Question

A, B are mutually exclusive events such that P(A)=1/3 & P(B)=1/2 then P(A' n B')=?

Answer 1

A∩B=0 ... as they are mutually exclusive
Now,
P(A'∩B')=P(A∪B)'
Now,
P(A∪B)=P(A) + P(B) - P(A∩B)
            =(1/3) + (1/2) - 0
            =5/6
⇒P(A∪B)' = 1-(5/6)
              =1/6
So,
P(A'∩B')=1/6

Answer 2

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Given, P(A) = 1/2 ,

P (A ∪ B) = 3/5

and P(B) = p.

(1) For Mutually Exclusive

Given that, the sets A and B are mutually exclusive.

Thus, they do not have any common elements

Therefore, P(A ∩ B) = 0

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Substitute the formulas in the above-given formula, we get

3/5 = (1/2) + p – 0

Simplify the expression, we get

(3/5) – (1/2) = p

(6 − 5)/10 = p

1/10 = p

Therefore, p = 1/10

Hence, the value of p is 1/10, if they are mutually exclusive.

(ii) For Independent events:

If the two events A & B are independent,

we can write it as P(A ∩ B) = P(A) P(B)

Substitute the values,

= (1/2) × p

= p/2

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

Now, substitute the values in the formula,

(3/5) = (1/2)+ p – (p/2)

(3/2)– (1/2)= p – (p/2)

(6 − 5)/10 = p/2

1/10 = p/2

p= 2/10

P = 1/5

Thus, the value of p is 1/5, if they are independent

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Hope It's Helpful.....:)