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Total number of ways in which any 3 random people can be chosen from 100 is $\binom{100}{3}$.

Now in our favourable cases, Michelle wins the prize, so in all favourable cases, Mitchelle is one of the 3 persons in the set which is chosen for winning. Now other 2 persons can be chosen in $\binom{99}{2}$ ways, so probability is

$P = \frac{\binom{99}{2}}{\binom{100}{3}} = \frac{\frac{99*98}{2}}{\frac{100*99*98}{6}} = 0.03$
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Michelle can wins first prize or second prize or third prize.

Probability of winning first prize =$\frac{1}{100}$

Probability of winning second prize = $\frac{1}{100}$

Probability of winning Third prize= $\frac{1}{100}$

Probability of Michelle can win any of these prize= $\frac{1}{100}$ +$\frac{1}{100}$  + $\frac{1}{100}$

=$\frac{3}{100}$ =0.03

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1 votes
Use binomial coefficients to compute the probability. Choose one of the prizes for Michelle and from the remaining 99 contestants choose the two other winners

(1 c 1)(99 c 2)/(100 c 3) = 3/100

= 0.3

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