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

Last updated on Mar 11, 2025

നേടുക Baye's Theorem ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Baye's Theorem MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Baye's Theorem MCQ Objective Questions

Top Baye's Theorem MCQ Objective Questions

Baye's Theorem Question 1:

The chances of defective nuts in two boxes A and B are \(\rm \frac 1 5, \frac 1 7\) respectively. A box is selected at random and a nut drawn from it at random is found to be defective. The probability that it came from the box 'B' is

  1. \(\frac {3}{ 12}\)
  2. \(\frac {7}{ 12}\)
  3. \(\frac {5}{ 12}\)
  4. \(\frac {1}{ 12}\)
  5. \(\frac {11}{ 12}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac {5}{ 12}\)

Baye's Theorem Question 1 Detailed Solution

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 that 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 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:

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

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

Baye's Theorem Question 2:

Suppose bag A contains 4 red and 5 black balls while another bag B contains 6 red and 7 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it is drawn from bag B ?

  1. 27/53
  2. 21/53
  3. 23/53
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 27/53

Baye's Theorem Question 2 Detailed Solution

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

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

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

Baye's Theorem Question 3:

A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the probability that the missing cards is black, is

  1. \(\frac{1}{3}\)
  2. \(\frac{2}{3}\)
  3. \(\frac{1}{2}\)
  4. \(\frac{{{{25}_{{C_{13}}}}}}{{{{51}_{{C_{13}}}}}}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{2}{3}\)

Baye's Theorem Question 3 Detailed Solution

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)\(P\left[ {A/E} \right] = \frac{{P\left[ A \right] \times P\left[ {E/A} \right]}}{{P\left[ A \right]P\left[ {E/A} \right] + P\left[ B \right]P\left[ {E/B} \right]}}\)

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

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

\(= \frac{{26}}{{26 + 13}} = \frac{2}{3}\)

Baye's Theorem Question 4:

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. What is the proportion of failures caused by X?

  1. \(\dfrac{9}{29}\)
  2. \(\dfrac{10}{29}\)
  3. \(\dfrac{19}{29}\)
  4. \(\dfrac{28}{29}\)

Answer (Detailed Solution Below)

Option 2 : \(\dfrac{10}{29}\)

Baye's Theorem Question 4 Detailed Solution

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 \(\dfrac{10}{29}\)

Baye's Theorem Question 5:

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

  1. \(\rm \frac{2}{5}\)
  2. \(\rm \frac{2}{7}\)
  3. \(\rm \frac{1}{7}\)
  4. \(\rm \frac{1}{5}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{2}{7}\)

Baye's Theorem Question 5 Detailed Solution

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

Baye's Theorem Question 6:

There are 3 bags each containing 5 white balls and 3 black balls. Also there are 2 bags each containing 2 white balls and 4 black balls. A white ball is drawn at random. The probability that this white ball is from a bag of the first group.

  1. \(\frac{{45}}{{16}}\)
  2. \(\frac{{15}}{{40}}\)
  3. \(\frac{{45}}{{61}}\)
  4. None of these

Answer (Detailed Solution Below)

Option 3 : \(\frac{{45}}{{61}}\)

Baye's Theorem Question 6 Detailed Solution

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,

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

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

\(= \frac{{\frac{8}{8}}}{{\frac{3}{8} + \frac{2}{{15}}}} = \frac{{\frac{3}{8}}}{{\frac{{45 + 16}}{{120}}}} = \frac{{\frac{3}{8}}}{{\frac{{61}}{{120}}}} = \frac{3}{8} \times \frac{{120}}{{61}} = \frac{{45}}{{61}}\)

Baye's Theorem Question 7:

Prabhu is known to speak truth 3 out of 4 times. he throws a die and reports that it is a six. Find the probability that it is actually a six?

  1. 5/8
  2. 7/8
  3. 3/8
  4. 1/8
  5. 1/4

Answer (Detailed Solution Below)

Option 3 : 3/8

Baye's Theorem Question 7 Detailed Solution

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

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

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

Baye's Theorem Question 8:

Bag I contains 5 red and 7 green balls while another Bag II contains 6 red and 8 green balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag I.

  1. 5/12
  2. 5/24
  3. 35/71
  4. 5/8
  5. None of these

Answer (Detailed Solution Below)

Option 3 : 35/71

Baye's Theorem Question 8 Detailed Solution

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

Hence, option (3) is correct.

Baye's Theorem Question 9:

In the last week due to coronavirus outbreak, 200 children were tested positive, 400 adults were tested positive and 600 senior citizens were tested positive. The probability of them dying are 0.01, 0.03 and 0.15 respectively. If one of the person tested positive dies due to coronavirus what is the probability that he/she was a senior citizen.

  1. 41/52
  2. 43/52
  3. 45/52
  4. 47/52

Answer (Detailed Solution Below)

Option 3 : 45/52

Baye's Theorem Question 9 Detailed Solution

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

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

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

Baye's Theorem Question 10:

An observed event B can occur after one of the three events A1, A2, A3. If

P(A1) = P(A2) = 0.4, P(A3) = 0.2 and P(B/A1) = 0.25, P(B/A2) = 0.4

P(B/A3) = 0.125, What is the probability of A1 after observing B?

  1. \(\frac{1}{3}\)
  2. \(\frac{6}{19}\)
  3. \(\frac{20}{57}\)
  4. \(\frac{2}{5}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{20}{57}\)

Baye's Theorem Question 10 Detailed Solution

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 ....Ethen 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, ....En  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.

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