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Given 

P(A) = 0.3 

P(B) = 0.6 

(i) P(A ∩ B) = P(A) × P(B) 

= 0.3 × 0.6 

= 0.18 

(ii) P(A ∪ \(\bar{B}\)) 

= 0.3 – 0.18 

= 0.12 

(iii) P(A ∪ B) = p(A) + P(B) – P(A ∩ B) 

= 0.3 + 0.6 – 0.18 

= 0.90 – 0.18 

(iv) P(\(\bar{A}\) ∩ \(\bar{B}\)) = [1 – P(A)] [1 – P(B)] 

= [1 – 0.3] [1 – 0.6] 

= 0. 7 × 0.4 

= 0.28

Let E₁, E₂ and E₃ be the events that boxes I, II and III are chosen, respectively

Then P(E₁) = P(E₂) = P(E₃) = 1/3

Also, let A be the event that ‘the coin drawn is of gold’ 

Then P(A|E₁) = P(a gold coin from bag I) = 2/2 =1 

P(A|E₂) = P(a gold coin from bag II) = 0 

P(A|E₃) = P(a gold coin from bag III) = 1/2 

Now, the probability that the other coin in the box is of gold = the probability that gold coin is drawn from the box I. 

= P(E₁|A) 

By Bayes' theorem, we know that

P(E₁/A) = (P(E₁)P(A/E₁))/)(P(E₁)P(A/E₁) + P(E₂)P(A/E₂) + P(E₃)P(A/E₃))

= (1/3 x 1)/(1/3 x 1 + 1/3 x 0 + 1/3 x 1/2) = 2/3

The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9.
Number of ways of selecting 4 different digits out of the 10 digits =¹⁰C₄
Now, each combination of 4 different digits can be arranged in 4! Ways.
∴ Number of four digits with no repetitions = 4! ¹⁰C₄  = 5040
There is only one number that can open the suitcase.

Thus, the required probability is 1/5040

Let the three red balls be R₁, R₂, R₃, three white balls be W₁, W₂, W₃ and three green balls be G₁ , G₂ , G₃ . 

Sample space,

S = {R₁, R₂, R₃, W₁, W₂, W₃,G₁ , G₂ , G₃} 

∴ n(S) = 9 

i. Let A be the event that the ball drawn is red. 

∴ A = {R₁ , R₂ , R₃ } 

∴ n(A) = 3

∴ P(A) = \(\frac{n(A)}{n(S)}\) = 3/9 

∴ P(A) = 1/3

ii. Let B be the event that the ball drawn is not red. 

B = {W₁,W₂,W₃,G₁,G₂,G₃} 

∴ n(B) = 6

∴ P(B) = \(\frac{n(B)}{n(S)}\) = 6/9

∴ P(B) = 2/3

iii. Let C be the event that the ball drawn is red or white.

∴ C = {R₁, R₂, R₃ ,W₁, W₂, W₃ } 

∴ n(C) = 6

∴ P(C) = \(\frac{n(C)}{n(S)}\) = 6/9

∴ P(C) = 2/3

∴ P(A) = 1/3; ∴ P(B) = 2/3; ∴ P(C) = 2/3

Total number of players in the hockey team 

= 6 + 4 + 1 = 11 

∴ n(S) = 11 

i. Let A be the event that the captain selected will be a goalie. 

There is only one goalie in the hockey team.

∴ n(A) = 1

∴ P(A) = \(\frac{n(A)}{n(S)}\)

∴ P(A) = 1/11

ii. Let B be the event that the captain selected will be a defender.

There are 6 defenders in the hockey team.

∴ n(B) = 6

∴ P(B) = \(\frac{n(B)}{n(S)}\)

∴ P(B) = 6/11

∴ P(A) = 1/11 ; P(B) = 6/11

The correct answer is : (A) 10

The probability that the ball is dropped in the basket by John = 4/5 = 0.80 

The probability that the ball is dropped in the basket by Vasim = 0.83

The probability that the ball is dropped in the basket by Akash = 58% =  58/100 = 0.58 

0.83 > 0.80 > 0.58

∴ Vasim has the greatest probability of success.

