Probability of Complementary Event MCQ Quiz in मराठी - Objective Question with Answer for Probability of Complementary Event - मोफत PDF डाउनलोड करा

Concept:

For any two independent events A and B, if P(A) and P(B) are their probability of occurring then:

• P(A ∩ B) = P(A) × P(B)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• P(A') = 1 - P(A)
• P(B') = 1 - P(B)

Calculation:

Given P(A) = 0.6 and P(B) = 0.2

P(A ∩ B) = P(A) × P(B)

⇒ P(A ∩ B) = 0.6 × 0.2 = 0.12

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A ∪ B) = 0.6 + 0.2 - 0.12

⇒ P(A ∪ B) = 0.68

P((A ∪ B)') = 1 -  P(A ∪ B)

⇒ P((A ∪ B)') = 1 -  0.68

⇒ P((A ∪ B)') = 0.32

Concept:

Formulas Used:

• P(A̅) = 1 - P(A)
• P(A ∪ B) = P(A)+ P(B) - P(A ∩ B)

Calculation:

Given: \(P\;\left( {A ∪ B} \right) = \frac{3}{4},\;P\;\left( {A ∩ B} \right) = \frac{1}{4}\) and P(A̅) \(= \frac{2}{3}\)

To find: P(B)

We know, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(B) = P(A ∪ B) + P(A ∩ B) - P(A)

⇒ P(B) = P(A ∪ B) + P(A ∩ B) - (1 - P(A))

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

\(\Rightarrow P(B)=1-\frac{1}{3}\)

\(\Rightarrow P(B)=\frac{2}{3}\)

Concept:

Complement of an event:

The complement of an event is the subset of outcomes in the sample space that are not in the event.

The probability of the complement of an event is one minus the probability of the event.

P (A̅) = 1 - P (A) ⇒ P (A) = 1 - P (A̅)

Formula: P (A ∪ B) =  P (A) + P (B) - P (A ∩ B)

Calculation: 

Given: 

\(\rm P(A \cup B) = \frac {3}{4},\;P(A \cap B) = \frac {1}{4},\; P(̅{A}) = \frac 2 3\)

To find: P (B)

As we know, P (A ∪ B) =  P (A) + P (B) - P (A ∩ B)

⇒ P (A ∪ B) =  1 - P (A̅) + P (B) - P (A ∩ B)

\(\Rightarrow \frac{3}{4} = 1-\frac{2}{3} + \rm P(B) - \frac{1}{4}\\\Rightarrow \frac{3}{4} +\frac{1}{4} = \frac{1}{3} + \rm P (B) \\ \therefore \rm P (B) = 1 -\frac{1}{3} =\frac{2}{3} \)

Concept:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

The probability of the complement of an event is one minus the probability of the event. P (A̅) = 1 - P (A)

To determine the probability of two independent events we multiply the probability of the first event by the probability of the second event. P(A ∩ B) = P(A).P(B)

Two events are mutually exclusive when two events cannot happen at the same time. P(A ∩ B) = 0

In equally likely events, the probabilities of each event are equal.

Calculation:

Here, \(P(\overline{A\cup B}) = \dfrac{1}{6}, P(A∩ B) = \dfrac{1}{4} \ \text{and} \ P(̅{A}) = \dfrac{1}{4}\)

P(A) = 1 - P(A̅) = 1 - 1/4 = 3/4

\(\rm P(A\cup B)=1-P(\overline{A\cup B}) = 1-\dfrac{1}{6}=\frac 56\)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(B) = 5/6 - 3/4 + 1/4 = 5/6 - 1/2 

= 2/6

P(B) = 1/3

Here, P(A ∩ B) = P(A).P(B) and P(A) ≠ P(A) so the events are independent but not equally likely.

Hence, option (1) is correct.

Calculation:

Let's break down the given information and use the properties of probability to find the value of k.

