The probability distribution of random variable X is given by :
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X) | K | 2K | 2K | 3K | K |
Let p = P(1 < X < 4 | X < 3). If 5p = $\lambda$K, then $\lambda$ equal to ___________.
Answer (integer)
30
Solution
$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$<br><br>$\Rightarrow k = {1 \over 9}$<br><br>Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$<br><br>$\Rightarrow p = {2 \over 3}$<br><br>Now, $5p = \lambda k$<br><br>$\Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$<br><br>$\Rightarrow \lambda = 30$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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