The probability that the ball is dropped in the basket by John = 4/5 = 0.80

The probability that the ball is dropped in the basket by Vasim = 0.83

The probability that the ball is dropped in the basket by Akash = 58% = 58/100 = 0.58

0.83 > 0.80 > 0.58

∴ Vasim has the greatest probability of success.

(b) \(\frac{1}{4}\)

Since P(A) and P(B) are two mutually exclusive and exhaustive events , P(A) + P(B) = 1

\(\Rightarrow\) P(A) + 3 P(A) = 1

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

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

\(\Rightarrow\) P(\(\bar B\)) = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\).

(b) \(\frac{2}{3}\)

P (A solving ) = \(\frac{1}{3}\) \(\Rightarrow\) P (A not solving) =1- \(\frac{1}{3}\) = \(\frac{2}{3}\)

P (B solving ) = \(\frac{1}{4}\) \(\Rightarrow\) P (B not solving) =1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

P (C solving) = \(\frac{1}{5}\) \(\Rightarrow\) P (C not solving) = 1- \(\frac{1}{5}\) = \(\frac{4}{5}\)

P (D solving) = \(\frac{1}{6}\) \(\Rightarrow\) P(D not solving) =  1- \(\frac{1}{6}\) = \(\frac{5}{6}\)

\(\therefore\) Probability (problem not solved)

= \(\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\) = \(\frac{1}{3}\)

\(\Rightarrow\) Probability (problem solved) =1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

Let E₁ be the event that A fails in an examination

and E₂ be the event that B fails in an examination

Then, we have

P (E₁) = 0.2 and P (E₂) = 0.3

Now, we have to find the P (either E₁ or E₂ fails)

∴ P(either E₁ or E₂ fails) = P(E₁) + P(E₂) – P(E₁⋂ E₂)

≤ P(E₁) + P(E₂)

≤ 0.2 + 0.3

≤ 0.5

Hence, the correct option is (C)

(a) 9 : 4

Let event A : a spade is drawn and event B : an ace is drawn. 

Probability of winning the bet = P (A or B) 

P (A or B) = P(A) + P(B) – P (A \(\cap\) B)

Note. Here A and B are not mutually exclusive events, hence the common part has to be taken into consideration.

= \(\frac{13}{52}\)+ \(\frac{4}{52}\) - \(\frac{1}{52}\) (There is one ace of spades)

= \(\frac{16}{52}\) = \(\frac{4}{13}\)

\(\therefore\) Probability of losing the bet = \(1- \frac{4}{13}\)= \(\frac{9}{13}\)

\(\therefore\) Odds against winning the bet = \(\frac{9}{13}\) \(\colon\) \(\frac{4}{13}\) = \(9\colon4\)

(c) 0.99997

P (aircraft will crash) = 0.03 × 0.02 × 0.05 

= 0.00003 

P (aircraft will not crash) = 1 – 0.00003 

= 0.99997.

If P_1, P₂, P_3, ..., P_n are the respective probabilities of the happening of certain n independent events, then the probability of the failure of all these events is given by: 

P = (1 – P₁) (1 – P₂)  ...  (1 – P_n)

If P₁, P₂, P₃, ..., P_n are the probabilities of the happening of ‘n’ independent events, then the (probability that at least one of the events must happen) 

                    = 1 – Probability of failure of all events 

                   = 1 – (1 – P₁) (1 – P₂) (1 – P₃) … (1 – P_n)

Given, P(A) = P(B) = P(C) p

P(occurrence of at least two of A,B,C) = P(A)P(B)P(\(\bar C\)) +P(A)P(\(\bar B\))P(C) + P(\(\bar A\))P(B)P(C) + P(A)P(B)P(C)

= p × p(1 - p) + p × (1 - p)p + (1 - p)p × p + p × p × p

=3p² (1 - p) + p³

= 3p² -3p³ +p³

= 3p² -2p³

We have, sample space 

= S = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT) 

Then 

A = {HHH} 

B = {HHT, HTH, THH} 

C = {T,T,T} 

D = {HHH, HTH, HHT, HTT} 

(i) Clearly A ∩ B = ϕ, A ∩ C = ϕ, C ∩ D = ϕ 

⇒ A and B ,A and C ; C and D are mutually exclusive. 

