mathematics-class-11-unit-06-chapter-04-probability-l-4-6-mq0uitlfaba

### <Success>Solution</Success>
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It is clear that $S \subseteq [0, 1] \times [0, 1]$
Now given that
$ 0.5 \leq P(A \cup B) \leq 1 $(obvious)
$ 0.125 \leq P(A \cap B) \leq 0.375 $
$\therefore$ $ 0.625 \leq P(A \cup B + P(A \cap B) \leq 1.375$
Now $P(A \cup B)+P(A \cap B)=P(A)+P(B)$
$\therefore 0.625 \leq P(A) + P(B) \leq 1.375$
or, $\frac{5}{8} \leq x + y \leq \frac{11}{8}$
$\therefore S$ in P X Y S R Q
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##### <Success>Solution</Success>
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Note that length of PQ = $\sqrt{(\frac{5}{8})^2 + (\frac{5}{8})^2} = \sqrt{2 \times (\frac{5}{8})^2} = \sqrt{2}(\frac{5}{8})$
& length of XY = $\sqrt{(1-\frac{3}{8})^2 + (1-\frac{3}{8})^2} = \sqrt{2 \times (\frac{5}{8})^2} = \sqrt{2}(\frac{5}{8})$
Parimeter of S is the sum of 6 line segment:
$ 2 \sqrt{2} (\frac{5}{8}) + \frac{3}{8} \times 4 = \sqrt{2} \frac{10}{8} + \frac{12}{8}$
$ = 1.41 \times \frac{10}{8} + \frac{12}{8} = \frac{26.1}{8} \approx 3.26$ perimeter length
Area of S
$\frac{1}{2} \times (\frac{11}{8})^2 - \frac{1}{2} (\frac{5}{8})^2 - \frac{1}{2}(\frac{3}{8})^2 - \frac{1}{2}(\frac{3}{6})^2$
$=\frac{1}{2 \times 64}(121 - 25 - 9 - 9) = \frac{78}{128} = 0.609$
$\therefore$ The answer in perimeter = 3.26, Area = 0.609.
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