Question 287952
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: {{{P=(3/10)^9= 1.9683*10^-5}}}!
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b) All are right-handed
Using the same approach as we did on (a), {{{P=(7/10)^9= 4035.3607*10^-5}}}, 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, {{{P=(3/10)^3*P(7/10)^5= 0.027+0.16807=0.19507}}}
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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