JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 14)
A random variable X has the following
probability distribution :
Then P(X > 2) is equal to :
| X: | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X): | K2 | 2K | K | 2K | 5K2 |
Then P(X > 2) is equal to :
$${1 \over {6}}$$
$${7 \over {12}}$$
$${1 \over {36}}$$
$${23 \over {36}}$$
Explanation
$$\sum\limits_{i = 1}^5 {P(X)} $$ = 1
$$ \Rightarrow $$ K2 + 2K + K + 2K + 5K2 = 1
$$ \Rightarrow $$ 6K2 + 5K – 1 = 0
$$ \Rightarrow $$ (6K - 1)(k + 1) = 0
$$ \Rightarrow $$ K = $${1 \over 6}$$ and K = -1(rejected)
$$ \therefore $$ P(X $$ > $$ 2)
= K + 2K + 5K2
= $${1 \over 6} + {2 \over 6} + {5 \over {36}}$$
= $${{23} \over {36}}$$
$$ \Rightarrow $$ K2 + 2K + K + 2K + 5K2 = 1
$$ \Rightarrow $$ 6K2 + 5K – 1 = 0
$$ \Rightarrow $$ (6K - 1)(k + 1) = 0
$$ \Rightarrow $$ K = $${1 \over 6}$$ and K = -1(rejected)
$$ \therefore $$ P(X $$ > $$ 2)
= K + 2K + 5K2
= $${1 \over 6} + {2 \over 6} + {5 \over {36}}$$
= $${{23} \over {36}}$$
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