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

(i) Given : P(A) = 0.25, P(A or B) = 0.5 and P(B) = 0.4 

To find : P(A and B) 

Formula used : P(A or B) = P(A) + P(B) – P(A and B) 

Substituting in the above formula we get, 

0.5 = 0.25 + 0.4 – P(A and B) 

0.5 = 0.65 – P(A and B) 

P(A and B) = 0.65 – 0.5 

P(A and B) = 0.15 

P(A and B) = 0.15 

(ii) Given : P(A) = 0.25, P(A and B) = 0.15 ( from part (i)) 

To find : P(A and \(\overline{B}\) ) 

Formula used : P(A and \(\overline{B}\) ) = P(A) – P(A and B) 

Substituting in the above formula we get,

P(A and \(\overline{B}\) ) = 0.25 – 0.15 

P(A and \(\overline{B}\) ) = 0.10 

P(A and \(\overline{B}\)) = 0.10

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 : P(A) = 0.60, P(A or B) = 0.85 and P(A and B) = 0.42 

To find : P(B) 

Formula used : P(A or B) = P(A) + P(B) – P(A and B) 

Substituting in the above formula we get,

0.85 = 0.60 + P(B) – 0.42 

0.85 = 0.18 + P(B) 

0.85 – 0.18 = P(B) 

0.67 = P(B) 

P(B) = 0.67

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

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

Correct Answer – (i) 0.15 (ii) 0.1
(ii) `P(A and bar(B)) = P(A nn bar(B)) = P(A) – P(A nn B).`

0.2

If A and B are two events associated with a random experiment such that p(A) = 0.3, p(B) = 0.2 and p(A ∩ B) = 0.1, then the value of p(A – B) is 0.2.

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

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