Question

For the three events $A,$ $B$ and $C,$ $P\text{(at least one occurring)}=\frac{3}{4},$ $P\text{(at least two occurring)}=\frac{1}{2}$ and $P\text{(exactly two occurring)}=\frac{2}{5}.$ Which of the following relations is (are) CORRECT ?

• A. $P(A\cap B\cap C)=\frac{1}{10}$
• B. $P(A\cap B)+P(B\cap C)+P(C\cap A)=\frac{7}{5}$
• C. $P\left(A\right)+P\left(B\right)+P\left(C\right)=\frac{27}{20}$
• D. $P(A\cap \overline{B}\cap \overline{C})+P(\overline{A}\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap \overline{C})=\frac{1}{4}$

Answer 1

Answer: (A), (C), (D)

$P(A\cap \overline{B}\cap \overline{C})+P(\overline{A}\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap \overline{C})=\frac{1}{4}$



Given $P(A\cup B\cup C)=\frac{3}{4}\dots \left(1\right)$
$P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)+P(A\cap B\cap C)=\frac{1}{2}\dots \left(2\right)$
And $P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)=\frac{2}{5}\dots \left(3\right)$
From $\left(2\right)-\left(3\right),$
$P(A\cap B\cap C)=\frac{1}{2}-\frac{2}{5}=\frac{1}{10}\Rightarrow P(A\cap B\cap C)=\frac{1}{10}\dots \left(4\right)$

$P(A\cap B)+P(B\cap C)+P(C\cap A)$
$=P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)+3P(A\cap B\cap C)$
$=\frac{2}{5}+3\times \frac{1}{10}\Rightarrow P(A\cap B)+P(B\cap C)+P(C\cap A)=\frac{7}{10}\dots \left(5\right)$

$P\left(A\right)+P\left(B\right)+P\left(C\right)$
$=P(A\cup B\cup C)+P(A\cap B)+P(B\cap C)+P(C\cap A)-P(A\cap B\cap C)$
From $\left(1\right),\left(4\right)$ and $\left(5\right),$
$P\left(A\right)+P\left(B\right)+P\left(C\right)=\frac{3}{4}+\frac{7}{10}-\frac{1}{10}=\frac{27}{20}$

Also, $P(A\cap \overline{B}\cap \overline{C})+P(\overline{A}\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap \overline{C})$
$=P\left(A\right)+P\left(B\right)+P\left(C\right)-2\left[P\right(A\cap B)+P(B\cap C)+P(C\cap A\left)\right]+3P(A\cap B\cap C)$
$=\frac{27}{20}-\frac{14}{10}+\frac{3}{10}=\frac{5}{20}=\frac{1}{4}$