Multiplication Theorem of Events MCQ Quiz - Objective Question with Answer for Multiplication Theorem of Events - Download Free PDF

Concept:

• For two events A and B, we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• If A and B are independent events, then P(A ∩ B) = P(A) × P(B).

Calculation:

Using the concept above, because A and B are independent events, we can write:

P(A ∪ B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.7 = 0.4 + P - 0.4 × P

⇒ 0.6P =0.3

⇒ P = 0.5.

CONCEPT:

• If P(A) = x then P(A̅) = 1 - x 
• If A and B are two independent events then P(A ∩ B) = P(A) × P(B)

CALCULATION:

Given: A and B appear for an interview for two posts such that the probability of A's selection is $\frac{1}{3}$ and that of B's selection is $\frac{2}{5}$.

Let E = event that A is selected

Let F = event that B is selected

⇒ P(E) = $\frac{1}{3}$ and P(F) = $\frac{2}{5}$

As we know that, if P(A) = x then P(A̅) = 1 - x 

⇒ P(E̅) = 1 - ($\frac{1}{3}$) = $\frac{2}{3}$ and P(F̅) = 1 - ($\frac{2}{5}$) = $\frac{3}{5}$

∴ P(event that one of them is selected) = P(E ∩ F̅) + P(E̅ ∩ F)

First let's find out P(E ∩ F̅) and P(E̅ ∩ F)

As we know that, if A and B are two independent events then P(A ∩ B) = P(A) × P(B)

⇒ P(E ∩ F̅) = P(E) × P(F̅) = ($\frac{1}{3}$) × ($\frac{3}{5}$) = $\frac{1}{5}$

⇒ P(E̅ ∩ F) = P(E̅) × P(F) = ($\frac{2}{3}$) × ($\frac{2}{5}$) = $\frac{4}{15}$

⇒ P(event that one of them is selected) = ($\frac{1}{5}$) + ($\frac{4}{15}$) = $\frac{7}{15}$

Hence, the correct option is 2.  

Concept:

If P(A) and P(B) is the probability of occurring any event independently, then P(A ⋃ B) = P(A) + P(B) - P(A ⋂ B) 

Calculation:

Probability of the problem being solved by A, P(A) = $\frac{1}{3}$

Probability of the problem being solved by B, P(B) = $\frac{2}{5}$

P(AB) = $\frac{1}{3}\times \frac{2}{5}$ = $\frac{2}{15}$

Then the probability that the problem is solved is

P(A ⋃ B) = P(A) + P(B) - P(AB)

= $\frac{1}{3}$ + $\frac{2}{5}$ - $\frac{2}{15}$

= $\frac{11}{15}$ - $\frac{2}{15}$

= $\frac{9}{15}$

= $\frac{3}{5}$

Concept:

For any events A and B

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• Events are independent if P(A ∩ B) = P(A) × P(B)
• Events are mutually exclusive if P(A ∩ B) = 0
• P($\overline{\mathrm{A}}$) = 1 - P(A)

Calculation:

Given $\mathrm{P}(\mathrm{A}\cup \mathrm{B})={\displaystyle \frac{5}{6}},\mathrm{P}(\mathrm{A}\cap \mathrm{B})={\displaystyle \frac{1}{3}}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{P}(\overline{\mathrm{A}})={\displaystyle \frac{1}{2}}$

P(A) = 1 - $\frac{1}{2}$ = $\frac{1}{2}$

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

$\frac{5}{6}=\frac{1}{2}+\mathrm{P}(\mathrm{B})-\frac{1}{3}$

P(B) = $\frac{2}{3}$

Given P(A ∩ B) ≠ 0

P(A ∩ B) = P(A) × P(B) = $\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}$

∴ A and B are independent events and are not mutually exclusive events.

Only statement 1 is correct.

Concept:

• P(not A) = 1 - P(A).
• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

  P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

  ('and' means '×' and 'or' means '+')

Calculation:

Probability of Atal speaking the truth is 70% = 0.7.

Probability of Atal NOT speaking the truth is 1 - 0.7 = 0.3.

Probability of George speaking the truth is 60% = 0.6.

Probability of George NOT speaking the truth is 1 - 0.6 = 0.4.

They both will contradict each other in the following case:

(Atal speaks the truth AND George does not speak the truth) OR (Atal does not speak the truth AND George speaks the truth).

∴ The required probability is:

[P(Atal speaking the truth) × P(George not speaking the truth)] + [P(Atal not speaking the truth) × P(George speaking the truth)]

= (0.7 × 0.4) + (0.3 × 0.6)

= 0.28 + 0.18

= 0.46.

∴ They both contradict each other in 46% of the cases.

Concept:

• P(not E) = 1 - P(E).
• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:
  P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]    ('and' means '×' and 'or' means '+')

Calculation:

Probability of Dhoni making a century = ${\displaystyle \frac{4}{5}}$.

Probability of Dhoni not making a century = $1-{\displaystyle \frac{4}{5}}={\displaystyle \frac{1}{5}}$.

Probability of Dhoni making no century in three matches:

P(not century) AND P(not century) AND P(not century)

P(not century) × P(not century) × P(not century)

= ${\displaystyle \frac{1}{5}}\times {\displaystyle \frac{1}{5}}\times {\displaystyle \frac{1}{5}}={\displaystyle \frac{1}{125}}$.

Explanation:

Mutually exclusive events:

Two events are said to be mutually exclusive if they cannot happen at the same time.

In another word, we can say that mutually exclusive is disjoint. If two events are disjoint, then the probability of them both occurring at the same time is 0

If A and B are mutually exclusive events then,

P(A ∩ B) = 0 and P (A ∪ B) = P (A) + P (B)

Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

Calculation:

A bag contain 5 black and 3 white balls, 

There is a  $\frac{3}{8}$ chance that a white marble will be chosen. The marble is chosen and not replaced. There are now 2 white marbles in a bag of 7 marbles, so there is now a  $\frac{2}{7}$ chance that a white marble will be chosen.

Now, P(both the balls are white) = $\frac{3}{8}\times \frac{2}{7}=\frac{3}{28}$

Hence, option (3) is correct.

Concept:

• P(not E) = 1 - P(E).
• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:
  P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]    ('and' means '×' and 'or' means '+')

Calculation:

It is given that P(Bird comes back) = ${\displaystyle \frac{2}{5}}$ and P(Bird stays) = ${\displaystyle \frac{1}{3}}$.

The probability that the bird does not stay (flies away) after coming back = $1-{\displaystyle \frac{1}{3}}={\displaystyle \frac{2}{3}}$.

The probability of the event in question is:

P(Bird comes back) AND P(Bird does not stay)

= P(Bird comes back) × P(Bird does not stay)

= ${\displaystyle \frac{2}{5}}\times {\displaystyle \frac{2}{3}}={\displaystyle \frac{4}{15}}$.

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. 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, P (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:

Probability of solving problem by A = P (A) =1/2

Probability of solving problem by B = P (B) =1/3

Probability of solving problem by C = P (C) =1/4

P (problem will not solved by A) = 1 – (1/2) = 1/2

P (problem will not solved by B) = 1 – (1/3) = 2/3

P (problem will not solved by C) = 1 – (1/4) = 3/4

P (Problem will be solved) = 1 – P (problem will not solved by A, B and C)

$=1-\left(\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\right)=1-\frac{1}{4}=\frac{3}{4}$