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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$
Determine
  1. $k$
  2. $P(X < 3)$
  3. $P(X > 6)$
  4. $P(0 < X < 3)$
✓

Answer

  1. Since, the sum of all the probabilities of a distribution is $1.$
$\therefore P(X = 0) + P(X = 1) + …. + P(X = 7) = 1$
$\Rightarrow 0 + k + 2k + 2k + 3k + k^2 + 2k^{2 }+ 7k^{2 }+ k = 1$
$\Rightarrow 10k^2 + 9k - 1 = 0$
$\Rightarrow (10k - 1) (k + 1) = 0$
$\Rightarrow 10k - 1 = 0$ or $k + 1 = 0$
$\Rightarrow\ \text{k}=\frac{1}{10}$ or $k = - 1$
Since, $k \geq 0,$ therefore $k = − 1$ is not possible.
$\therefore\ \text{k}=\frac{1}{10}$
  1. $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$
$= 0 + k + 2k$
$=3\text{k}=3\times\frac{1}{10}=\frac{3}{10}$
  1. $P(X > 6) = P(X = 7)$
$=7\text{k}^2+\text{k}=7\Big(\frac{1}{10}\Big)^2+\frac{1}{10}=\frac{17}{100}$
  1. $P(0 < X < 3) = P(X = 1) + P(X = 2)$
$= k + 2k = 3k = \frac{3}{10}$

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