(ii) A, C are simple events. 

(iii) B, D are compounds events.

As E₁ and E₂ are independent events.

∴ P((E₁∪ E₂)∩(E₁’ ∩ E₂’))

As by De Morgan’s law. 

E₁’ ∩ E₂’ = (E₁∪ E₂)’

∴ P((E₁∪ E₂)∩(E₁’ ∩ E₂’)) = P((E₁∪ E₂)∩(E₁∪ E₂)’) 

⇒ P((E₁∪ E₂)∩(E₁∪ E₂)’) = P(empty set) = 0

When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Accordingly,
A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}
We now observe that
A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠ Φ
B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠ Φ
C ∩ D = Φ
(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.

(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event.
Thus, B and D are compound events.

Given, A and B are independent.

\(\therefore\) A and \(\bar B\) are also independent.

P(A ∩ \(\bar B\)) = P(A)P(\(\bar B\))

= P(A)(1 - P(B))

= P(A) - P(A)P(B)

Let S = (1, 2, 3,………..,100)

Let A be the set of cubes from the set of first 100 natural numbers,

A = (1, 8, 27, 64)

P (selected no. being cube) = \(\cfrac{n(A)}{n(S)}\)

= \(\cfrac4{100}\)

= \(\cfrac1{100}\) = 0.04

Given   P(\(\bar A\)) = 0.3.P(B) = 0.4 and P(A ∩ \(\bar B\)) = 0.5

We know that P(A ∩ \(\bar B\)) = 0.7 - 0.5 = 0.2

Now, P(A∪B)=P(A) + P(B) - P(A∩B)

= 0.7 +  0.4 - 0.4

= 0.9

Therefore,

\(P(\overline{A\cup B})\)  = 1 - P(A ∪ B)

=1 - 0.9

= 0.1

⇒ \(P(\cfrac{B}{\bar A \cap\bar B})\) = \(\cfrac14\)

Here, P(A) + P(B) + P(C) = 1 …(1)

For mutually exclusive events A, B, and C,

P(A and B) = P(B and C) = P(A and C) = 0

Given: P(B) = (3/2) P(A) and P(C) = (1/2) P(B)

(1) => P(A) + (3/2) P(A) + (1/2) P(B) = 1

=> P(A) + (3/2) P(A) + (1/2){(3/2) P(A)} = 1

=> P(A) + (3/2) P(A) + (3/4) P(A) = 1

=> 13/4 P(A) = 1

or P(A) = 4/13

Given

P(A ∩ B) = P(B ∩ C) = P(A ∩ C) = 0

⇒ P(A ∩ B ∩ C) = 0

Also given P(A ∪ B ∪ C) = 1

By formula we know, 

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ B ∩ C)

Substituting the values we get

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

⇒ P(A) + P(B) + P(C) = 1

Given: A and B are two mutually exclusive events.

P (A) = 0.4 and P (B) = 0.5

By definition of mutually exclusive events we know that:

P (A ∪ B) = P (A) + P (B)

Now, we have to find

(i) P (A ∪ B) = P (A) + P (B) = 0.5 + 0.4 = 0.9

(ii) P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}

P (A′ ∩ B′) = 1 – P (A ∪ B)

= 1 – 0.9

= 0.1

(iii) P (A′ ∩ B) [This indicates only the part which is common with B and not A.

