Conditional Probability MCQ Quiz - Objective Question with Answer for Conditional Probability - Download Free PDF

Calculation:

Given,

\( P(A) = \frac{1}{3} \)

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

\( P(A \cap B) = \frac{1}{4} \)

We need to calculate\( P(A^C \cap B^C) \), the probability that neither A  nor B  occurs.

Using the complement rule:

\( P(A^C \cap B^C) = 1 - P(A \cup B) \)

Now, apply the inclusion-exclusion formula to find \(( P(A \cup B) \):

\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

Substitute the given values:

\( P(A \cup B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{4} \)

To simplify:

\( P(A \cup B) = \frac{4}{12} + \frac{6}{12} - \frac{3}{12} = \frac{7}{12} \)

Now, calculate \(P(A^C \cap B^C) \):

\( P(A^C \cap B^C) = 1 - \frac{7}{12} = \frac{5}{12} \)

Hence, the correct answer is Option 2.

Calculation:

Given,

The probability that A , B , and C  become managers are:

\( P(A) = \frac{3}{10}, \, P(B) = \frac{1}{2}, \, P(C) = \frac{4}{5} \)

The conditional probabilities that the bonus scheme is introduced are:

\( P(D|A) = \frac{4}{9}, \, P(D|B) = \frac{2}{9}, \, P(D|C) = \frac{1}{3} \)

We need to find the probability that B  is the manager given that the bonus scheme has been introduced using Bayes' Theorem:

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

First, calculate the total probability P(D)  that the bonus scheme is introduced:

\( P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \)

Substitute the values:

\( P(D) = \left( \frac{4}{9} \times \frac{3}{10} \right) + \left( \frac{2}{9} \times \frac{1}{2} \right) + \left( \frac{1}{3} \times \frac{4}{5} \right) \)

Simplifying:

\( P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15} \)

Find a common denominator (LCD = 90):

\( P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} = \frac{46}{90} = \frac{23}{45} \)

Now, use Bayes' Theorem to find P(B|D) :

\( P(B|D) = \frac{P(D|B)P(B)}{P(D)} = \frac{\left( \frac{2}{9} \times \frac{1}{2} \right)}{\frac{23}{45}} \)

Simplifying:

\( P(B|D) = \frac{\frac{2}{18}}{\frac{23}{45}} = \frac{\frac{1}{9}}{\frac{23}{45}} = \frac{5}{23} \)

The probability that the manager appointed was B , given that the bonus scheme has been introduced, is 5/23.

Hence, the correct answer is Option 1.

Calculation:

Given,

The probability that A , B , and C  become managers are:

\( P(A) = \frac{3}{10}, \, P(B) = \frac{1}{2}, \, P(C) = \frac{4}{5} \)

The conditional probabilities that the bonus scheme is introduced are:

\( P(D|A) = \frac{4}{9}, \, P(D|B) = \frac{2}{9}, \, P(D|C) = \frac{1}{3} \)

The total probability that the bonus scheme will be introduced is given by:

\( P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \)

Substituting the values:

\( P(D) = \left(\frac{4}{9} \times \frac{3}{10}\right) + \left(\frac{2}{9} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{4}{5}\right) \)

Now, simplifying the terms:

\( P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15} \)

The least common denominator (LCD) is 90. So:

\( P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} \)

Adding the fractions gives:

\( P(D) = \frac{46}{90} = \frac{23}{45} \)

Hence, the correct answer is Option 3.

Calculation: 

E₁ : Smokers 

⇒ \(\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{4}\)

E₂ : non-smokers 

⇒ \(\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{3}{4}\)

E : diagnosed with lung cancer 

⇒ \(\mathrm{P}\left(\mathrm{E} / \mathrm{E}_{1}\right)=\frac{27}{28} \)

⇒ \(\mathrm{P}\left(\mathrm{E} / \mathrm{E}_{2}\right)=\frac{1}{28}\)

⇒ \(\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_{1}\right)}{\mathrm{P}(\mathrm{E})} \)

\(=\frac{\frac{1}{4} \times \frac{27}{28}}{\frac{1}{4} \times \frac{27}{28}+\frac{3}{4} \times \frac{1}{28}}=\frac{27^{9}}{30_{10}}=\frac{9}{10}\)

K = 9

Hence, the correct answer is 9. 

