A random variable X has the following
probability distribution :
| X: | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X): | K2 | 2K | K | 2K | 5K2 |
Then P(X > 2) is equal to :
Solution
$\sum\limits_{i = 1}^5 {P(X)}$ = 1
<br><br>$\Rightarrow$ K<sup>2</sup>
+ 2K + K + 2K + 5K<sup>2</sup>
= 1
<br><br>$\Rightarrow$ 6K<sup>2</sup>
+ 5K – 1 = 0
<br><br>$\Rightarrow$ (6K - 1)(k + 1) = 0
<br><br>$\Rightarrow$ K = ${1 \over 6}$ and K = -1(rejected)
<br><br>$\therefore$ P(X $>$ 2)
<br><br>= K + 2K + 5K<sup>2</sup>
<br><br>= ${1 \over 6} + {2 \over 6} + {5 \over {36}}$
<br><br>= ${{23} \over {36}}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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