Question

14) In a class of 125 students. 70 passed in Mathematics,
55 passed in Statistics and 30 passed in both What is the
probability that a student selected at random from the class
has passed in only one subject?
13
A)
25
3
B)
25
17
C)
25
8
D)
25​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

Given

• Total strength of class = 125

• Number of students passed in Mathematics = 70

• Number of students passed in Statistics = 55

• Number of students passed in both = 30

Let A represents the students passed in Mathematics

and

Let B represents the students passed in Statistics.

So,

$\rm :\implies\:P(A) = \dfrac{70}{125}$

$\rm :\implies\:P(B) = \dfrac{55}{125}$

$\rm :\implies\:P(A \: \cap \: B) = \dfrac{30}{125}$

Now,

• We have to find the probability that student selected at random has passed only one subject.

So 2 cases arises

Case 1.

• A student passed Mathematics but failed in Statistics.

The probability of this is given by

$\rm :\implies\:P(A \: \cap \: \overline {B}) = P(A) - P(A \cap \: B)$

$\rm :\implies\:P(A \: \cap \: \overline {B}) =\dfrac{70}{125} - \dfrac{30}{125}$

$\rm :\implies\:P(A \: \cap \: \overline {B}) =\dfrac{40}{125}$

Case 2.

• A student failed in Mathematics but passed in Statistics.

The probability of this is given by

$\rm :\implies\:P(B \: \cap \: \overline {A}) =P(B) - P(A \: \cap \: B)$

$\rm :\implies\:P(B \: \cap \: \overline {A}) =\dfrac{55}{125} - \dfrac{30}{125}$

$\rm :\implies\:P(B \: \cap \: \overline {A}) =\dfrac{25}{125}$

So, required probability that student selected at random has passed only one subject is given by

$\rm :\implies\:P(A \: \cap \: \overline {B}) + P(B \: \cap \: \overline {A})$

$\rm :\implies\:\dfrac{40}{125} + \dfrac{25}{125}$

$\rm :\implies\:\dfrac{65}{125}$

$\rm :\implies\:\dfrac{13}{25}$