Probability For Competitive Exam | Buy Now |
In an interview the probability of Praveen to got selected is 0.4. The probability of Geetha to got selected is 0.5. The probability of Sam to got selected is 0.6. The probability of Suresh to got selected is 0.8. What is the probability that at least 2 of them got selected on that day?
Required probability=1 – no one got selected – 1 got selected No one got selected = (1-0.4) x (1-0.5) x (1-0.6) x (1-0.8) = 0.024 1 got selected= 0.4 x ((1-0.5) x (1-0.6) x (10.8) ) + 0.5 x ((1-0.4) x (1-0.6) x (1-0.8)) +0.6 x ((1-0.4) x (1-0.5) x (1-0.8)) + 0.8 x ((1-0.4) x (1-0.5) x (1-0.6)) = 0.016 + 0.024 + 0.036 + 0.096 = 0.17 So, Required probability = 1 – 0.024 – 0.17 = 0.806
A basket contains 10 red ball and 15 white ball. out of which 3 red and 4 white balls are damaged. If two balls selected at random, what is the probability that either both are white balls or both are not damaged?
P(A) = 15c2 / 25c2, P(B) = 18c2 / 25c2 P(A∩B) = 11c2 / 25c2 P(A∪B) = P(A) + P(B) – P(A∩B) => (15c2 / 25c2)+( 18c2 / 25c2)-( 11c2 / 25c2)=406/600==>203/300
Two dice are thrown simultaneously. What is the probability of getting the sum of the numbers as even?
Throw two dice n(s)=36 E is nos sum is even. Hence E={( 1,1 ),(1,3 ) (1,5 ) (2,2 ) , (2,4 ) , (2,6 ), …………( 6,2 ), (6,4 ), ( 6,6 )} n(E)= 18 Thus required probability= 18/36= 1/2
12 persons are seated around a round table.What is the probability that two particular persons sit together?
In a circle of n different persons, the total number of arrangements possible = (n – 1)! n(S) = (12 – 1) = 11 ! Taking two persons as a unit, total persons = 11 Therefore no. of ways for these 11 persons to around the circular table = (11 – 1)! = 10! In any unit, 2 particular person can sit in 2! ways. Hence total number of ways that any three person can sit, =n(E) = 10! * 2! Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = (10! * 2!)/11! = 2/11.
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