Explanation -

x + y = 24, x, y ∈ N 

AM > GM

⇒ xy ≤ 144

xy ≤ 108 

Favorable pairs of (x, y) are

(13, 11), (12, 12), (14, 10), (15, 9), (16, 8),

(17, 7), (18, 6), (6, 18), (7, 17), (8, 16), (9, 15),

(10, 14), (11, 13)

i.e. 13 cases 

Total choices for x + y = 24 is 23

Probability = \(\frac{13}{23}\) = \(\frac{\mathrm{m}}{\mathrm{n}}\)

n – m = 10  

Hence Option (4) is correct.

Concept:

• \(\rm P(A|B) = \frac {P(A \;∩ \; B)}{P(B)}\)
• \(\rm P(B|A) = \frac {P(A \;∩ \; B)}{P(A)}\)
• A ⊂ B = Proper Subset: every element of A is in B, but B has more elements.
• ϕ = Empty set = {}

Calculation:

Given: P(B/A) = 1

⇒ \(\rm P(B|A) = \frac {P(A \;∩ \; B)}{P(A)} = 1\)

⇒ P(A ∩ B) = P(A)

⇒ (A ∩ B) = A

So, every element of A is in B, but B has more elements.

∴ A ⊂ B

Concept:

Let A and B be any two events associated with a random experiment. Then, the probability of occurrence of an event A under the condition that B has already occurred such that P(B) ≠ 0, is called the conditional probability and denoted by P(A | B)

i.e \(P\;\left( {A|\;B} \right) = \frac{{P\;\left( {A\; ∩ B} \right)}}{{P\left( B \right)}}\)

Similarly, \(P\left( {B\;|\;A} \right) = \frac{{P\left( {A\; ∩ B} \right)}}{{P\left( A \right)}},\;where\;P\left( A \right) \ne 0\)

Calculation:

Let b stand for boy and g for girl.

The sample space S = {(b, b), (b, g), (g, g), (g, b)}

Let A = Both the children are boys

Let B = At least one of the child is boy.

i.e A = {(b, b)}, B = {(b, b), (b, g), (g, b)} and A ∩ B = {(b, b)}

⇒ P (B) = 3/4 and P (A ∩ B) = 1/4

As we know that, \(P\;\left( {A|\;B} \right) = \frac{{P\;\left( {A\; ∩ B} \right)}}{{P\left( B \right)}}\)

⇒ P (A | B) = 1/3

Explanation

If the event B occurs does not change the probability that event A occurs, and event A and B are the independent event then

⇒ P(A I B) = P(A)

According to bayes’s theorem It states the relation between the probabilities of A and B, P(A) and P(B) and the conditional probabilities of A gives B and B gives A.

⇒ P(A I B) and P(B I A)

⇒ P(A I B) = P(B I A).P(A)/P(B)

⇒ P(B I A) = P(A ∩ B)/P(A)

⇒ P(A ∩ B) = P(A).P(B)

⇒ P(A I B) = P(A ∩ B).P(A)/P(A). P(B)

⇒ P(A I B) = P(A). P(B).P(A)/P(A).P(B)

⇒ P(A I B) = P(A)

Concept:

For two events A and B:

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• The conditional probability of B given A is defined as: 

              P(B|A) = \(\rm \dfrac{P(A\cap B)}{P(A)}\), when P(A) > 0

Calculation:

Using the relation P(B|A) = \(\rm \dfrac{P(A\cap B)}{P(A)}\), we get:

0.6 = \(\rm \dfrac{P(A\cap B)}{0.4}\)

⇒ P(A ∩ B) = 0.24

Now using the relation P(A ∪ B) = P(A) + P(B) - P(A ∩ B), we get:

P(A ∪ B) = 0.4 + 0.8 - 0.24 = 0.96.

