Consider the probability distributionbegin{ar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$\sum P(X=x) = K + 2K + K^2 + 2K + 5K^2 = 1$.Combining l

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) The sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$\sum P(X=x) = K + 2K + K^2 + 2K + 5K^2 = 1$.Combining like terms,we get $6K^2 + 5K - 1 = 0$.Factoring the quadratic equation: $(6K - 1)(K + 1) = 0$.This gives $K = \frac{1}{6}$ or $K = -1$.Since probability cannot be negative,we must have $K = \frac{1}{6}$.We need to find $P(X > 2) = P(X=3) + P(X=4) + P(X=5)$.$P(X > 2) = K^2 + 2K + 5K^2 = 6K^2 + 2K$.Substituting $K = \frac{1}{6}$: $P(X > 2) = 6(\frac{1}{6})^2 + 2(\frac{1}{6}) = 6(\frac{1}{36}) + \frac{2}{6} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

Consider the probability distribution
$\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline P(X=x) & K & 2K & K^2 & 2K & 5K^2 \\ \hline \end{array}$
Then the value of $P(X > 2)$ is

Similar Questions

$A$ player tosses $2$ fair coins. He wins $Rs. 5$ if $2$ heads appear,$Rs. 2$ if $1$ head appears,and $Rs. 1$ if no head appears. Then,the variance of his winning amount is

Find the probability distribution of the number of tails in the simultaneous tosses of three coins.

In a book of $500$ pages, it is found that there are $250$ typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of $2$ pages will contain no error, is:

If the probability of a bad reaction from a vaccination is $0.01$, then the probability that exactly two out of $300$ people will get a bad reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Consider the probability distribution$\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline P(X=x) & K & 2K & K^2 & 2K & 5K^2 \\ \hline \end{array}$Then the value of $P(X > 2)$ is