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Maths-

General

Easy

Question

A random variable X has the following distribution

The value of $k$ and P(X<3) are equal to

$k = \frac{1}{10}, P (X < 3) = \frac{3}{5}$
$k = \frac{1}{10}, P (X < 3) = \frac{3}{10}$
$k = \frac{3}{10}, P (X < 3) = \frac{1}{10}$
$k = \frac{1}{24}, P (X < 3) = \frac{5}{12}$

Hint:

Sum of all probabilities is 1

The correct answer is: $k equals fraction numerator 1 over denominator 10 end fraction comma P left parenthesis X less than 3 right parenthesis equals fraction numerator 3 over denominator 10 end fraction$

Given :

Sum of all probabilities is 1
P(x=1) + P(x = 2) + P(x = 3) + P(x =4) = 1
k + 2k + 3k + 4k = 1
k = $1 over 10$
Now, P(x < 3) = P(x=1) + P(x =3)
P(x < 3) = k + 2k = 3k = $3 over 10$
Thus, the value of $k$ is $1 over 10$ and P(X<3) is $3 over 10$

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