JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 14)
The probability distribution of random variable X is given by :
Let p = P(1 < X < 4 | X < 3). If 5p = $$\lambda$$K, then $$\lambda$$ equal to ___________.
| 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
30
Explanation
$$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$$
$$ \Rightarrow k = {1 \over 9}$$
Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$
$$ \Rightarrow p = {2 \over 3}$$
Now, $$5p = \lambda k$$
$$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$$
$$ \Rightarrow \lambda = 30$$
$$ \Rightarrow k = {1 \over 9}$$
Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$
$$ \Rightarrow p = {2 \over 3}$$
Now, $$5p = \lambda k$$
$$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$$
$$ \Rightarrow \lambda = 30$$
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