Question:
Of the students in a college,it is known that \(60\%\) reside in hostel and \(40\%\) are day scholers.(not residing in hostel).Previous year result report that \(30\%\) of all students who reside in hostel attain A grade and \(20\%\) of day scholars attain A grade in their annual examination.At the end of the year,one student is choosen at random from the college and he has an A grade,What is the probability that the student is a hostler?
Of the students in a college,it is known that \(60\%\) reside in hostel and \(40\%\) are day scholers.(not residing in hostel).Previous year result report that \(30\%\) of all students who reside in hostel attain A grade and \(20\%\) of day scholars attain A grade in their annual examination.At the end of the year,one student is choosen at random from the college and he has an A grade,What is the probability that the student is a hostler?
Updated On: Sep 20, 2023
Show Solution
Verified By Collegedunia
Solution and Explanation
The correct answer is: \(\frac{9}{13}\)
Let \(E_1\)=Students residing in the hostel, \(E_2\)=day scholars(not residing in the hostel) and A=Student who attain grade A,
\(P(E_1)=\frac{60}{100}, P(E_2)=\frac{40}{100}\)
\(P(A|E_1)=0.3,P(A|E_2)=0.2\)
The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by \(P(E_1|A)\)
Therefore,by Bayes'theorem,
\(P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{0.6\times 0.3}{0.6\times 0.3+0.4\times 0.2}\)
\(=\frac{0.18}{0.26}\)
\(=\frac{18}{26}\)
\(=\frac{9}{13}\)
Let \(E_1\)=Students residing in the hostel, \(E_2\)=day scholars(not residing in the hostel) and A=Student who attain grade A,
\(P(E_1)=\frac{60}{100}, P(E_2)=\frac{40}{100}\)
\(P(A|E_1)=0.3,P(A|E_2)=0.2\)
The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by \(P(E_1|A)\)
Therefore,by Bayes'theorem,
\(P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{0.6\times 0.3}{0.6\times 0.3+0.4\times 0.2}\)
\(=\frac{0.18}{0.26}\)
\(=\frac{18}{26}\)
\(=\frac{9}{13}\)
Was this answer helpful?
0
0
Top CBSE CLASS XII Mathematics Questions
- 2, b, c are in A.P. and the range of determinant $\begin{vmatrix}1&1&1\\ 2&b&c\\ 4&b^{2}&c^{2}\end{vmatrix}$ is [2, 16]. Then find range of c is
- A committee of 11 members is to be formed from 8 males and 5 females. Let m denotes the number of ways of selecting committee having atleast 6 males and n denotes the number of ways of selecting atleast 3 females then
Determine whether each of the following relations are reflexive, symmetric, and transitive.
- Relation R in the set A={1,2,3,...13,14} defined as R={(x,y): 3x-y=0}.
- Relation R in the set of N natural numbers defined as R={(x,y): y=x+5 and x<4}.
- Relation R in the set A={1,2,3,4,5,6} defined as R={(x,y): y is divisible by x}.
- Relation R in the set of Z integers defined as R={(x,y): x-y is an integer}
- Relation R in the set of human beings in a town at a particular time given by
- R={(x,y): x and y work at same place}
- R={(x,y): x and y live in the same locality}
- R={(x,y): x is exactly 7 am taller that y}
- R={(x,y): x is wife of y}
- R={(x,y):x is father of y}
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
View More Questions
Top CBSE CLASS XII Probability Questions
- A urn contains 5 red and 5 black balls.A ball is drawn at random,its colour is noted and is returned to the urn.Morever,2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random.What is the probability that the second ball is red?
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find:
- P(A∩B)
- P(A|B)
- P(A∪B)
- A bag contains 4 red and 4 black balls,another bag contains 2 red and 6 black balls.One of the two bags is selected at random and a ball is drawn from the bag which is found to be red.Find the probability that the ball is drawn from the first bag.
\(Evaluate \ P(A∩B)\ if \ 2P(A) = P(B) =\) \(\frac {5}{13}\) \(and \ P(A|B)=\) \(\frac 25\)
Determine P: A coin is tossed three times, where
- E: heads on third toss,F:heads on first two tosses.
- E: at least two heads,F:at most two heads.
- E: at most two tails,F:at least one tail.
View More Questions
Top CBSE CLASS XII Questions
- 2, b, c are in A.P. and the range of determinant $\begin{vmatrix}1&1&1\\ 2&b&c\\ 4&b^{2}&c^{2}\end{vmatrix}$ is [2, 16]. Then find range of c is
- A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is
- A capillary tube of radius r is dipped inside a large vessel of water. The mass of water raised above water level is M. If the radius of capillary is doubled, the mass of water inside capillary will be
- A circular disc is rotating about its own axis. An external opposing torque 0.02 Nm is applied on the disc by which it comes rest in 5 seconds. The initial angular momentum of disc is
- A circular disc is rotating about its own axis at uniform angular velocity \(\omega.\) The disc is subjected to uniform angular retardation by which its angular velocity is decreased to \(\frac {\omega}{2}\) during 120 rotations. The number of rotations further made by it before coming to rest is
View More Questions
Concepts Used:
Bayes Theorem
Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
- Events A and B are mutually exhaustive events.
- P(A) and P(B) are the probabilities of events A and B, respectively.
- P(A|B) is the conditional probability of the happening of event A, given that event B has happened.
- P(B|A) is the conditional probability of the happening of event B, given that event A has already happened.
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.