Let X be a random variable which takes values | vedclass

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

Question

(C) The sum of probabilities for a random variable is $1$. Given $P(X=r) = K r^3$ for $r \in \{1, 2, 3, 4\}$. Summing these: $K(1^3) + K(2^3) + K(3^3)

Vedclass · Accepted Answer

$K = \frac{1}{100}$ and $P\left(\left.\frac{1}{2}  1\right) = \frac{8}{99}$

Vedclass · Answer

(C) The sum of probabilities for a random variable is $1$. Given $P(X=r) = K r^3$ for $r \in \{1, 2, 3, 4\}$. Summing these: $K(1^3) + K(2^3) + K(3^3) + K(4^3) = 1$. $K(1 + 8 + 27 + 64) = 1 \Rightarrow 100K = 1 \Rightarrow K = \frac{1}{100}$. We need to find $P\left(\left.\frac{1}{2}  1\right)$. This is equivalent to $P(X=2 \mid X \in \{2, 3, 4\})$. By the definition of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. Here $A = \{X=2\}$ and $B = \{X=2, 3, 4\}$. $P(A \cap B) = P(X=2) = 8K$. $P(B) = P(X=2) + P(X=3) + P(X=4) = 8K + 27K + 64K = 99K$. Thus,$P(A \mid B) = \frac{8K}{99K} = \frac{8}{99}$. Therefore,$K = \frac{1}{100}$ and the conditional probability is $\frac{8}{99}$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

Let $X$ be a random variable which takes values $1, 2, 3, 4$ such that $P(X=r) = K r^3$ where $r = 1, 2, 3, 4$. Then:

Similar Questions

Two dice are thrown and the sum of the numbers appearing on the dice is observed to be a multiple of $4$. If $p$ is the conditional probability that number $4$ has appeared at least once, then $3p + 2 =$

Two planes $I$ and $II$ drop bombs on a target. The probabilities of hitting the target by $I$ and $II$ are $0.3$ and $0.2$ respectively. The second plane drops the bomb only if the first plane fails to hit the target. What is the probability that the second plane hits the target?

If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$,then $P(A^{\prime} \cap B^{\prime})$ is

If $P(A \cap B) = \frac{7}{10}$ and $P(B) = \frac{17}{20}$,where $P$ stands for probability,then $P(A \mid B)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module