Events and Sample Spaces MCQ Quiz - Objective Question with Answer for Events and Sample Spaces - Download Free PDF

Concept:

• P(A') = 1 - P(A)
• P(A' ∩ B') = P((A ∪ B)') {Probability of not A and not B}
• P(A and not B) = P(A ∩ B') = P(A) - P(A ∩ B)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given:  P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.3, 

⇒  P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒  P(A ∪ B) = 0.5 + 0.7 - 0.3 = 0.9__(i)

Now the probability of not A and not B,

P(A' ∩ B') = P((A ∪ B)') = 1 - P(A ∪ B) 

⇒ P(A' ∩ B') = 1 - 0.9 { From (i)}

⇒ P(A' ∩ B') = 0.1 __(ii)

The probability of B and not A,

P(A' ∩ B) = P(B) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.7 - 0.3 = 0.4 __(iii)

The probability of A and not B,

P(A ∩ B') = P(A) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.5 - 0.3 = 0.2 __(iv)

From (ii), (iii) and (iv),

P(A' ∩ B') + P(A' ∩ B) + P(A ∩ B') = 0.1 + 0.4 + 0.2 = 0.7

∴ The correct option is (2).

Calculation:

Given, set {1, 2, 3, 4, 5}

Let A and B be two subsets.

For each x ∈ {1, 2, 3, 4, 5} , there are four possibilities:

• x ∈ A ∩ B
• x∈ A’ ∩ B
• x ∈ A ∩ B’
• x ∈ A’ ∩ B’

∴ The number of elements in sample space = 45

∴ Required probability

= (5C2 × 33)/45

= (10 × 27)/210

= 135/29 

∴ The required probability is 135/29.

The correct answer is Option 2.

Calculation:

Given, ax2 + bx + c = 0 

For real roots, D > 0 

⇒ b² > 4ac 

When b = 3 : (a, c) = (1, 1), (1, 2), ( 2,1) 

When b = 4 : (a, c) = (1, 1), (1, 2), (2,1), (1,3), (3,1) 

When b = 5 : (a, c) = (1,1), (1,2), (2,1), (1,3), (3, 1), (1,4), (4,1), (1,5), (5,1), (1,6), (6,1), (2,3), (3,2), (2,2) 

b = 6 : (a, c) = (1,1), (1,2), (2,1), (1,3), (3,1), (1,4), (4,1), (1,5), (5,1), (1,6), (6,1), (2,3), (3,2), (2,4), (4,2), (2,2) 

∴ Total favourable cases = 38 

∴ Required probability, p = $\frac{38}{6\times 6\times 6}$

⇒ 216p = 38

∴ The value of 216p is 38.

The correct answer is Option 2. 

Calculation:

Given, ax² + bx + c = 0, where a, b, c ∈ {1, 2, 3, 4, 5, 6,7, 8}   

Now, ax2 + bx + c = 0 have repeated roots

⇒ D = 0

⇒ b² – 4ac = 0

⇒ b² = 4ac

Possible cases are:  

(1, 2, 1) ; (2, 4, 2) ; (1, 4, 4) ; (4, 4, 1) ; (3, 6, 3) ; (2, 8, 8) ; (8, 8, 2) ; (4, 8, 4)

⇒ 8 cases​​

$\text{ Prob }=\frac{8}{8\times 8\times 8}=\frac{1}{64}$

∴ The probability of this equation having repeated roots is $\frac{1}{64}$

The correct answer is Option 3.

Concept:

• If two events A and B are independent, P(A ∩ B) = P(A) ⋅ P(B) 
• P(A / B) = $\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}$.

