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

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.

Given:

A, B, and C are three mutually exclusive and exhaustive events such that if P(B) = \(\frac{3}{2}\)P(A) and P(C) = \(\frac{1}{2}\)P(B).

Concept:

If A, B, and C are three mutually exclusive and exhaustive events then,

P (A U B U C) = P(A) + P(B) + P(C) = 1

Solution:

According to the question,

A, B, and C are three mutually exclusive and exhaustive events such that if P(B) = \(\frac{3}{2}\)P(A) and P(C) = \(\frac{1}{2}\)P(B).

P (A U B U C) = P(A) + \(\frac{3}{2}P(A)\) + \(\frac{1}{2}P(B)\)

P (A U B U C) = P(A) + \(\frac{3}{2}P(A)\) + \(\frac{1}{2}\times \frac{3}{2}P(A)\)

P (A U B U C) = P(A) + \(\frac{3}{2}P(A)\) + \(\frac{3}{4}P(A)\)

P (A U B U C) =  \(\frac{13}{4}P(A)\)

Also,

P (A U B U C) = 1

\(\frac{13}{4}P(A)=1 \\ P(A)=\frac{4}{13}\)

Hence, option 4 is correct.

Concept:

Conditional Probability:

• Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A | B), which represents the probability of event A occurring given that event B has already occurred.
• The formula for conditional probability is given by:
  • P(A | B) = P(A ∩ B) / P(B)
• Where:
  • P(A ∩ B) is the probability of both events A and B occurring.
  • P(B) is the probability of event B occurring.
• In this problem, we are given the following events:
  • A: A 6 on the third throw.
  • B: A 4 on the first throw and 5 on the second throw.
• We need to find the probability of event A occurring given that event B has already occurred.

Calculation:

Step 1: Formula for Conditional Probability

We use the formula for conditional probability:

P(A | B) = P(A ∩ B) / P(B)

Step 2: Find P(B)

Event B occurs when:

• A 4 appears on the first throw.
• A 5 appears on the second throw.

The probability of each event happening is:

• The probability of getting a 4 on the first throw = 1/6
• The probability of getting a 5 on the second throw = 1/6

Since the throws are independent, the probability of event B is:

P(B) = (1/6) × (1/6) = 1/36

Step 3: Find P(A ∩ B)

Event A ∩ B occurs when:

• A 4 appears on the first throw (for event B).
• A 5 appears on the second throw (for event B).
• A 6 appears on the third throw (for event A).

The probability of each event happening is:

• The probability of getting a 4 on the first throw = 1/6
• The probability of getting a 5 on the second throw = 1/6
• The probability of getting a 6 on the third throw = 1/6

Since the events are independent, the probability of A ∩ B is:

P(A ∩ B) = (1/6) × (1/6) × (1/6) = 1/216

Step 4: Calculate P(A | B)

Now, we can calculate the conditional probability:

P(A | B) = P(A ∩ B) / P(B) = (1/216) / (1/36) = 1/6

Step 5: Conclusion

The probability of A given that B has already occurred is:

P(A | B) = 1/6

∴ The correct answer is Option (1): 1/6

Concept:

Probability of Union of Two Events:

• In probability theory, the probability of the union of two events A and B is given by the formula:
  • P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• Where:
  • P(A ∪ B) is the probability of either event A or event B occurring.
  • P(A) is the probability of event A occurring.
  • P(B) is the probability of event B occurring.
  • P(A ∩ B) is the probability of both events A and B occurring together (i.e., their intersection).
• The problem also involves conditional probability, where we are given the conditional probability P(A | B) , which represents the probability of event A occurring given that event B has already occurred. The formula for conditional probability is:
  • P(A | B) = P(A ∩ B) / P(B)

Calculation:

Given:

P(A) = 0.4

P(B) = 0.8

P(A | B) = 0.6

Step 1: Calculate P(A ∩ B) using the formula for conditional probability:

P(A | B) = P(A ∩ B) / P(B)

Rearrange the formula to solve for P(A ∩ B):

P(A ∩ B) = P(A | B) × P(B)

P(A ∩ B) = 0.6 × 0.8 = 0.48

Step 2: Use the formula for the probability of the union of two events:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

P(A ∪ B) = 0.4 + 0.8 - 0.48 = 1.2 - 0.48 = 0.72

∴ The probability P(A ∪ B) = 0.72.

The correct answer is Option (2): 0.72

Calculation

\(\mathrm{p}\left(\mathrm{~S}_{5}\right)= \frac{4}{36}=\frac{1}{9}\)

\(\mathrm{p}\left(\mathrm{~S}_{\mathrm{8}}\right)=\frac{5}{36}\)

Required probability = \(\frac{1}{9}+\frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9}+\left(\frac{8}{9} \cdot \frac{31}{36}\right)^{2} \cdot \frac{1}{9}+\ldots \infty\)

= \(\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19}\)

Hence option 2 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) = \(\frac{1}{10}\).

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 \(\rm \frac{1}{2}\)

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.

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

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