Search any question & find its solution
Question:
Answered & Verified by Expert
A random variable $X$ has the following probability distribution
For the event $E=\{X / X$ is a prime number $\}$ and the event $F=\{X / X < 4\}$, the probability $P(E \cup F)=$
Options:
For the event $E=\{X / X$ is a prime number $\}$ and the event $F=\{X / X < 4\}$, the probability $P(E \cup F)=$
Solution:
2827 Upvotes
Verified Answer
The correct answer is:
0.77
Given,
$\begin{aligned} & \Sigma P\left(x_i\right)=1 \\ & \begin{aligned} \therefore 0.15+ & 0.23+k+0.10+0.20+0.08+0.07+0.05=1 \\ K & =1-0.88=0.12 \\ E & =\{x: x \text { is prime number }\}=\{2,3,5,7\} \\ P(E) & =0.23+0.12+0.20+0.07=0.62 \\ F & =\{x: x < 4\}=\{1,2,3\}\end{aligned}\end{aligned}$
$\begin{aligned} P(F)=0.15 & +0.23+0.12=0.50, E \cap F=\{2,3\} \\ P(E \cap F) & =0.23+0.12=0.35 \\ P(E \cup F) & =P(E)+P(F)-P(E \cap F) \\ & =0.62+0.50-0.35=0.77\end{aligned}$
$\begin{aligned} & \Sigma P\left(x_i\right)=1 \\ & \begin{aligned} \therefore 0.15+ & 0.23+k+0.10+0.20+0.08+0.07+0.05=1 \\ K & =1-0.88=0.12 \\ E & =\{x: x \text { is prime number }\}=\{2,3,5,7\} \\ P(E) & =0.23+0.12+0.20+0.07=0.62 \\ F & =\{x: x < 4\}=\{1,2,3\}\end{aligned}\end{aligned}$
$\begin{aligned} P(F)=0.15 & +0.23+0.12=0.50, E \cap F=\{2,3\} \\ P(E \cap F) & =0.23+0.12=0.35 \\ P(E \cup F) & =P(E)+P(F)-P(E \cap F) \\ & =0.62+0.50-0.35=0.77\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.