probability

Probability Level 3

A committee of four members is to be selected at random from among six labor delegates and four management representatives. What is the probability that the committee will have at least two labor representatives?


The answer is 0.88.

Mohammad Ahmad by Mohammad Ahmad , 26, Pakistan

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2 solutions

Krit Phuengphan
Jun 1, 2014

P ( E ) = N ( E ) N ( S ) P(E)=\frac{N(E)}{N(S)}

P ( E ) = C 6 , 2 C 4 , 2 + C 6 , 3 C 4 , 1 + C 6 , 4 C 10 , 4 P(E)=\frac{C_{6,2}C_{4,2}+C_{6,3}C_{4,1}+C_{6,4}}{C_{10,4}}

P ( E ) = 90 + 80 + 15 210 = 37 42 0.881 P(E)=\frac{90+80+15}{210}=\frac{37}{42}~\boxed{0.881}

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I thought it in this way. Total possibilities=10 choose 4=210. Ways with 0 labor = 1. Ways with 1 labor = 6. So ans =(210-1-6)/210=203/210. Please point out my fault.Anyway your ans is correct.

Chandrachur Banerjee - 6 years, 12 months ago
Unstable Chickoy
Jun 15, 2014

P = ∑ x = 2 4 ( C 6 , x ) ( C 4 , 4 − x ) C 10 , 4 P = \frac{\sum_{x=2}^4{(C_{6,x})(C_{4,4-x})}}{C_{10,4}}

P = 37 42 = 0.88095 P = \frac{37}{42} = \boxed{0.88095}

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