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Probability Ncert Solutions Maths class 11th Maths Class 11th Probability

44. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted fo rboth NCC and NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.

(ii) The student has opted neither NCC nor NSS.

(iii) The student has opted NSS but not NCC.

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Answered by

alok kumar singh | Contributor-Level 10

4 months ago

44. Given that, total number of student, n (S) = 60

Let A: student opted for NCC

n (A) = 30

B: student opted for NSS

n (B) = 32

And student who opted both NCC and NSS, n (A∩B) = 24

(i) Probability that student opted for NCC or NSS,

P (A∪B) = P (A) + P (B) – P (A∩B)

$= \frac{n ()}{n ()} + \frac{n ()}{n ()} - \frac{n (A \capB)}{n ()}$

$= \frac{30}{60} + \frac{32}{60} - \frac{24}{60}$

$= \frac{30 + 32 - 24}{60} = \frac{38}{60} = \frac{19}{30}$

(ii) Probability that student opted neither NCC or NSS

P (not A and not B) = P (A'∩B') = P (A∪B)' = 1 – P (A∪B)

$= 1 - \frac{19}{30} = \frac{30 - 19}{30} = \frac{11}{30}$

(iii) Probabilities that student opted NSS but not NCC

P (B but not A) = P (B) – P (A∩B)

$= \frac{32}{30} - \frac{24}{30}$

$= \frac{32 - 24}{30} = \frac{8}{30} = \frac{2}{15}$

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