We are given:
P(A̅) = 0.7
P(B̅) = k
P(A ∪ B) = 0.8

First, let's consider the complement of the union of A and B, which is (A ∪ B)'

Using the property

P((A ∪ B)' = 1 - P(A ∪ B),

we can write:

P((A ∪ B))' = 1 - P(A ∪ B)
P((A ∪ B))' = 1 - 0.8
P((A ∪ B))'  = 0.2

Next, we can express (A ∪ B)' in terms of A̅ and B̅ using the De Morgan's law:

(A ∪ B)' = A̅ ∩ B̅

Now, we can equate this expression with the probability P((A ∪ B)':

A̅ ∩ B̅ = 0.2

Since A and B are independent events, the probability of their intersection is the product of their individual probabilities:

P(A̅ ∩ B̅) = P(A̅) * P(B̅)

Substituting the given values:

0.2 = 0.7 * k

To solve for k, divide both sides of the equation by 0.7:

0.2 / 0.7 = k

k = 2/7

The correct answer is option (3)

Concept:

Formulas Used:

• P(A̅) = 1 - P(A)
• P(A ∪ B) = P(A)+ P(B) - P(A ∩ B)

Calculation:

Given: \(P\;\left( {A ∪ B} \right) = \frac{3}{4},\;P\;\left( {A ∩ B} \right) = \frac{1}{4}\) and P(A̅) \(= \frac{2}{3}\)

To find: P(B)

We know, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒ P(B) = P(A ∪ B) + P(A ∩ B) - P(A)

⇒ P(B) = P(A ∪ B) + P(A ∩ B) - (1 - P(A))

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

\(\Rightarrow P(B)=1-\frac{1}{3}\)

\(\Rightarrow P(B)=\frac{2}{3}\)

Concept:

If A is any event, then P(A) represents the probability of the occurrence of event A.

P(A') = 1 - P(A), where P(A') represents the probability of not occurrence of A.

Calculation:

If four coins are tossed, the total cases

= 2⁴

= 16

In a toss, there are two possibilities, T or H

So, the probability of getting head = 1/2

∴ The probability that at least one head turns up

= 1 - the probability that no head appears in four toss

= 1 - the probability of getting tail in four toss

= 1 - \(\left(\frac{1}{2}\right)^4\)

= 1- (1/16)

= 15/16

∴ The probability of getting at least one head is 15/16.

The correct answer is Option 3.

Concept:

Complement of an event:

The complement of an event is the subset of outcomes in the sample space that are not in the event.

The probability of the complement of an event is one minus the probability of the event.

P (A̅) = 1 - P (A) ⇒ P (A) = 1 - P (A̅)

Formula: P (A ∪ B) =  P (A) + P (B) - P (A ∩ B)

Calculation: 

Given: 

\(\rm P(A \cup B) = \frac {3}{4},\;P(A \cap B) = \frac {1}{4},\; P(̅{A}) = \frac 2 3\)

To find: P (B)

As we know, P (A ∪ B) =  P (A) + P (B) - P (A ∩ B)

⇒ P (A ∪ B) =  1 - P (A̅) + P (B) - P (A ∩ B)

\(\Rightarrow \frac{3}{4} = 1-\frac{2}{3} + \rm P(B) - \frac{1}{4}\\\Rightarrow \frac{3}{4} +\frac{1}{4} = \frac{1}{3} + \rm P (B) \\ \therefore \rm P (B) = 1 -\frac{1}{3} =\frac{2}{3} \)

Concept:

Events A and B are independent if P(A ∩ B) = P(A)P(B)

Calculation:

Given, \(P(A ∩ B)=\frac{1}{4}\), \(P(\overline{A})=\frac{1}{4}\)

Also, \(P(\overline{A ∪ B})=\frac{1}{6}\)

⇒ \(1-P(A)-P(B)+P(A ∩ B)=\frac{1}{6}\)

⇒ P(A̅) - P(B) + \(\frac{1}{4}\) = \(\frac{1}{6}\)

⇒ P(B) = \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\)

⇒ P(B) = \(\frac{1}{3}\)

Also, P(A) = 1 - P(A̅) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

Now, P(A ∩ B) = \(\frac{1}{4}\) = P(A)P(B)

∴ The events A and B are independent but not equally likely.

The correct answer is Option 3.

Concept:

Formulas used:

• P(A) = 1 – P(A̅)
• P(A̅ ∩ B) = P(A) – P(A ∩ B)

Calculation:

Let,

X: Event that X will qualify the examination.

Y: Event that Y will qualify the examination.

Given: P(X) = 0.05, P(Y) = 0.10, P(X ∩ Y) = 0.02

P (only one of them will qualify) = P(X and not Y) + P(Y and not X)

= P(X ∩ Y̅) + P(X̅ ∩ Y)

= P(X) – P(X ∩ Y) + P(Y) – P(X ∩ Y)

= P(X) + P(Y) – 2P(X ∩ Y)

= 0.05 + 0.10 – 2(0.02)

= 0.15 – 0.04

= 0.11