A random variable X has the following probabi | vedclass

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

Question

(A) For any probability distribution, the sum of all probabilities must be $1$.$\sum P(X = x_i) = 10k + 9k + 8k + 8k + 6k + 5k + 4k + 3k + k = 54k = 1

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) For any probability distribution, the sum of all probabilities must be $1$.$\sum P(X = x_i) = 10k + 9k + 8k + 8k + 6k + 5k + 4k + 3k + k = 54k = 1$.Therefore, $k = \frac{1}{54}$.The event $A$ consists of prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, which are $\{2, 3, 5, 7\}$.$P(A) = P(2) + P(3) + P(5) + P(7) = 9k + 8k + 6k + 4k = 27k$.The event $B$ consists of values $x_i > 5$, which are $\{6, 7, 8, 9\}$.$P(B) = P(6) + P(7) + P(8) + P(9) = 5k + 4k + 3k + k = 13k$.The intersection $A \cap B$ consists of values that are both prime and greater than $5$, which is $\{7\}$.$P(A \cap B) = P(7) = 4k$.Using the addition theorem of probability, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = 27k + 13k - 4k = 36k$.Substituting $k = \frac{1}{54}$, we get $P(A \cup B) = 36 	imes \frac{1}{54} = \frac{36}{54} = \frac{2}{3}$.

$X = x_i$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$P(X = x_i)$	$10k$	$9k$	$8k$	$8k$	$6k$	$5k$	$4k$	$3k$	$k$

$X=x$	$1$	$2$	$3$
$P(X=x)$	$1/5$	$2/5$	$2/5$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

Similar Questions

For the following probability distribution of a random variable $X$, the Expected value $E(X)$ and Variance $Var(X)$ of $X$ are respectively:
$X=x$ $1$ $2$ $3$
$P(X=x)$ $1/5$ $2/5$ $2/5$

$A$ random variable $X$ has the following probability distribution. For events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,what is the probability $P(E \cup F)$?

$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

In a communication network, $98 \%$ of messages are transmitted with no error. If a random variable $X$ denotes the number of incorrectly transmitted messages, then the probability that at most one message is transmitted incorrectly out of $500$ messages sent, is

Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable $X$ to be the number of rotten apples in a draw of two apples,the variance of $X$ is

If $X$ is a Poisson variate with $P(X=0)=0.8$,then the variance of $X$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ random variable $X$ has the following probability distribution:$X = x_i$$1$$2$$3$$4$$5$$6$$7$$8$$9$$P(X = x_i)$$10k$$9k$$8k$$8k$$6k$$5k$$4k$$3k$$k$where $k$ is a real number. If $A = \{ x_i : x_i \text{ is a prime number} \}$ and $B = \{ x_i : x_i > 5 \}$ are two events, then $P(A \cup B) = $