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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 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 > 4)
✓

Answer

i. Since $P(x)$ is a probability distribution of $X$,
$ \Sigma_{x=0}^7 P(x)=1$
$\therefore P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)=1$
$\therefore 0+ k +2 k +2 k +3 k + k ^2+2 k ^2+7 k ^2+ k =1$
$\therefore 10 k ^2+9 k -1=0$
$\therefore 10 k ^2+10 k - k -1=0$
$\therefore 10 k ( k +1)-1( k +1)=0$
$\therefore( k +1)(10 k -1)=0$
$\therefore 10 k -1=0$
$\therefore 10 k -1=0 \quad \ldots \ldots . .( k \neq-1)$
$\therefore k =\frac{1}{10} $
$ \text { ii. } P(X<3)=P(0)+P(1)+P(2)$
$=0+k+2 k$
$=3 k$
$=3\left(\frac{1}{10}\right)$
$=\frac{3}{10} $
iii. $P (0< X <3)=+ P (1)+ P (2)$
$ = k +2 k$
$=3 k$
$=3\left(\frac{1}{10}\right)$
$=\frac{3}{10} . $

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