Question

Traditional definition of independence says that events A and B are
independent if and only if P(A n B)=P(A)×P(B). Show that P(A n B)=P(A)×P(B) if
and only if
a. P(A n B’) = P(A) × P(B’)b. P(A’ n B’) = P(A’) × P(B’)
You may use these without proof in your solutions:
o P(A n B’) = P(A) – P(A n B). [This can be proven by first showing that
(A ∩ B ′ ) ∪̇ (A ∩ B) = A and using the Addition Rule.]
o P(A n B) = 1 – P(A’ n B’). [This can be proven by noting that under de
Morgan’s Law, (A n B)’ = A’ n B’.]

Answer 1

Solution

Back-up Theory

P(A ∩ B') = P(A) – P(A ∩ B) ..................................................................................................... (1)

P(A ∪ B) = 1 – P(A' ∩ B') ..................................................................................................... (2)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) ...................................................................................... (2b)

Now, to work out the solution,

Part 1 sub-part (a)

Given P(A ∩ B) = P(A) x P(B), .............................................................................................. (3a)

to show P(A ∩ B') = P(A) x P(B') ......................................................................................... (3b)

Now, vide (1), LHS of (3b)

P(A ∩ B')  

= P(A) - P(A ∩ B)

= P(A) – {P(A) x P(B)} [vide (3a)]

= P(A){1 – P(B)}

= P(A) x P(B')

= RHS of (3b).

Hence, (3b) is proved. Answer 1

Part 1 sub-part (b)

Given P(A ∩ B) = P(A) x P(B), .............................................................................................. (4a)

to show P(A' ∩ B') = P(A') x P(B') ......................................................................................... (4b)

Now, vide (2), LHS of (4b)

P(A' ∩ B')  

= 1 - P(A ∪ B)

= 1 – {P(A) + P(B) - P(A ∩ B)} [vide (2b)]

= 1 – {P(A) + P(B) - P(A) x P(B)} [vide (4a)]

= {1 – P(A)} + P(B){1 - P(A)}

= {1 – P(A)}{1 - P(B)}

= P(A') x P(B')

= RHS of (4b).

Hence, (4b) is proved. Answer 2

Part 2 sub-part (a)

Given P(A ∩ B') = P(A) x P(B'), .............................................................................................. (5a)

to show P(A ∩ B) = P(A) x P(B)......................................................................................... (5b)

Now, vide (1), LHS of (5b)

P(A ∩ B)  

= P(A) - P(A ∩ B')

= P(A) – {P(A) x P(B')} [vide (5a)]

= P(A){1 – P(B')}

= P(A) x P(B)

= RHS of (5b).

Hence, (5b) is proved. Answer 3

Part 2 sub-part (b)

Given P(A' ∩ B') = P(A') x P(B'), .............................................................................................. (6a)

to show P(A ∩ B) = P(A) x P(B)......................................................................................... (6b)

Now,

Vide (2),

P(A' ∩ B') = 1 - P(A ∪ B)

= 1 – { P(A) + P(B) - P(A ∩ B)} [vide (2b)]

= {1 – P(A)} + P(B) - P(A ∩ B)

= P(A') + P(B) - P(A ∩ B)............................................................................................ (6c)

(6c) and (6a) =>

P(A') x P(B') = P(A') + P(B) - P(A ∩ B)

Or, P(A ∩ B) = P(A') + P(B) – { P(A') x P(B')}

= P(A'){1 - P(B')} + P(B)

= {P(A') x P(B)} + P(B)

= P(B){1 - P(A')}

= P(B) x P(A)

Hence, (6b) is proved. Answer 4

DONE