = P (Drawing a red ball from bag A) = 4/8 = 1/2

Similarly, P (X | E₂) = P (Drawing a red ball from bag B) = 2/8 = 1/4

Here, we have to find the probability of drawing a ball from bag A given that the ball is red in color i.e P (E₁ | X)

Applying Bayes Theorem, we get:

\(P{\rm{(}}{E_1}\;{\rm{|}}X) = \frac{{P\left( {{E_1}} \right) \cdot P\;\left( {X\;|\;{E_1}} \right)}}{{\left[ {P\left( {{E_1}} \right) \cdot P\;\left( {X\;|\;{E_1}} \right) + \;P\left( {{E_2}} \right) \cdot P\;\left( {X\;|\;{E_2}} \right)} \right]}}\;\)

\(P{\rm{(}}{E_1}\;{\rm{|}}X) = \frac{{\frac{1}{2} \cdot \frac{1}{{2}}}}{{\left[ {\frac{1}{2} \cdot \frac{1}{2} + \;\frac{1}{2} \cdot \frac{1}{{4}}} \right]}} = \frac{{2}}{{3}}\;\)

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,

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

Concept:

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

\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

Calculation:

Let the events B1, B2 and B3 be defined as:

B1 : The insured person is light vehicle driver

B2 : The insured person is medium vehicle driver

B3 : The insured person is heavy vehicle driver

Let the event E be 'The insured person meeting an accident'

⇒P (B₁) = 2000/12000 = 1/6, P  (B₂) = 4000/12000 = 1/3 and P (B₃) = 6000/12000 = 1/2

Similarly, P (E | B1) = 0.01, P (E | B2) = 0.03 and P (E | B3) = 0.15 ------(Given)

Here, we have to find the value of P (B₂ | E)

As we know that according to bayes' theorem: \(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

\( \Rightarrow P{\rm{(}}{B_2}\;{\rm{|}}E) = \frac{{P\left( {{B_2}} \right) \cdot P\;\left( {E\;|\;{B_2}} \right)}}{{\left[ {P\left( {{B_1}} \right) \cdot P\;\left( {E\;|\;{B_1}} \right) + \;P\left( {{B_2}} \right) \cdot P\;\left( {E\;|\;{B_2}} \right) + P\left( {{B_3}} \right) \cdot P\;\left( {E\;|\;{B_3}} \right)} \right]}}\;\)

\( \Rightarrow P{\rm{(}}{B_2}\;{\rm{|}}E) = \frac{3}{{26}}\)

Concept:

E = the event a man reports that it is even

E1 = the event of an even number is picked

E2 = the event of an odd number is picked

To find : The probability that it is actually even i.e. P(E1|E)

P(E1|E) = \(\rm \dfrac {P(E|E_1).P(E_1)}{P(E|E_1).P(E_1)+P(E|E_2).P(E_2)}\)

Calculations:

Given, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even.

E₁ = the event of an even number is picked

E₂ = the event of an odd number is picked

⇒P(E1) = \(\dfrac 3 7\)

⇒P(E2) = \(\dfrac 4 7\)

E = the event a man reports that it is even

 A man speaks truth 2 out of 3 times.

⇒P(E|E₁) = \(\dfrac 2 3\)

⇒P(E|E2) = \(\dfrac 1 3\)

To find : The probability that it is actually even i.e. P(E₁|E)

P(E1|E) = \(\rm \dfrac {P(E|E_1).P(E_1)}{P(E|E_1).P(E_1)+P(E|E_2).P(E_2)}\)

⇒P(E1|E) = \(\rm \dfrac {(\dfrac 2 3).(\dfrac 3 7)}{(\dfrac 2 3).(\dfrac 3 7)+(\dfrac 1 3).(\dfrac 4 7)}\)

⇒P(E1|E) = \(\rm \dfrac 6 {10}\)

⇒⇒P(E1|E) = \(\rm \dfrac 3 5\)

Hence, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. The probability that it is actually even is \(\rm \dfrac 3 5\)