Q76 / 160 · AP EAMCET 2015 · Mathematics

The probability distribution of a random variable is given below Figure
Then, $P(0 < \mathrm{x} < 5)$ is equal to

  • A
    $\frac{1}{10}$
  • B
    $\frac{3}{10}$
  • Correct answer
    C
    $\frac{8}{10}$
  • D
    $\frac{7}{10}$

Solution

As we know, $\sum_{i=1}^8 P\left(x_i\right)=1$ $\begin{aligned} & 0+K+2 K+2 K+3 K+K^2+2 K^2+7 K^2+K=1 \\ & 9 K+10 K^2=1 \\ & 10 K^2+9 K-1=0 \\ & 10 K^2+10 K-K-1=0 \\ & 10 K(K+1)-1(K+1)=0 \\ & (K+1)(10 K-1)=0 \\ & \therefore \quad K=-1, \frac{1}{10} \end{aligned}$ As the probability cannot be negative. So $\mathrm{K}$ must be greater than 0 . $\begin{aligned} & \therefore \quad K=\frac{1}{10} \\ & P(0 < x < 5)=P(X=1)+P(X=2) \\ & +P(X=3)+P(X=4) \\ & =K+2 K+2 K+3 K=8 K \\ & =\frac{8}{10} \end{aligned}$

Hence, the correct answer is option: 'C'.

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