The probabilities of three events A, B and C are given by P(A) = 0.6, P(B) = 0.4 and P(C) …

Question

The probabilities of three events A, B and C are
given by
P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.

If P(A$$ \cup $$B) = 0.8, P(A$$ \cap $$C) = 0.3, P(A$$ \cap $$B$$ \cap $$C) = 0.2,
P(B$$ \cap $$C) = $$\beta $$
 and P(A$$ \cup $$B$$ \cup $$C) = $$\alpha $$, where
0.85 $$ \le \alpha \le $$ 0.95, then $$\beta $$ lies in the interval :

Accepted Answer

P(A $$ \cup $$ B) = P(A) + P(B) -- P(A $$ \cup $$ B)

$$ \Rightarrow $$ 0.8 = 0.6 + 0.4 -- P(A $$ \cap $$ B)

$$ \Rightarrow $$ P(A $$ \cap $$ B) = 0.2

P(A$$ \cup $$B$$ \cup $$C) = P(A) + P(B) + P(C) -- P(A $$ \cap $$ B) -- P(B $$ \cap $$ C) --P(C $$ \cap $$ A) + P(A $$ \cap $$ B $$ \cap $$ C)

$$ \Rightarrow $$ $$\alpha $$ = 0.6 + 0.4 + 0.5 - 0.2 - $$\beta $$ - 0.3 + 0.2

$$ \Rightarrow $$ $$\alpha $$ + $$\beta $$ = 1.2

$$ \Rightarrow $$ $$\alpha $$ = 1.2 - $$\beta $$

Given, 0.85 $$ \le \alpha \le $$ 0.95

$$ \Rightarrow $$ 0.85 $$ \le $$ 1.2 - $$\beta $$ $$ \le $$ 0.95

$$ \Rightarrow $$ 0.25 $$ \le \beta \le $$ 0.35

Solutions · 1