ST115 2017-2018 Assignment 1 Solutions (1) (1).pdf

# Therefore p b r g 1 p b c r c g c 1 131 476 345 476

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Therefore, P ( B R G ) = 1 - P ( B c R c G c ) = 1 - 131 476 = 345 476 . ALTERNATIVE SOLUTION: If among the five chosen sheets of paper every colour appears at least once, then either one colour appears three times and other two colours once, or two colours appear twice each and the remaining colour appears once. More precisely, consider the events A = { at least one sheet of paper of each colour is obtained } B = { one colour appears three times and other two colours once } C = { two colours appear twice each and the remaining colour appears once } . Then the events B and C are disjoint, A = B C and, by the finite additivity, P ( A ) = P ( B ) + P ( C ) . To compute P ( B ), we reason in the following way: there are three choices for the colour that appears three times, ( 6 3 ) choices of the sheets of that colour, ( 6 1 ) choices of the sheet in the second colour and 3
( 6 1 ) choices of the sheet in the third colour. Also, the total number of choices of choosing five sheets out of 18 is ( 18 5 ) . Thus, P ( B ) = 3 × ( 6 3 )( 6 1 )( 6 1 ) ( 18 5 ) = 3 × 6 × 6 × 5 × 4 18 × 17 × 14 × 2 = 6 × 5 17 × 7 = 30 17 × 7 . For P ( C ), we observe that there are ( 3 1 ) choices of the colour that appears once, ( 6 1 ) choices of the sheet in that colour, ( 6 2 ) choices of the sheets in the second colour and ( 6 2 ) choices of the sheets in the third colour. Hence, P ( C ) = ( 3 1 )( 6 1 )( 6 2 )( 6 2 ) ( 18 5 ) = 3 × 6 × 5 × 3 × 5 × 3 18 × 17 ×

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