Let X be a random variable which takes values 1 , 2 , 3 , 4 such that P ( X = r ) = K r 3 where r = 1 , 2 , 3 , 4 then

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Aicharya · Accepted Answer

According to the given condition, we have random variables

$11, 22, 33 and 44$

As, $P (X = r) = K r^{3}$ where, $r = 1, 2, 3, 4$

Hence,

$P (X = 1) = K {(1)}^{3} = K$ ,

$P (X = 2) = K {(2)}^{3} = 8 K$ ,
$P (X = 3) = K {(3)}^{3} = 27 K$ and

$P (X = 4) = K {(4)}^{3} = 64 K$
As, we know, here

$\sum_{1}^{4} P (X = r) = 1$ (sum of probabilities)

$\Rightarrow K + 8 K + 27 K + 64 K = 1 \Rightarrow K = \frac{1}{100}$

Now,

$P (\frac{1}{2} < X < \frac{5}{2} ∣ X > 1) = \frac{P [\{1, 2\} \cap \{2, 3, 4\}]}{P \{2, 3, 4\}} = \frac{P (X = 2)}{P (X = 2) + P (X = 3) + P (X = 4)}$

$= \frac{P (X = 2)}{P (X = 2) + P (X = 3) + P (X = 4)} = \frac{8 K}{8 K + 27 K + 64 K} = \frac{8}{99}$

Let X be a random variable which takes values 1 , 2 , 3 , 4 such that P ( X = r ) = K r 3 where r = 1 , 2 ,…

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