Concept:

\(\rm P(A|B)=\dfrac{P(A∩ B)}{P(B)}\)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given P(A) = L and P(B) = M

Here P(A ∪ B) ≤ 1 (∵ Max value of probability is 1)

P(A) + P(B) - P(A ∩ B) ≤ 1

L + M - P(A ∩ B) ≤ 1

P(A ∩ B) ≥ L + M - 1

Now \(\rm P(A|B)=\dfrac{P(A∩ B)}{P(B)}\)

\(\rm P(A|B)\geq\dfrac{L+M-1}{M}\)

Given:

Urn A consists 3 blue and 4 green balls 

Urn B consists 5 blue and 6 green balls

One ball drawn at random from one of the urns was blue

Concept:

Baye's Theorem :

Solution:

Let B be the event that the ball drawn is blue and E₁ be the event that the ball is drawn from urn 1 and E₂ be the event that the ball is drawn from urn 2.

∴ P(B) = P(B ∩ E1) + P(B ∩ E₂)

⇒ P(B) = \(\frac{1}{2}\times\frac{3}{7} + \frac{1}{2}\times\frac{5}{11} \)

= 34/77

∴ P(E₂/B) = \(\frac{P(E_1 \cap B)}{P(B)} = \frac{P(E_2) P(B/E_2))}{P(B)}\)

= \(\frac{\frac{1}{2}\times \frac{5}{11}}{\frac{34}{77}}\)

= 35/68

Concept:

The probability of the occurrence of an event A out of a total possible outcomes N, is given by: P(A) = \(\rm \frac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur.

Calculation:

When three dice are thrown simultaneously, there are 216 elements in the total sample space.
Let us define event A such that the sum of the numbers on the dice is 6.
A = (2,2,2), (2,1,3), (2,3,1), (1,3,2), (1,2,3), (3,1,2), (3,2,1), (1,1,4), (4,1,1), (1,4,1)

Let B be the event of getting a number 2 in all three dice ⇒ (2,2,2).
Since it is given that event A has already occurred, i.e. the sum of the numbers on each die is 6, we have 10 cases out of which only one (2,2,2) is favourable.

P(B) = \(\frac{1}{10}\).

Concept:

Consider two events A and B

\(\rm P(\frac{A}{B})=\frac{P(A∩ B)}{P(B)}\)

\(\rm P(A)=1-P(A')\)

\(\rm P(A'∩ B)=P(B)-P(A∩ B)\)

Calculation:

Here, \(\rm P(A')=\frac13, P(B')=\frac23\) and \(\rm P(A∩ B)=\frac15\)

\(\rm P(B)=1-P(B')=1-\frac23=\frac13\)  ------ (∵ P(A) = 1 - P(A'))

\(\rm P(\frac{\bar A}{B})=\frac{P(\bar A∩ B)}{P(B)}\)   ------ (∵ \(\rm P(\frac{A}{B})=\frac{P(A∩ B)}{P(B)}\))

\(\rm =\frac{P(B)-P(A∩ B)}{P(B)}\)   ------   (∵ P(A' ∩ B)= P(B) - P(A ∩ B)  )

\(=\rm \frac{\frac13-\frac15}{\frac13}\)

= 2 × 3/15

= 2/5

Hence, option (2) is correct.

Given:

P(not A) = \(\rm \frac{7}{10}\), and P(not B) = \(\rm \frac{3}{10}\)

P(A|B) = \(\rm \frac{3}{14}\)

Formula used:

• \(\rm P(B|A) = \frac{P(A\cap B)}{P(A)}\)
• \(\rm P(A|B) = \frac{P(A\cap B)}{P(B)}\)
• P(A) = 1 - P(not A)
• P(B) = 1 - P(not B)

Calculation:

We have,

\(\rm P(\bar{B}) = 0.3, P(\bar{A}) = 0.7\)

⇒ P(B) = 1 - 0.3 = 0.7 and

P(A) = 1 - 0.7 = 0.3

⇒ P(A) = 0.3      ----(1)

We know that,

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

⇒ \(\frac{3}{14} \) = \(\frac{P(A\cap B)}{0.7}\)       (∵ P(A|B) = \(\rm \frac{3}{14}\))

⇒ \(\rm P(A\cap B)\) = 0.15      ----(2)   

We know that,

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

⇒ P(B|A) = \(\rm \frac{0.15}{0.3}\)        [From equation (1)& (2)]

⇒ P(B|A) = 0.5 = \(\rm \frac{1}{2}\)

∴  P(B|A) is equal to \(\rm \frac{1}{2}\)

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

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