Question 287952: Assume that 30% of the population is left-handed and the remainder is right-handed (there are no ambidextrous individuals). If you stop the next nine people you meet, what is the probability of the following situations? For the purposes of this problem, assume independence in the selection of the nine individuals.
(a) all will be left-handed
(b) all will be right-handed
(c) exactly three will be left-handed
(d) at least two will be left-handed
Answer by amnd(23)   (Show Source):
You can put this solution on YOUR website! Since 30% (3/10) of the population is left-handed, 70% (7/10) will be right-handed. Thus, the probability of meeting a person who is left-handed is 0.3 (or 3/10) and right-handed 0.7 (or 7/10).
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a) All are left-handed
Assuming that all of the 9 meetings ("events") are independent of each other, the multiplication rule applies [^ = AND]:
P(1^2^3 ... ^9)=P(1)*P(2)*...*P(9)
Since all of them have the same probability of happening (0.3), P would be 3/10*3/10...*3/10 nine times, or:  !
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b) All are right-handed
Using the same approach as we did on (a),  , yielding a quite significantly bigger amount.
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c) Exactly three will be left-handed
Since the order doesn't matter, the multiplication rule still applies. Say that P(1)=P(2)=P(3)=0.3 (for the left-handed) while P(4) to P(9)=0.7 (for the rest, who would be right-handed).
Therefore, 
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d) At least two will be left-handed
We only need two individuals to be left-handed; as for the rest, it matters not whether they're left or right-handed. Therefore, the probability of the seven other people being either left or right-handed would be 1.
P=(0.3)*(0.3)*1...*1=0.09
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