A random variable X has the following probability distribution: X : 1 2 3 4 5 P X : k 2 2 k k 2 k 5 k 2 Then, P X > 2 is equal to:

W e know ∑ P i = 1 ⇒ 6 k 2 + 5 k = 1 6 k 2 + 5 k - 1 = 0 6 k 2 + 6 k - k - 1 = 0 6 k - 1 k + 1 = 0 ⇒ k = - 1 (rejected) ; k = 1 6 P X > 2 = k + 2 k + 5 k 2 = 1 6 + 2 6 + 5 36 = 6 + 12 + 5 36 = 23 36

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MathematicsProbabilityRandom Variable and its Probability DistributionMedium2 minPYQ_2020

MathematicsMediumsingle choice

A random variable $X$ has the following probability distribution:
$\begin{matrix} \begin{matrix} X : & 1 & \begin{matrix} 2 & 3 & \begin{matrix} 4 & 5 \end{matrix} \end{matrix} \end{matrix} \\ P (X) : \begin{matrix} k^{2} & \begin{matrix} 2 k & k & \begin{matrix} 2 k & 5 k^{2} \end{matrix} \end{matrix} \end{matrix} \end{matrix}$
Then, $P (X > 2)$ is equal to:

Options:

Answer:

Solution:

W e know $\sum P_{i} = 1 \Rightarrow 6 k^{2} + 5 k = 1$

$6 k^{2} + 5 k - 1 = 0$

$6 k^{2} + 6 k - k - 1 = 0$

$(6 k - 1) (k + 1) = 0 \Rightarrow k = - 1$ (rejected) ; $k = \frac{1}{6}$

$P (X > 2) = k + 2 k + 5 k^{2}$

$= \frac{1}{6} + \frac{2}{6} + \frac{5}{36} = \frac{6 + 12 + 5}{36} = \frac{23}{36}$

Stream:JEESubject:MathematicsTopic:ProbabilitySubtopic:Random Variable and its Probability Distribution

⏱ 2mℹ️ Source: PYQ_2020

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