Hence this indicates only B]
P (only B) = P (B) – P (A ∩ B)

As A and B are mutually exclusive so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A′ ∩ B) = P (B) = 0.5

(iv) P (A ∩ B′) [This indicates only the part which is common with A and not B.

Hence this indicates only A]

P (only A) = P (A) – P (A ∩ B)

As A and B are mutually exclusive so they don’t have any common parts.

P (A ∩ B) = 0

∴ P (A ∩ B′) = P (A) = 0.4

Given: A and B are two events.

P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35

By definition of P (A or B) under axiomatic approach we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

Now we have to find:

(i) P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= 0.54 + 0.69 – 0.35

= 0.88

(ii) P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s Law}

P (A′ ∩ B′) = 1 – P (A ∪ B)

= 1 – 0.88

= 0.12

(iii) P (A ∩ B′) [This indicates only the part which is common with A and not B.

Hence this indicates only A]

P (only A) = P (A) – P (A ∩ B)

∴ P (A ∩ B′) = P (A) – P (A ∩ B)

= 0.54 – 0.35

= 0.19

(iv) P (A′ ∩ B) [This indicates only the part which is common with B and not A.

Hence this indicates only B]

P (only B) = P (B) – P (A ∩ B)

∴ P (A′ ∩ B) = P (B) – P (A ∩ B)

= 0.69 – 0.35

= 0.34

Given A and B are two mutually exclusive events And, 

P(A) = 0.4 P(B) = 0.5 

By definition of mutually exclusive events we know that: 

P(A ∪ B) = P(A) + P(B) 

We have to find- 

(i) P(A ∪ B) = P(A) + P(B) = 0.5 + 0.4 = 0.9 

(ii) P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s Law} 

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.9 = 0.1

(iii) P(A’ ∩ B) = This indicates only the part which is common with B and not A 

⇒ This indicates only B. 

P(only B) = P(B) – P(A ∩ B) 

As A and B are mutually exclusive So they don’t have any common parts 

⇒ P(A ∩ B) = 0 

∴ P(A’ ∩ B) = P(B) = 0.5 

(iv) P(A ∩ B’) = This indicates only the part which is common with A and not B 

⇒ This indicates only A. 

P(only A) = P(A) – P(A ∩ B) 

As A and B are mutually exclusive So they don’t have any common parts 

⇒ P(A ∩ B) = 0 

∴ P(A ∩ B’) = P(A) = 0.4

Given A and B are two events 

And, P(A) = 0.54 P(B) = 0.69 P(A ∩ B) = 0.35 

By definition of P(A or B) under axiomatic approach we know that: 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

We have to find

(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

= 0.54 + 0.69 – 0.35 = 0.88 

(ii) P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s Law} 

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.88 = 0.12

(iii) P(A ∩ B’) = This indicates only the part which is common with A and not B ⇒ This indicates only A. 

P(only A) = P(A) – P(A ∩ B) 

∴ P(A ∩ B’) = P(A) - P(A ∩ B) = 0.54 – 0.35 = 0.19 

(iv) P(A’ ∩ B) = This indicates only the part which is common with B and not A ⇒ This indicates only B. 

P(only B) = P(B) – P(A ∩ B) 

∴ P(A’ ∩ B) = P(B) – P(A ∩ B) = 0.69 – 0.35 = 0.34

Given: A and B are two events.

P (A) = 0.5, P (B) = 0.3 and P (A ∩ B) = 0.2

Now we need to find P (A ∪ B).

By definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

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

P (A ∪ B) = 0.5 + 0.3 – 0.2

= 0.8 – 0.2

= 0.6

∴ P (A ∪ B) is 0.6

Given: A and B are two events.

P (A′) = 0.5, P (A ∩ B) = 0.3 and P (A ∪ B) = 0.8

Since, P (A′) = 1 – P (A) 

P (A) = 1 – 0.5

= 0.5

Now we need to find P (B).

By definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

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

P (B) = 0.8 + 0.3 – 0.5

= 1.1 – 0.5

= 0.6

∴ P (B) is 0.6

Given: A and B are two events.

P (A) = 0.3, P (B) = 0.5 and P (A ∪ B) = 0.5

Now we need to find P (A ∩ B).

By definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

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

P (A ∩ B) = 0.3 + 0.4 – 0.5

= 0.7 – 0.5

= 0.2

∴ P (A ∩ B) is 0.2

We need to fill the table:

(i) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

Using data from table, we get: 

∴ P(A ∪ B) = \(\frac{1}{3}+\frac{1}{5}-\frac{1}{15} = \frac{8}{15}-\frac{1}{15}=\frac{7}{15}\) 

(ii) By definition of P(A or B) under axiomatic approach we know that: 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

⇒ P(B) = P(A ∪ B) + P(A ∩ B) – P(A) 

Using data from table, we get: 

∴ P(B) = 0.6 + 0.25 – 0.35 = 0.5 

(iii) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

⇒ P(A ∩ B) = P(B) + P(A) - P(A ∪ B) 

Using data from table, we get: 

∴ P(A ∩ B) = 0.5 + 0.35 – 0.7 = 0.15 

Filled table is:

Given A and B are two events 

And, P(A’) = 0.5 P(A ∩ B) = 0.3 P(A ∪ B) = 0.8

∵ P(A’) = 1 – P(A) ⇒ P(A) = 1 – 0.5 = 0.5 

We need to find P(B). 

By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that: 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

∴ P(B) = P(A ∪ B) + P(A ∩ B) – P(A) 

⇒ P(B) = 0.8 + 0.3 – 0.5 = 1.1 – 0.5 = 0.6 

∴ P(B) = 0.6

Correct option is A. 1/4

P(A) = 1/3 P(B) → Given

3 P(A) = P(B) → (1)

Now, since it’s given that A and B are mutually exclusive:

P(A ⋂ B) = 0 → (2)

Now then,

P(S) = P(A⋃ B) = 1 → (3)

P(A⋃ B) = P(A) + P(B) + P(A⋂B)

P(A) + P(B) = 1 → From (2) & (3)

P(A) + 3 P(A) = 1 → From (1)

4 P(A) = 1

P(A) = 1/4

Given A and B are two events 

And, P(A) = 0.5 P(B) = 0.3 P(A ∩ B) = 0.2 

We need to find P(A ∪ B). 

By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that: 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

⇒ P(A ∪ B) = 0.5 + 0.3 – 0.2 = 0.8 – 0.2 = 0.6 

∴ P(A ∪ B) = 0.6

Given A and B are two events

And, P(A) = 0.3 P(B) = 0.5 P(A ∪ B) = 0.5 

We need to find P(A ∩ B). 

By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that: 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B) 

⇒ P(A ∩ B) = 0.3 + 0.4 – 0.5 = 0.7 – 0.5 = 0.2 

∴ P(A ∩ B) = 0.2

P(B) = \(\frac{3}{5}\)and P(A’ ∩ B) = \(\frac{1}{2}\) are given.

∴ P(A’ ∩ B) = P(B) – P(A ∩ B)

∴ \(\frac{1}{2} = \frac{3}{5}\) – P(A ∩ B)

∴ P(A ∩ B) = \(\frac{3}{5} – \frac{1}{2} = \frac{6−5}{10 }= \frac{1}{10}\)

Now, P(A|B) = \(\frac{P(A∩B)}{P(B)}\)

= \(\frac{1}{10}\frac{3}{5} = \frac{1}{10}×\frac{5}{3} = \frac{1}{6}\)

∴ P(A|B) = \(\frac{1}{6}\)

We know that the coin has two sides head (H) and tail (T)

So the possible outcomes are X^m. (where x is the number of outcomes when a coin is tossed and m is number of coins)

∴ 2¹= 2 i.e. head and tail

∴ The possible outcomes are H and T.