MCQ
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:
  • A
    $K.$
  • B
    $P(X < 3).$
  • C
    $P(X > 6).$
  • $P(0 < X < 3).$

Answer

Correct option: D.
$P(0 < X < 3).$
Here $k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k =1 $
$\Rightarrow 10k^2 + 9k –1 = 0$
$\Rightarrow (10k – 1) (k + 1) = 0$
$\Rightarrow k = \frac{1}{10}$
$\therefore\text{ (i) k =}\frac{1}{10}$
$(ii) P(x < 3) = 0 + k + 2k = 3k = \frac{3}{10}$
$(iii) P(x > 6) = 7k^2 + k = \frac{7}{100}+\frac{1}{10}=\frac{17}{100}$
$(iv) P(0 < x < 3) = k + 2k = 3k = \frac{3}{10}$

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