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100 Students Appeared for Two Examination, 60 Passed the First, 50 Passed the Second and 30 Passed Both. Find the Probability that a Student Selected at Random Has Passed at Least One Examination.

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Question

100 students appeared for two examination, 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has passed at least one examination.

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Solution

Let S be the sample space associated with the experiment of students who appeared for two examination.
Then n(S) = 100
∴ Total number of elementary events = 100
Consider the following events:
A = students passed in first examination
B = students passed in second examination
Then n(A) = 60 and n(B) = 50 and n(A ∩ B) = 30

\[\therefore P\left( A \right) = \frac{60}{100}, P\left( B \right) = \frac{50}{100}\]  and
\[P\left( A \cap B \right) = \frac{30}{100}\]
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) -P (A ∩ B)
               = \[\frac{50}{100} + \frac{60}{100} - \frac{30}{100} = \frac{50 + 60 - 30}{100} = \frac{80}{100} = \frac{4}{5}\]
Hence, the probability that a student selected at random has passed at least one examination is \[\frac{4}{5} .\]
 

 

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Concept of Probability - Probability of 'Not', 'And' and 'Or' Events
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Chapter 33: Probability - Exercise 33.4 [Page 68]

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R.D. Sharma Mathematics [English] Class 11
Chapter 33 Probability
Exercise 33.4 | Q 16 | Page 68

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