Maths Questions

Solution : Since the probability of solving the problem by A, B and C is \(1\over 2\), \(1\over 3\) and \(1\over 4\) respectively. \(\therefore\)  Probability that problem is not solved is = P(A’)P(B’)P(C’) = (\(1 – {1\over 2}\))(\(1 – {1\over 3}\))(\(1 – {1\over 4}\)) = \(1\over 2\)\(2\over 3\)\(3\over 4\) = \(1\over 4\) = Hence, the […]

Solution : The probability that Mr A selected the lossing horse = \(4\over 5\) \(\times\) \(3\over 4\) = \(3\over 5\) The probability that Mr A selected the winning horse = 1 – \(3\over 5\) = \(2\over 5\) Similar Questions Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an

Solution : Sample space of rolling two dice = 36 Now, Event of sum of 5 = { (1,4) (2,3) (3,2) (4,1) } = 4 \(\implies\) probability of getting sum of 5 is 4/36 = 1/9 Now, Event of even sum = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6)

Solution : Given, probabilities of speaking truth are P(A) = \(4\over 5\) and P(B) = \(3\over 4\) And their corresponding probabilities of not speaking truth are P(A’) = \(1\over 5\)  and P(B’) = \(1\over 4\) The probability that they contradict each other = P(A).P(B’) + P(B).P(A’) = \(4\over 5\) \(\times\) \(1\over 4\) + \(1\over 5\)

Solution : \(\therefore\)  Total number of cases = \(3^3\) Now, favourable cases = 3(as either all has applied for house 1 or 2 or 3) \(\therefore\) Required Probability = \(3\over 3^3\) = \(1\over 9\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to