Question

In a class of $ 125 $ students, $ 70 $ of them passed in English, $ 55 $ passed in mathematics and $ 30 $ passed in both. Find the probability that the student selected at random from the class has passed in at least one subject.
A. $ \dfrac{{19}}{{25}} $
B. $ \dfrac{3}{{25}} $
C. $ \dfrac{{17}}{{25}} $
D. $ \dfrac{8}{{25}} $

Answer 1

Hint: When the probability of event is to be calculated in which at least is mentioned then the probability should be calculated by subtracting in 1 the probability of nonoccurrence of an event. . $ P(Atleast) = 1 - P\left( 0 \right) $ .
Where, P(0) is the nonoccurrence of an event and $ P(atleast) = $ Probability of occurrence of an event at least once.

Complete step-by-step answer:
Total number of students in the class is, $ N = 125 $
Number of students passed in English is, $ E = 70 $ .
Number of students passed in mathematics is, $ M = 55 $
Number of students passed in English and mathematics both is, $ B = 30 $
The probability of an event that student selected at random has passed in at least one subject is given by,
 $ P(A) = $ Probability of number of students passed in only one subject + Probability of students passed in both the subjects
It is also equal to,
P(A)=1-P(F)
Where, $ P(F) = $ Number of students failed
The number of students passed in English only is ,
 $
  {E_n} = 70 - 30 \\
  {E_n} = 40 \\
  $
The number of students passed in mathematics only is ,
 $
  {M_n} = 55 - 30 \\
  {M_n} = 25 \\
  $
In order to calculate the number of students failed , subtract the number of students passed in English only + Number of students passed in mathematics only + number of students passed in both subjects from the total number of students.
 $
  F = 125 - 30 - 40 - 25 \\
  F = 30 \\
  $
Probability of number of students failed, $ P(F) = \dfrac{{30}}{{125}} $
Substitute the value of $ P(F) = \dfrac{{30}}{{125}} $ in equation (1), to calculate the number of students that passed in at least one subject as,
 $
  P(A) = 1 - \dfrac{{30}}{{125}} \\
  P(A) = \dfrac{{125 - 30}}{{125}} \\
  P(A) = \dfrac{{95}}{{125}} \\
  P(A) = \dfrac{{19}}{{25}} \\
  $
So, the correct answer is “Option A”.

Note: The important thing is the understanding and calculation of the number of students who have failed.
Complementary events are those which have only two occurrences. For instance, getting passed in an examination and getting failed in an examination.
 $ P(p) + P(\bar p) = 1 $
 $ P(\bar p) = 1 - P(p) $
Where,
 $ P(p) = $ Probability of getting passed in an examination.
 $ P(\bar p) = $ Probability of getting failed in an examination.