Calculation:

Given E₁ and E₂ are independent events  with P(E1) = $\frac{1}{2}$ and P(E1 ∪ E2) = $\frac{2}{3}$

∴ P(E₁ ∩ E₂) = P(E₁) ⋅ P(E₂)

⇒ P(E1 ∩ E2) = $\frac{1}{2}\mathrm{P}({\mathrm{E}}_2)$     ...(1)

We know that P(E₁ ∪ E₂) = P(E₁) + P(E₂) - P(E₁ ∪ E₂) 

⇒ P(E1 ∪ E2) = P(E1) + P(E2) -  P(E1) ⋅ P(E2)

⇒ $\frac{2}{3}$ = $\frac{1}{2}$ + P(E2) - $\frac{1}{2}$⋅P(E2) 

⇒ P(E₂)(1 - $\frac{1}{2}$) = $\frac{2}{3}-\frac{1}{2}$

⇒ $\frac{1}{2}\mathrm{P}({\mathrm{E}}_2)=\frac{1}{6}$

⇒ P(E₂) = $\frac{1}{3}$

∴ Statement 1 is correct.

P(E1 | E2) = $\frac{\mathrm{P}({\mathrm{E}}_1\cap {\mathrm{E}}_2)}{\mathrm{P}({\mathrm{E}}_2)}$

⇒ P(E1 | E2) = $\frac{\frac{1}{2}\mathrm{P}({\mathrm{E}}_2)}{\mathrm{P}({\mathrm{E}}_2)}$ = $\frac{1}{2}$.

∴ Statement 2 is correct.

Both the statements are correct.

Concept:

Permutations with Repetition = nr

Where n is the number of things to choose from r different things when repetition is allowed, and order matters.

Favorable cases = Total cases - Unfavorable cases

Calculation:

According to the question

Four dies are rolled 

So, Total Possible number of outcomes = 64

Now, Total outcomes when no 2 appears = 54

Now, From the concept used

Favorable cases = 64 - 54

⇒ 1296 - 625

⇒  671

∴ The number of possible outcomes in which at least one die shows 2 is 671.

Concept:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

For mutually exclusive events A and B, P(A ∩ B) = 0

Calculation:

We know, for mutually exclusive events A and B,  P(A ∩ B) = 0

∴ P(A ∪ B) = P(A) + P(B) - 0

=  P(A) + P(B)

Hence, option (2) is correct.

Concept:

Probability ≤ 1

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

For mutually exclusive events A and B, P(A ∩ B) = 0

P(B̅) = 1 - P(B)

Calculation:

We know, for mutually exclusive events A and B,  P(A ∩ B) = 0

P(B̅) = 1 - P(B)

P(A ∪ B) ≤ 1

P(A) + P(B) ≤ 1

P(A) ≤ 1 - P(B)

P(A) ≤ P(B̅)

Hence, option (1) is correct.

Concept:

Mutually exclusive events:

• Two events are said to be mutually exclusive if they cannot happen at the same time.
• Example:  if we toss a coin, either heads or tails might turn up, but not heads and tails at the same time.
• If A and B are mutually exclusive events then, P (A∩B) =0 and P (A ∪ B) = P (A) + P (B)

Complement of an event:

• The complement of an event is the subset of outcomes in the sample space that are not in the event. A complement is itself an event.
• The probability of the complement of an event is one minus the probability of the event.
• P (A’) or P (A^c) or $\mathrm{P}\left(\overline{\mathrm{A}}\right)$ = 1 - P (A) Where (A) be probability of Event A and P (A’) or P (A^c) or $\mathrm{P}\left(\overline{\mathrm{A}}\right)$  be probability of the complement of Event A

Calculation:

Given: P(A) = 1/3 and P(B) = 1/4

As we know that $\mathrm{P}\left(\overline{\mathrm{A}}\cap \overline{\mathrm{B}}\right)=\mathrm{P}\left(\overline{\mathrm{A}\cup \mathrm{B}}\right)=1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)$

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

A and B are two mutually exclusive events so P(A ∩ B) = 0

∴ P(A ∪ B) = P(A) + P(B) = (1/3) + (1/4) = 7/12

Now,

$\mathrm{P}\left(\overline{\mathrm{A}}\cap \overline{\mathrm{B}}\right)=1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=1-\frac{7}{12}=\frac{5}{12}$

Given:

A and B are two independent events.

