Class 11 Maths Probability Equally likely outcome

Probabilities of equally likely outcome: Let P(ωi) = p, for all ωi ∈ S where 0 ≤ p ≤ 1, then  p=1/n   where n= number of elements.

Probability of the event ‘A or B’

P(A∪B) =  P(A) + P(B) − P(A∩ B)

Probability of the event ‘A and B’

P(A∩B) =  P(A) + P(B) − P(AU B)

Probability of the event ‘Not A’

P( A′ ) = P(not A) = 1 – P(A)

Numerical: A bag contains 9 similar discs of which 4 are red, 3 are blue and 2 are green.. A disc is drawn at random from the bag. Calculate the probability that it will be  (i) red, (ii) green, (iii) blue, (iv) not blue, (v) either red or blue Solution: Total number of disc is 9

P(Red Disc) = Number of red Disc/ total number of Disc  = 4/9

P(Blue Disc) = Number of Blue Disc/ total number of Disc  = 3/9

P(Green Disc) = Number of Green Disc/ total number of Disc  = 2/9

P(Not Blue Disc) = 1 – P(Blue Disc) = 1- 3/9 = 6/9

P(Either red or Blue)  = P(red U Blue) = P(red) + P(Blue) – P(Red ∩ Blue)

= 4/9 + 3/9 – 0 = 7/9

Numerical: Two students A & B appeared in an examination. The probability that A will qualify examination is 0.05 & that B will qualify examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that (a) Both A and B will not qualify the examination. (b) At least one of them will not qualify the examination (c) Only one of them will qualify the examination. Solution: Given P(A) = .05 , P(B) = .10   & P(A ∩ B) = .02

Probability that Both A and B will not qualify the examination = P( AUB)’ = 1- P( AUB)

Probability that Both A and B will qualify the examination = P( AUB)

P(A∪B) =  P(A) + P(B) − P(A∩ B)

= 0.05 + 0.1 -0.02

= 0.13

P( AUB)’ = 1- P( AUB)  = 1- 0.13 = 0.87

Probability that at least one of them will not qualify the examination = 1- P(A ∩ B) = 1 -0.02 = 0.98

Probability that only one of them will qualify the examination = P( AUB) - P(A ∩ B)  = 0.13 – 0.02  = 0.11

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