Baye's Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Baye's Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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 that occurs together with any one of E₁ or E₂ or … or E_n 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 be the event of choosing the bag A, E₂ be the event of choosing the bag B and X be the event of drawing defective nut.

Therefore, P (E₁) = P (E₂) = \(\frac 1 2\) 

Given:

P (Drawing defective nut from bag A) = P (X | E₁)  = \(\frac 1 5\)

P (Drawing defective nut from bag B) = P (X | E₂)  = \(\frac 1 7\)

Here, we have to find the probability of drawing defective nut from bag B i.e P (E₂ | X)

As we know that according to Bayes' theorem:

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

∴ \( \rm P{\rm{(}}{E_2}\;{\rm{|}}X) = \frac{{\frac{1}{2} \cdot \frac{1}{{7}}}}{{\left[ {\frac{1}{2} \cdot \frac{1}{5} + \;\frac{1}{2} \cdot \frac{1}{{7}}} \right]}} = \frac{{5}}{{12}}\;\)

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

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

Calculation:

Let E₁ be the event of choosing the 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) = 4/9

Similarly, P (X | E₂) = P (Drawing a red ball from bag B) = 6/13

Here , we have to find probability of drawing a ball from bag B given that ball is red in colour i.e P (E₂ | X)

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

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

\( \Rightarrow P{\rm{(}}{E_2}\;{\rm{|}}X) = \frac{{\frac{1}{2} \cdot \frac{6}{{13}}}}{{\left[ {\frac{1}{2} \cdot \frac{4}{9} + \;\frac{1}{2} \cdot \frac{6}{{13}}} \right]}} = \frac{{27}}{{53}}\;\)

Let us define the events

A: Missing cared is black

B: Missing card is red

E: The first 13 cards are red.

\(P\left( A \right) = P\left( B \right) = \frac{1}{2}\)

\(P\left( {E/A} \right)\; = \frac{{{{26}_C}_{13}}}{{{{51}_C}_{13}}}\)

\(P\left( {E/B} \right) = \frac{{{{25}_C}_{13}}}{{{{51}_C}_{13}}}\)

The probability that the missing cards is black = P(A/E)\(P\left[ {A/E} \right] = \frac{{P\left[ A \right] \times P\left[ {E/A} \right]}}{{P\left[ A \right]P\left[ {E/A} \right] + P\left[ B \right]P\left[ {E/B} \right]}}\)

\(= \frac{{\frac{1}{2}{ \times ^{26}}{C_{13}}}}{{\frac{1}{2}\left[ {{{26}_{{C_{13}}}} + {{25}_{{C_{13}}}}} \right]}}\)

\(= \frac{{\frac{{26!}}{{13!13!}}}}{{\frac{{26!}}{{13!13!}} + \frac{{25!}}{{13!12!}}}}\)

Concept:

Consider, 

F : The event that the cake fails to raise.

A : The event that Cook X  bake a special kind of cake

B: The event that Cook Y  bake a special kind of cake

C : The event that Cook Z  bake a special kind of cake

The proportion of failures caused by X is given by P(A|F).

P(A|F) = \(\rm \dfrac {P(F|A)P(A)}{P(F|A)P(A)+P(F|B)P(B)+P(F|C)P(C)}\)

Calculations:

Consider, 

F : The event that the cake fails to raise.

A : The event that Cook X  bake a special kind of cake

B: The event that Cook Y  bake a special kind of cake

C : The event that Cook Z  bake a special kind of cake

Three cooks X, Y and Z bake a special kind of cake, and with respective probabilities 0.02, 0.03 and 0.05, it fails to rise.

P (F|A) = 0.02

P (F|B) = 0.03

P (F|C) = 0.05

In the restaurant where they work, X bakes 50%, Y bakes 30% and Z bakes 20% of cakes. 

P(A) = 0.5

P(B) = 0.3

P(C) = 0.2

To find : The proportion of failures caused by X is given by P(A|F).

P(A|F) = \(\rm \dfrac {P(F|A)P(A)}{P(F|A)P(A)+P(F|B)P(B)+P(F|C)P(C)}\)

Substitute the values, we get

P(A|F) = \(\rm \dfrac {(0.02)(0.5)}{(0.02)(0.5)+(0.03)(0.3)+(0.05)(0.2)}\)

P(A|F) = 34.48 = \(\dfrac{10}{29}\)

Hence, three cooks X, Y and Z bake a special kind of cake, and with respective probabilities 0.02, 0.03 and 0.05, it fails to rise. In the restaurant where they work, X bakes 50%, Y bakes 30% and Z bakes 20% of cakes. The proportion of failures caused by X is \(\dfrac{10}{29}\)

Concept:

Let E1, E2,…, En be a set of events associated with a sample space S, where all the events E1, E2,…, En 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(Ei/A) = \({P(E_i)P(A|E_i)\over\sum_{k=1}^nP(E_k)P(A|E_k)}\)

Calculation

P(4W4B/2W2B) = \(\begin{aligned} & \frac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)} \\ & +\ldots \ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B) \end{aligned}\)

= \(\frac{\frac{1}{5}\times\frac{^4\mathbf{C}_2\times^4\mathbf{C}_2}{^8\mathbf{C}_4}}{\frac{1}{5}\times\frac{^2\mathbf{C}_2\times^6\mathbf{C}_2}{^8\mathbf{C}_4}+\frac{1}{5}\times\frac{^3\mathbf{C}_2\times^5\mathbf{C}_2}{^8\mathbf{C}_4}+...+\frac{1}{5}\times\frac{^6\mathbf{C}_2\times^2\mathbf{C}_2}{^8\mathbf{C}_4}}\)

= \(\frac{2}{7}\)

∴ Option 2 is correct

Let E₁, E₂, A be the events.