Concept:

For any two independent events A and B : P(A ∩ B) = P(A)·P(B)

Explanation:

The probability of occurrence of at least one of A and B means " A but not B or B but not A or A and B both ".

Indirectly it means '1 - probability of occurrence of neither A and B'.

⇒ P(at least one of A and B) = 1 - P(neither A and B)

⇒ P(at least one of A and B) = 1 - P(A' ∩ B')

Now, P(A' ∩ B') = P(A’) P(B’) as A and B are independent events.

⇒ P(at least one of A and B) = 1 - P(A')P(B')

Formula used: 

Selection of r things out of total n things:

${}^n{C}_r=\frac{n!}{r!\times (n-r)!}$

Probability of occurrence of the event:

P(E) =  $\frac{n(E)}{n(S)}$

Where,

n(E) = Number of favorable outcome

n(S) = Number of possible outcome

Calculation:

Possible pair will be: 

(1, 2), (2, 3), (3, 4), (4, 5)

(1, 3), (2, 4), (3, 5)

(1, 4), (2, 5)

(1, 5)

⇒ n(S) = 10

To make the unit digit zero in the product, possible pairs will be 

(4, 5), (2, 5)

⇒ n(E) = 2

Hence, the required probability

P(E) = n(E)/n(S) = 2/10

⇒ P(E) =  1/5

∴ Required probability is $\frac{1}{5}$.

Alternate Method

Given digits are 1, 2, 3,  4, and 5. 

As discussed above we know, that two digits can be chosen out of 5 in ⁵C₂ ways.

⇒ n(S) = ⁵C₂ = $\frac{5!}{2!\times 3!}$

 ⇒ n(S) = $\frac{5\times 4\times 3!}{2\times 1\times 3!}=10$

⇒ n(S) = 10

Since the last digit in the product appears as 0 which is only possible when any of 2 or 4 is multiplied by 5 so possible required result,

(E) = {(2, 5), (4, 5)}

⇒ n(E) = 2

Hence, the required probability

P(E) = n(E)/n(S) = 2/10

⇒ P(E) =  1/5

∴ Required probability is $\frac{1}{5}$.

If we toss a coin 2 times, the sample space is (HH, HT, TH, TT)

Then the number of sample points is = 4 = 22

Now, if we toss a coin 3 times, the sample space is (HHH, HHT, HTT, HTH, THT, THH, TTT, TTH)

Then the number of sample points is  = 8 =23

Similarly, for 8 times toss:

Total cases = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 or 28

Concept:

Simple Events:- If an event E has only one sample point of a sample space, i.e., a single outcome of an experiment it is called a Simple or Elementary event.

Formula Used:

Probability of an event occurring = (No. of favorable outcomes)/(Total no. of outcomes)

Concept:

• P(A') = 1 - P(A)
• P(A' ∩ B') = P((A ∪ B)') {Probability of not A and not B}
• P(A and not B) = P(A ∩ B') = P(A) - P(A ∩ B)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given:  P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.3, 

⇒  P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒  P(A ∪ B) = 0.5 + 0.7 - 0.3 = 0.9__(i)

Now the probability of not A and not B,

P(A' ∩ B') = P((A ∪ B)') = 1 - P(A ∪ B) 

⇒ P(A' ∩ B') = 1 - 0.9 { From (i)}

⇒ P(A' ∩ B') = 0.1 __(ii)

The probability of B and not A,

P(A' ∩ B) = P(B) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.7 - 0.3 = 0.4 __(iii)

The probability of A and not B,

P(A ∩ B') = P(A) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.5 - 0.3 = 0.2 __(iv)

From (ii), (iii) and (iv),

P(A' ∩ B') + P(A' ∩ B) + P(A ∩ B') = 0.1 + 0.4 + 0.2 = 0.7

∴ The correct option is (2).

Concept:

For independent events, outcome of an event is not affected by any other event.

Calculation:

If A and B are two independent events, then

P(A ∩ B) = P(A)⋅P(B) .

The correct answer is option 3.