The probability of A, B, C solving a problem are 1/3, 2/7 and 3/8 respectively. -Maths 9th

The probability of A, B, C solving a problem are 1/3, 2/7 and 3/8 respectively. -Maths 9th
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Let E1, E2, E3 be the eventsthat the problem issolved by A, B, C respectively and let p1, p2, p3 be corresponding probabilities. Then,p1 = P(E1) = \(rac{1}{3}\), p2 = P(E2) = \(rac{2}{7}\),  p3 = P(E3) = \(rac{3}{8}\), q1 = P(\(\bar{E}_1\)) = 1 - \(rac{1}{3}\) = \(rac{2}{3}\),q2 = P(\(\bar{E}_2\)) = 1 - \(rac{2}{7}\) = \(rac{5}{7}\), q3 = P(\(\bar{E}_3\)) = 1 - \(rac{3}{8}\) = \(rac{5}{8}\).The problem will be solved by exactly one of them if it happens in the following mutually exclusive ways: (1) A solves and B, and C do not solve; (2) B solves and A, and C do not solve; (3) C solves and A, and B do not solve; Required probability = p1 q2 q3 + q1 p2 q3 + q1 q2 p3= \(rac{1}{3}\) x \(rac{5}{7}\) x \(rac{5}{8}\) + \(rac{2}{3}\) x \(rac{2}{7}\) x \(rac{5}{8}\) + \(rac{2}{3}\) x \(rac{5}{7}\) x \(rac{3}{8}\) = \(rac{25}{168}\) + \(rac{5}{42}\) + \(rac{5}{28}\) = \(rac{25}{56}.\)
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Last Answer : The chosen consists of players (6 + 5). ∴ Number of ways of selecting 11 players out of 15 players = n(S) = 15C11 Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers Then, n(A) = 8C6 ... 7 imes6}{2}\) x \(rac{4 imes3 imes2 imes1}{15 imes14 imes13 imes12}\) = \(rac{28}{65}.\)

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Last Answer : This answer was deleted by our moderators...

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Last Answer : This answer was deleted by our moderators...

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