Question

(a) Let A, B and C be mutually independent events. Show that C and A ∪ B are independent.

(b) A box contains 4 cards numbered 1 to 4 respectively. Two cards are chosen successively and with replacement from the box and their numbers are noted. Consider the following events: D = “the sum of the two numbers drawn is even”, and E = “the first number drawn is even”.

i). Find P(D) and P(E).

ii). Determine if D and E are independent events. Justify your answer

Answer 1

solution:

a) it is given that , A, B and C be mutually independent events

when two events are mutually independent event,then P(A⋂B⋂C) = P(A)P(B)P(C)

We have to prove that P(A ⋂ (B⋃C)) = P(A) P(B⋃C)

So, P(A ⋂ (B⋃C)) = P((A⋂B) ⋃ (A⋂C))

= P(A⋂B) + P(A⋂C) - P(A⋂B) ⋂ P(A⋂C)

= P(A) P(B) + P(A) P(C) - P(A⋂B⋂C)..................because ABC are independent event

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

= P(A) [P(B) + P(C) - P(B) P(C)]

= P(A) [P(B) + P(C) - P(B ⋂C)]

= P(A) P(B⋃C)

so, P(A ⋂ (B⋃C)) = P(A) P(B⋃C)

b)

box contains four cards numbered 1,2,3,4

P(D) be the the probability of sum of two card number is even if two cards are drawn with replacement.

there are two cases which are favourable for event D

case i - drawing first card even and second card even.

case ii drawing first card odd and second card also odd.

probability of drawing first card even and second card even = 2/4*2/4 = 1/4

probability of drawing first card odd and second card also odd = 2/4*2/4 = 1/4

so P(D) = 1/4+1/4 = 2/4 = 1/2

P(E) be the event of drawing first number is even

- drawing first card even then second card may be any of the card

so , P(E) = 2/4*4/4 = 2/4 = 1/2

(ii) event D and E are independent event , because happening of D and E does not effect the outcome of each other.

because the cards are drawn with replacement so there is no effect of a event on other event so event D and E are independent events.