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A committee of five will be chosen from 3 referees, 4 umpires and 6 scorers. What is the probability that there will be at least 3

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A committee of five will be chosen from 3 referees, 4 umpires and 6 scorers. What is the probability that there will be at least 3 scorers in the committee?

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Maria

(2cos^2A-1)^2÷cos^4A-sin^4=1-2sin^2A
(2 \cos(2) a – 1) {}^{2} \div \cos(4) – \sin(4) = 1 – 2 \sin(2) a(2cos(2)a−1)2÷cos(4)−si

There is an auditorium with 25 rows of seats there are 10seat in the first row, 12 in the second row, 14 seat in the third row and

1 thought on “A committee of five will be chosen from 3 referees, 4 umpires and 6 scorers. What is the probability that there will be at least 3”

  1. Answer :

    • 5 member committee make out of
    • 3 referees+ 4 umpires+ 6 scorers = 13 member

    # total selection = 13c5 = 1287

    at least 3 scorer (x = 3,4,5) valid

    case 1 = 3 scorer + 2 (referee+umpire)

    case 2 = 4 scorer + 1 ( referee / umpire)

    caee 3 = all 5 scorer

    = 6c3×7c2 + 6c2×7c1 + 6c5

    = 420 + 105 + 6

    = 531

    Answer = 531/1287

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