Two real numbers a and b are selected uniformly at random from the interval -0-4- Let R -a- b- a2-4- b - 2 -a- The probability that the chosen numbers lie in R is1A-1B-C-D-

The correct option is A 13 nbsp; Area of shaded region, R= ∫ _0^4(2 √(x) x^24)dx =163 Area of OABC=4 × 4=16 Required probability =16/316=13 ...

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Standard XII

Mathematics

Method of Intervals

Two real numb...

Question

Two real numbers $a$ and $b$ are selected uniformly at random from the interval $[0, 4] .$ Let $R = {(a, b) : \frac{a^{2}}{4} \leq b \leq 2 \sqrt{a}} .$ The probability that the chosen numbers lie in $R$ is

\frac{1}{3}

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\frac{1}{2}

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\frac{2}{3}

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\frac{3}{4}

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Solution

The correct option is A $\frac{1}{3}$

Area of shaded region, $R = \int_{0}^{4} (2 \sqrt{x} - \frac{x^{2}}{4}) d x$
$= \frac{16}{3}$
Area of $O A B C = 4 \times 4 = 16$
Required probability $= \frac{16 / 3}{16} = \frac{1}{3}$

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Two real numbers a and b are selected uniformly at random from the interval [0,4]. Let R={(a,b):a24≤b≤2√a}. The probability that the chosen numbers lie in R is

The correct option is A 13 Area of shaded region, R=∫40(2√x−x24)dx =163 Area of OABC=4×4=16 Required probability =16/316=13

Two real numbers $a$ and $b$ are selected uniformly at random from the interval $[0, 4] .$ Let $R = {(a, b) : \frac{a^{2}}{4} \leq b \leq 2 \sqrt{a}} .$ The probability that the chosen numbers lie in $R$ is

The correct option is A $\frac{1}{3}$

Area of shaded region, $R = \int_{0}^{4} (2 \sqrt{x} - \frac{x^{2}}{4}) d x$
$= \frac{16}{3}$
Area of $O A B C = 4 \times 4 = 16$
Required probability $= \frac{16 / 3}{16} = \frac{1}{3}$