
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
164.7k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Advanced 2025 Notes
