AP EAMCET · Maths · Probability

For the random variable $X$ with probability distribution is given by the table

$X = x$ $0$ $1$ $2$ $3$

$P (X = x)$ $K$ $K + \frac{1}{7}$ $2 K$ $\frac{2}{5}$

The mean of $X$ is

A $\frac{31}{35}$
B $\frac{57}{35}$
C $\frac{63}{35}$
D $\frac{67}{35}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{67}{35}$

Step-by-step Solution

Detailed explanation

Given the distribution, X=x 0 1 2 3 P(X=x) K K+17 2 K 25 As we know sum of all probabilities of an event is unity, ⇒K+K+17+2K+25=1 4K=1-17+25=1-1935 4K=1635 ⇒K=435 X=x 0 1 2 3 P(X=x) 435 935 835 25 Mean…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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$X = x$	$0$	$1$	$2$	$3$
$P (X = x)$	$K$	$K + \frac{1}{7}$	$2 K$	$\frac{2}{5}$

For the random variable X with probability distribution is given by the table X=x 0 1 2 3 P(X=x) K K+17 2 K 25 The mean of X is

Step-by-step Solution

For the random variable $X$ with probability distribution is given by the table

$X = x$ $0$ $1$ $2$ $3$

$P (X = x)$ $K$ $K + \frac{1}{7}$ $2 K$ $\frac{2}{5}$

The mean of $X$ is