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MHT CET · Maths · Probability

If the p.m.f. of a r.v. \(\mathrm{X}\) is given by \(\mathrm{P}(\mathrm{X}=x)=\frac{\left(\begin{array}{l}5 \ x\end{array}\right)}{2^{5}}\) if \(x=0,1,2, \ldots \ldots 5\) \(=0\) otherwise then which of the following is not true?

  1. A \(P(X \leq 1)=P(X \geq 4)\)
  2. B \(P(X \leq 2) \geq P(X \geq 4)\)
  3. C \(P(X \leq 3) \leq P(X \geq 3)\)
  4. D \(P(X \leq 2)=P(X \geq 3)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(P(X \leq 3) \leq P(X \geq 3)\)

Step-by-step Solution

Detailed explanation

(B)
\(P(X \leq 2) =P(X=0)+P(X=1)+P(X=2) \)
\( =\frac{{ }^{5} C_{0}}{2^{5}}+\frac{{ }^{5} C_{1}}{2^{5}}+\frac{{ }^{5} C_{2}}{2^{5}}=\frac{1+5+10}{2^{5}}=\frac{16}{2^{5}} \)
\( P(X \geq 4) =P(X=4)+P(X=5) \)
\( =\frac{C_{4}}{2^{4}}+\frac{C_{6}}{2^{6}}=\frac{5+1}{2^{5}}=\frac{6}{2^{5}}\)
\(\mathrm{P}(\mathrm{X} \leq 3) =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)~+\) \(\mathrm{P}(\mathrm{X}=3) \)
\( =\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}=\frac{1+5+10+10}{2^{5}}=\frac{26}{2^{5}} \)
\( \mathrm{P}(\mathrm{X} \geq 3) =\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \)
\( =\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}=\frac{10+5+1}{2^{5}}=\frac{16}{2^{5}} \)
\( \mathrm{P}(\mathrm{X} \leq 1) =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \)
\( =\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}=\frac{1+5}{2^{5}}=\frac{6}{2^{5}}\)