If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$ find $\mathrm{P}$ $($ not $\mathrm{A}$ and not $\mathrm{B})$
If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$ find $\mathrm{P}$ $($ not $\mathrm{A}$ and not $\mathrm{B})$
It is given that, $\mathrm{P}(\mathrm{A}) \frac{1}{4}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$
$\mathrm{P}$ $($ not on $\mathrm{A} $ and not on $\mathrm{B})$ $=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B^{\prime}}\right)$
$\mathrm{P}$ $($ not on $\mathrm{A} $ and not on $\mathrm{B})$ $=\mathrm{P}\left((\mathrm{A}^{\prime} \cup \mathrm{B})\right)$ $\left[A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}\right]$
$=1-P(A \cup B)$
$=1-[P(A)+P(B)-P(A \cap B)]$
$=1-\frac{5}{8}$
$=\frac{3}{8}$
Similar Questions
Fill in the blanks in following table :
$P(A)$
$P(B)$
$P(A \cap B)$
$P (A \cup B)$
$0.35$
...........
$0.25$
$0.6$
Fill in the blanks in following table :
| $P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
| $0.35$ | ........... | $0.25$ | $0.6$ |
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- [IIT 1975]
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