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Please answer ASAP. I cannot get this right no matter how carefully I try to solve it lol

Please answer ASAP. I cannot get this right no matter how carefully I try to solve-example-1

1 Answer

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When we get a question with "probability of at least two the same" we should think of it as 1 minus the probability of all different.


Rather than dazzle you with chooses and factorials let's work our way up to it.


If there was only one lady, there's probability 1 no two will choose the same style.


With two ladies, the second has a 13/14 th chance of choosing a different coat.


With three ladies , the probability of all different is
(13)/(14) \cdot (12)/(14)


With eight ladies, we get seven fractions. We'll do 1 minus to get "two or more the same":




p = 1 - (13)/(14) \cdot (12)/(14)\cdot \frac {11}{14} \cdot (10)/(14) \cdot (9)/(14) \cdot (8)/(14) \cdot (7)/(14) = 1 - (13!/6!)/14^7 \approx 0.918


Second choice


5 boys 12 watch styles


P(Two boys different),=11/12


P(Three boys different),=11/12 * 10/12


P(Five boys different) = 11*10*9*8/12^4 = .3819


P(two of five match) = 1 - 11*10*9*8/12^4 = .6181


last choice




User Yianna
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