Question

Problem 1. A boy has a fever after coming home in the afternoon. His mother thinks that it could be related to the following three possible reasons: A : He plays football in the rain, B : He takes a cold water shower after playing, C : He eats too many ice creams.

(iii) The mother has 80% confidence that her son’s fever is caused by at least one of the three reasons. She further estimates that the probabilities of the three individual reasons are 0.5, 0.5, 0.2 respectively, and she believes that they are pair- wisely independent. Are the three reasons mutually independent?

(iv) Suppose that the mother is 100% sure that her son’s fever is caused by at least one of the three reasons. Moreover, she believes that they are mutually inde- pendent although she doesn’t know the exact probabilities of any of the individual reasons. After a moment’s thought, she tells her son that one of the three reasons must be certain (that is, one of P(A) = 1 or P(B) = 1 or P(C) = 1 must be true)! Should the boy believe his mother’s assertion?

Answer 1

Solution

Back-up Theory

P(at least one of A, B and C) = P(A ∪ B ∪ C) ........................................................................................... (1)

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

If A and B independent, P(A ∩ B) = P(A) x P(B) ...................................................................................... (3)

If 3 events A, B and C are mutually independent, they are necessarily pair-wise independent................. (4)

[but converse need not be true.]

Now, to work out the solution,

Part (iii)

We have:

P(at least one of A, B and C) = P(A ∪ B ∪ C) = 0.8 ................................................................................... (5)

P(A) = 0.5; P(B) = 0.5; P(C) = 0.2 ............................................................................................................ (6)

A, B and C are pair-wise independent and so vide (3),

P(A ∩ B) = P(A) x P(B); P(B ∩ C) = P(B) x P(C); P(C ∩ A) = P(C) x P(A) ............................................... (7)

(2), (5), (6) and (7) =>

0.8 = 0.5 + 0.5 + 0.2 – (0.5 x 0.5) – (0.5 x 0.2) - (0.2 x 05) + P(A ∩ B ∩ C)

Or, P(A ∩ B ∩ C) = 0.05 .............................................................................................................................. (8)

Now, P(A) x P(B) x P(C) = 0.5 x 0.5 x 0.2 = 0.05 ......................................................................................... (8a)

Extending (2) to three events, (8) and (8a) => A, B and C are mutually independent. i.e.,

the three reasons mutually independent. Answer 1

Part (iv)

The mother is 100% sure that her son’s fever is caused by at least one of the three reasons, vide (1) =>

P(A ∪ B ∪ C) = 1 ......................................................................................................................................... (9)

Given, one of P(A) = 1 or P(B) = 1 or P(C) = 1 must be true, let P(A) = 1 .................................................. (10)

Also given, ‘Moreover, she believes that they are mutually independent’, vide (4) and

extending (2) to three events, P(A ∩ B ∩ C) = P(A) x P(B) x P(C) and (..................................................... (11)

(2), (9), (10) and (11) =>

1 = 1 + P(B) + P(C) - P(B) – {P(B) x P(C)} - P(C) + P(B) x P(C)

Or, 0 = 0.

This is a very valid statement.

So, there is NO reason why the boy should not believe his mother’s assertion. Answer 2

DONE