Question

A random variable \[X\] has the probability distribution

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


For the events \[E=\{X\text{ is prime number }\!\!\}\!\!\text{ }\] and \[F=\{X<4\}\], the probability of \[P(E\cup F)\] is
A. \[0.50\]
B. \[0.77\]
C. \[0.35\]
D. \[0.87\]

Answer 1

Hint: In the above question, we are asked to calculate the probability of occurrence of certain outcomes in the events $E$ and $F$. In order to get the result of the given expression we should know the concept of a probability distribution.



Formula Used:Consider $S$ be a sample space of a random experiment. A real-valued function $X: S\to R$ is called a random variable. The random variable \[X\] is said to be discrete or discontinuous if the range of \[X\] is countable and is said to be continuous if the range of \[X\] is not countable.
If $X:S\to R$ is a discrete random variable with range $\{{{x}_{1}},{{x}_{2}},{{x}_{3}},...\}$ then $\sum\limits_{r=1}^{\infty }{P(X={{x}_{r}})=1}$
If $X:S\to R$ is a discrete random variable with the range $\{{{x}_{1}},{{x}_{2}},{{x}_{3}},...\}$ then $\{P(X={{x}_{r}}):r=1,2,...\}$ is called probability distribution of \[X\].



Complete step by step solution:We are given a probability distribution table that contains probabilities of different outcomes of \[X\], which we can use to answer the given question.
Given that,
\[\begin{align}
  & P(X=1)=0.15 \\
 & P(X=2)=0.23 \\
 & P(X=3)=0.12 \\
 & P(X=4)=0.10 \\
 & P(X=5)=0.20 \\
 & P(X=6)=0.08 \\
 & P(X=7)=0.07 \\
 & P(X=8)=0.05 \\
\end{align}\]
We are given in the statements that for the event $E$, \[X\] is a prime number. On observing the probability distribution table, we get for the event $E$, the values of \[X\] will be 2, 3, 5, and 7.
\[\begin{align}
  & P(E)=P(X=2)+P(X=3)+P(X=5)+P(X=7) \\
 & \text{ }=0.23+0.12+0.20+0.07 \\
 & \text{ }=0.62\text{ }...\text{(1)} \\
\end{align}\]
Similarly, for the event \[F\], we are given that \[X\] is less than 4. On taking the probability distribution table we get that the values of \[X\] for event \[F\] will be 1, 2, and 3.
\[\begin{align}
  & P(F)=P(X=1)+P(X=2)+P(X=3) \\
 & \text{ }=0.15+0.23+0.12 \\
 & \text{ }=0.50\text{ }...\text{(2)} \\
\end{align}\]
The common outcomes of the events $E$ and \[F\] are 2 and 3.
\[\begin{align}
  & P(E\cap F)=P(X=2)+P(X=3) \\
 & \text{ }=0.23+0.12 \\
 & \text{ }=0.35\text{ }...\text{(3)} \\
\end{align}\]
We know the formula of the relationship between union and intersection of events, and putting the values from equation (1), (2), and (3),
\[\begin{align}
  & P(E\cup F)=P(E)+P(F)-P(E\cap F) \\
 & \text{ }=0.62+0.50-0.35 \\
 & \text{ }=0.77 \\
\end{align}\]



Option ‘B’ is correct

Note:We can see that by combining the data from the probability distribution table and the formula of the relationship between the union and intersection of two events, we can obtain the required value of the expression.