The probabilities of three events
Question:

The probabilities of three events $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are given by $P(A)=0.6, P(B)=0.4$ and $P(C)=0.5$. If $P(A \cup B)=0.8$, $\mathrm{P}(\mathrm{A} \cap \mathrm{C})=0.3, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0.2, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\beta$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\alpha$, where $0.85 \leq \alpha \leq 0.95$, then $\beta$ lies in the interval:

  1. (1) $[0.35,0.36]$

  2. (2) $[0.25,0.35]$

  3. (3) $[0.20,0.25]$

  4. (4) $[0.36,0.40]$


Correct Option: , 2

Solution:

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

$=1-0.8=0.2$

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

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

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

$\Rightarrow \beta=1.2-\alpha$

$\because \alpha \in[0.85,0.95]$ then $\beta \in[0.25,0.35]$

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