Conditional Probability MCQ Quiz in मल्याळम - Objective Question with Answer for Conditional Probability - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

There are three rooms, I, II, III

The room I → 2 gift box and 3 empty box

Room II → 3 gift box and 2 empty box

Room III → 4 gift box and 1 empty box

Probability of choosing either, Room I or Room II or Room III = \(\frac{1}{3}\)

Mr Jhon select one box from the chosen room, then the probability that Mr John wins a box having gift items = Chooses room I and selects gift box +  Chooses room II and selects gift box + Chooses room III and selects gift box

Required probability:

\(P=\frac{1}{3}\times \frac{2}{5}+\frac{1}{3}\times \frac{3}{5}+\frac{1}{3}\times \frac{4}{5}=\frac{1}{3}\times\frac{9}{5}=\frac{3}{5}\)

Concept:

Independent Events

Let E and F be two events associated with a sample space s. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we say that the two events are independent. thus, two events E and F will be independent, if 

     (a) P(F | E)= P(F) , provided P(E) ≠ 0

     (b) P(E | F) =P(E) , provided P(F) ≠ 0

Using the multiplication theorem on probability, we have 

     (c) P(E ⋂ F) = P(E) P(F)

There events A, B and C are said to be mutually independent if all the following condition hold:

• P(A ⋂ B) = P(A) P(B)
• P(A ⋂ C) = P(A) P(C)
• P(B ⋂ C) = P(B) P(C)
• P(A ⋂ B ⋂ C) = P(A) P(B) P(C)

Calculation:

Given:

P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\),

and \(P(\overline{A \cup B}) = \dfrac{1}{6}\)

⇒ 1 - P(A ∪ B) = 1/6 {Such that p(A) + p(\(\bar{A}\)) = 1)}

⇒ 1 - P(A) - P(B) + P(A ∩ B) = 1/6

Such that P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(\(\bar{A}\)) - P(B) + 1/4 =1/6

⇒ P(B) = 1/4 + 1/4 - 1/6

⇒ P(B) = 1/3 and P(A) = 3/4 

Now  P(A ∩ B) = 1/4 = 3/4 × 1/3 = P(A) P(B)

Hence, events A and B are independent events but not equally likely.

Data:  

\(p\left( A \right) = \frac{3}{5}\)

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

Formula

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

Calculation:

\(\frac{2}{3} = \;\frac{{P\left( {A \cap B} \right)}}{{\frac{3}{5}}}\)  , 

\(P\left( {A \cap B} \right) = \;\frac{2}{{5}}\)

Also, \(P\left( {\frac{A}{B}} \right) = \frac{{A\;\left( {A \cap B} \right)}}{{P\left( B \right)}},\;\;\)

\(\frac{4}{5} = \frac{{\frac{2}{{5}}}}{{P\left( B \right)}}\) ,

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

Required probability, \(P\left( {\frac{{A'}}{{B'}}} \right) = \frac{{P\left( {A' \cap B'} \right)}}{{P\left( {B'} \right)}}\)

\(= \frac{{P{{\left( {A \cup B} \right)}'}}}{{1 - P\left( B \right)}}\; = \frac{{1 - P\left( {A \cup B} \right)}}{{1 - P\left( B \right)}}\)

\(= \frac{{1 - \left( {P\left( A \right) + P\left( B \right) - P\;\left( {A \cap B} \right)} \right)}}{{1 - P\left( B \right)}}\;\)

\(= \frac{{1 - \left( {\frac{3}{5} + \frac{1}{2} - \frac{2}{{5}}} \right)}}{{1 - \frac{1}{2}}}\)

Explanation:

P (A⋃B) = P (A) + P (B) – P (A⋂B)

\(\begin{array}{l} \therefore \frac{5}{6} = P\left( A \right) + \frac{1}{2}-\frac{1}{3}\\ \Rightarrow P\left( A \right) = \frac{5}{6}-\frac{1}{2} + \frac{1}{3}\\ = \frac{{5-3 + 2}}{6} \end{array}\)

= 4/6

= 2/3

We have, P(A).P(B) = P(A⋂B), for independent events.

P(A).P(B) = (2/3) × (1/2) = 1/3

This is equal to P(A⋂B).

Thus events A and B are independent events.

The correct answer is option 4.

Concept:

The given data,

A box contains five balls of the same size and shape. Three of them are green colored balls and two of them are orange-colored balls

G = green

O= Orange

If a green ball is drawn, it is not replaced. If an orange ball is drawn, it is replaced with another orange ball. The first ball is drawn,

The event is,

P(E)= \({3 \over 5} \times {2 \over 4} \times {2 \over 5} \times {2 \over 5}\)

P(E)=\({3 \over 10}+ {4 \over 25}\)

P(E)= \({23 \over 50}\)

Hence the correct answer is 23/50.

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 = \(\frac{6}{{11}}\)

Explanation:

P (A⋃B) = P (A) + P (B) – P (A⋂B)

∴ \(\frac{5}{6} = P\left( A \right) + \frac{1}{2}-\frac{1}{3}\)

⇒ \(P\left( A \right) = \frac{5}{6}-\frac{1}{2} + \frac{1}{3}\)

= \( \frac{{5-3 + 2}}{6}\)

= 4/6

= 2/3

We have, P(A).P(B) = P(A⋂B), for independent events.

P(A).P(B) = (2/3) × (1/2) = 1/3

This is equal to P(A⋂B).

Thus events A and B are independent events.

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

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, \(P\left( {\frac{{P'}}{{Q'}}} \right) = \frac{{P\left( {P' \cap Q'} \right)}}{{P\left( {Q'} \right)}}\)

\(= \frac{{P{{\left( {P \cup Q} \right)}'}}}{{1 - P\left( Q \right)}}\; = \frac{{1 - P\left( {P \cup Q} \right)}}{{1 - P\left( Q \right)}}\)

\(= \frac{{1 - \left( {P\left( P \right) + P\left( Q \right) - P\;\left( {P \cap Q} \right)} \right)}}{{1 - P\left( Q \right)}}\;\)

\(= \frac{{1 - \left( {\frac{1}{4} + \frac{1}{6} - \frac{1}{{12}}} \right)}}{{1 - \frac{1}{6}}}\)

P(A∪B) = P(A) + P(B) – P(A∩B)

Given the two events are independent, so:

P(A∩B) = P(A).P(B)

Let the probability of event A happening be x. The probability of event B happening will also be x. (Given, the two events are equally likely)

P(A∪B) = x + x – x.x

∴ 2x – x² = 0.75

⇒ x² – 2x + 0.75  = 0

10x² – 20x + 7.5 = 0

10x² – 15x – 5x + 7.5 = 0

(5x – 2.5) (2x - 3) = 0

\(\Rightarrow x = \frac{1}{2}~or~\frac{3}{2}\)

As x can not be greater than 1, hence x = 0.5