E₁ be the selecting a bag from the first group

E₂ be the selecting a bag from the second group.

And A be the ball drawn is white.

Since there are 5 bags out of which 3 bags belong to the first group and 2 bags to the second group.

\(P\left( {{E_1}} \right) = \frac{3}{5},\:P\left( {{E_2}} \right) = \frac{2}{5}\)

f E₁ has already occurred, then a bag containing from the first group is chosen. The bag chosen contains 5 white balls and 3 black balls. Therefore the probability of drawing white balls from it is .

So \(P\left( {A/{E_1}} \right) = \frac{5}{8}\)

Similarly \(P\left( {A/{E_2}} \right) = \frac{2}{6} = \frac{1}{3}\)

We have to find \(P\left( {{E_1}/A} \right)\) i.e given that the ball is drawn is white, the probability that it is drawn from a bag of the first group

By Bayes rule,

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

\(= \frac{{\frac{3}{5} \times \frac{5}{8}}}{{\frac{3}{5} \times \frac{5}{8} + \frac{2}{5} \times \frac{1}{3}}}\)

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 X be the event that 'Prabhu reports that six occurs in throwing of the die and let E₁ be the event that six occurs and E₂ be the event that six does not occur.

⇒ P (E₁) = 1/6 and P (E₂) = 1 - P (E1) = 5/6

P (X | E₁) = Probability 'Prabhu speaks truth' = 3/4

P (X | E₂) = Probability 'Prabhu does not speaks truth' = 1 - P (X | E1) = 1/4

Here, we have to find the value of P (E₁ | X).

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

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

\( \Rightarrow P{\rm{(}}{E_1}\;{\rm{|}}X) = \frac{{\frac{1}{6} \cdot \frac{3}{{4}}}}{{\left[ {\frac{1}{6} \cdot \frac{3}{4} + \;\frac{5}{6} \cdot \frac{1}{{4}}} \right]}} = \frac{{3}}{{8}}\;\)

Concept:

• Bayes’ theorem is a way of finding a probability when we know certain other probabilities.

Formula: \({\rm{P}}\left( {{\rm{A}}|{\rm{B}}} \right) = \frac{{{\rm{P}}\left( {\rm{A}} \right){\rm{\;P}}\left( {{\rm{B}}|{\rm{A}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

• And = Multiply (A and B = AB)

Or = Addition (A or B = A + B)

Calculation:

Let A be the event of choosing the bag I, B the event of choosing the bag II and C be the event of drawing a red ball.

Then, P(A) = P(B) = 1/2

Total balls in bag I = 5 + 7 = 12

Also, P(C/A) = P (drawing a red ball from Bag I) = 5/12

Total balls in bag II = 6 + 8 = 14

P(C/B) = P (drawing a red ball from Bag II) = 6/14 = 3/7

To find the probability of drawing a red ball from bag I out of two bags,

Use Bayes’ formula,

P(A|C) = \(\frac{{{\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {{\rm{C}}|{\rm{A}}} \right)}}{{{\rm{P}}\left( {\rm{A}} \right){\rm{P}}({\rm{C}}|{\rm{A}}) + {\rm{P}}\left( {\rm{B}} \right){\rm{P}}({\rm{C}}|{\rm{B}})}} = \frac{{\frac{1}{2} \times \frac{5}{{12}}}}{{\frac{1}{2} \times \frac{5}{{12}} + \frac{3}{7} \times \frac{1}{2}}} = \frac{{\frac{5}{{24}}}}{{\frac{5}{{24}} + \frac{3}{14}}} = \frac{5}{{24}} \times \frac{{336}}{{142}} = \frac{35}{{71}}\)

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 : Children who were found coronavirus positive in the last week

B2 : Adults who were found coronavirus positive in the last week

B3 : Senior citizens who were found coronavirus positive in the last week

Let the event E be 'The person tested coronavirus positive died'

⇒P (B₁) = 200/1200 = 1/6, P  (B₂) = 400/1200 = 1/3 and P (B₃) = 600/1200 = 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 (B₃ | E)

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

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

\( \Rightarrow P{\rm{(}}{B_3}\;{\rm{|}}E) = \frac{45}{{52}}\)

Given :

P(A1) = P(A2) = 0.4, P(A3) = 0.2

P(B/A1) = 0.25, P(B/A2) = 0.4

P(B/A3) = 0.125

Formula used :

Bayes' theorem: Let A be an event which occurs with E1 or E2 or ....En then according to Bayes theorem

 \(P(\frac{E_i}{A})= \frac{P(A|E_i)\ P(E_i)}{\sum _{k = 1}^nP(E_k)\ P(A|E_k)}\) , i = 1, 2, 3, ....n

Where, E₁, E₂, ....E_n  be a set of events associated with a sample space S, where all events E1, E2, ....En have a non-zero probability of occurrence.

Calculations :

Using the equation Bayes theorem, we get

⇒ \(P(\frac{A_1}{B}) = \frac{P(A_1).P(\frac{B}{A_1})}{P(A_1)P(\frac{B}{A_1})+P(A_2)P(\frac{B}{A_2})+P(A_3)P(\frac{B}{A_3})}\)

⇒ \(\frac{0.1}{0.1+0.16+0.025}\)  = \(\frac{20}{57}\)

∴ The probability of A1 after observing B is 20/57.