Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).
Assertion (A): If a random variable X assumes values 0, 1, 2, 3 such that $2P(X = 0) = 3 P(X = 1) = 4 P(X = 2) = 5P(X = 3)$, then the probability distribution of X is
|
X |
0 |
1 |
2 |
3 |
|
P(X) |
$\frac{30}{77}$ |
$\frac{20}{77}$ |
$\frac{15}{77}$ |
$\frac{12}{77}$ |
Reason (R): If a random variable X assumes values $x_1, x_2,...,x_n$, then $P(X = x_1)+P(X = x_2)+...+P(X = x_n) = 1$.
Select the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Let $2P(X = 0) = 3P(X = 1) = 4P(X = 2) = 5P(X = 3) = k$
$⇒P(X = 0) = \frac{k}{2}, P(X = 1) = \frac{k}{3}, P(X = 2) =\frac{k}{4}$ and $P(X = 3) =\frac{k}{5}$
$∵P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1$
$⇒\frac{k}{2}+\frac{k}{3}+\frac{k}{4}+\frac{k}{5}=1$
$⇒\frac{30k+20k+15k+12k}{60}=1⇒k=\frac{60}{77}$
Hence, $P(X = 0) =\frac{k}{2}⇒P(X = 0) =\frac{30}{77}$
$P(X = 1) =\frac{k}{3}⇒P(X = 1) =\frac{20}{77}$
$P(X = 2) =\frac{k}{4}⇒P(X = 2) =\frac{15}{77}$
$P(X = 3) =\frac{k}{5}⇒P(X = 3) =\frac{12}{77}$
So, the probability distribution of X is
|
X |
0 |
1 |
2 |
3 |
|
P(X) |
$\frac{30}{77}$ |
$\frac{20}{77}$ |
$\frac{15}{77}$ |
$\frac{12}{77}$ |
∴ Assertion is true.
Also, Reason is true.
Reason is the correct explanation of Assertion.
∴ Option (1) is the correct answer.