The probability distribution of a random variable x is given below.
|
X |
0 |
1 |
2 |
3 |
|
P(x) |
k/3 |
k/2 |
k/4 |
k/7 |
Then the value of k is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{84}{103}$
Given:
$P(x) = \frac{k}{3}, \frac{k}{2}, \frac{k}{4}, \frac{k}{7}$
Total probability = 1
$\Rightarrow \frac{k}{3} + \frac{k}{2} + \frac{k}{4} + \frac{k}{7} = 1$
Take LCM = 84
$\Rightarrow k\left(\frac{28 + 42 + 21 + 12}{84}\right) = 1$
$\Rightarrow k\left(\frac{103}{84}\right) = 1$
$\Rightarrow k = \frac{84}{103}$
Required value of $k = \frac{84}{103}$