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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Probability

NDA 2010 · Mathematics · Probability

The probability distribution of random variable X with two missing...

The probability distribution of random variable X with two missing probabilities p1p_1 and p2p_2 is given below:
It is further given that P(X≤2)=0.25\mathrm{P}(\mathrm{X} \leq 2)=0.25 and P(X≥4)=0.35\mathrm{P}(\mathrm{X} \geq 4)=0.35. Consider the following statements: $\begin{array}{ll}\text{ I. } p_1=p_2 & \text{ II. } p_1+p_2=\mathrm{P}(\mathrm{X}=3)\end{array} Which of the statements given above is/are correct?$

Options

  1. AI only
  2. BII only
  3. CBoth I and II
  4. DNeither I nor II

Correct answer

D. Neither I nor II

Step-by-step solution

Subtracting equation (ii) from equation (i) k+P1=0.25...(i)2k+P2=0.35...(ii)−−−−k+P1−P2=−0.10\begin{aligned} k+\mathrm{P}_1 & =0.25 ...(i)\\ 2 k+\mathrm{P}_2 & =0.35...(ii) \\ -\quad- & - \\ \hline-k+\mathrm{P}_1-\mathrm{P}_2 & =-0.10 \end{aligned}k+\mathrm{P}_2-\mathrm{P}_1=0.10...(iii)...(iii)k+P1+4k+P2+2k=17k+P1+P2=1...(iv)\begin{aligned} & k+\mathrm{P}_1+4 k+\mathrm{P}_2+2 k=1 \\ & 7 k+\mathrm{P}_1+\mathrm{P}_2=1...(iv) \end{aligned}Addingtheequations(iii)and(iv)Adding the equations (iii) and (iv)7k+P2+P1=1k+P2−P1=0.108k+2P2=1.1\begin{aligned} 7 k+\mathrm{P}_2+\mathrm{P}_1 & =1 \\ k+\mathrm{P}_2-\mathrm{P}_1 & =0.10 \\ \hline 8 k+2 \mathrm{P}_2 & =1.1 \end{aligned}$\begin{aligned} & 4 k+\mathrm{P}_2=0.55 \\ & \mathrm{P}_2=0.55-4 k \end{aligned}From(ii)From (ii)\begin{gathered} 2 k+0.55-4 k=0.35 \\ k=0.10 \end{gathered}NowNowP2=0.55−4×.10P2=0.15\begin{aligned} & \mathrm{P}_2=0.55-4 \times .10 \\ & \mathrm{P}_2=0.15 \end{aligned}From(i),From (i),\mathrm{P}_1=0.25-k\Rightarrow P_1=0.15$
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