JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 15)
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 :
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 :
[0.35, 0.36]
[0.20, 0.25]
[0.25, 0.35]
[0.36, 0.40]
Explanation
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
$$ \Rightarrow $$ 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
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