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junior college 2 | H2 Maths
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Date Posted: 1 year ago
Views: 216
C = 1 implies 1B2R or 1W1B2R
P(C=1) = 2/6 x 3/5 x 2/4 x 2! + 1/6 x 2/5 x 3/4 x 2/3 x 3! (factorial term is for permutation of all balls less the final red one)
= 1/10 x 2! + 1/30 x 3!
= 2/5
C = 0 implies 2R or 1W2R
P(C=0) = 3/6 x 2/5 + 1/6 x 3/5 x 2/4 x 2!
= 1/5 + 1/20 x 2!
= 3/10
C = 2 implies 2B2R or 1W2B2R
Fast method
P(C=0) + P(C=1) + P(C=2) = 1
P(C=2) = 1 - P(C=0) - P(C=1) = 3/10
Long method
P(C=2) = 2/6 x 1/5 x 3/4 x 2/3 x 3!/2! + 1/6 x 2/5 x 1/4 x 3/3 x 2/2 x 4!/2!
= 1/30 x 3 + 1/60 x 12
= 3/10
P(C=1) = 2/6 x 3/5 x 2/4 x 2! + 1/6 x 2/5 x 3/4 x 2/3 x 3! (factorial term is for permutation of all balls less the final red one)
= 1/10 x 2! + 1/30 x 3!
= 2/5
C = 0 implies 2R or 1W2R
P(C=0) = 3/6 x 2/5 + 1/6 x 3/5 x 2/4 x 2!
= 1/5 + 1/20 x 2!
= 3/10
C = 2 implies 2B2R or 1W2B2R
Fast method
P(C=0) + P(C=1) + P(C=2) = 1
P(C=2) = 1 - P(C=0) - P(C=1) = 3/10
Long method
P(C=2) = 2/6 x 1/5 x 3/4 x 2/3 x 3!/2! + 1/6 x 2/5 x 1/4 x 3/3 x 2/2 x 4!/2!
= 1/30 x 3 + 1/60 x 12
= 3/10
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Date Posted:
1 year ago
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