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 :
Solution
P(A $\cup$ B) = P(A) + P(B) – P(A $\cup$ B)
<br><br>$\Rightarrow$ 0.8 = 0.6 + 0.4 – P(A $\cap$ B)
<br><br>$\Rightarrow$ P(A $\cap$ B) = 0.2
<br><br>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)
<br><br>$\Rightarrow$ $\alpha$ = 0.6 + 0.4 + 0.5 - 0.2 - $\beta$ - 0.3 + 0.2
<br><br>$\Rightarrow$ $\alpha$ + $\beta$ = 1.2
<br><br>$\Rightarrow$ $\alpha$ = 1.2 - $\beta$
<br><br>Given, 0.85 $\le \alpha \le$ 0.95
<br><br>$\Rightarrow$ 0.85 $\le$ 1.2 - $\beta$ $\le$ 0.95
<br><br>$\Rightarrow$ 0.25 $\le \beta \le$ 0.35
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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