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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

$P = \frac{\binom{99}{2}}{\binom{100}{3}} = \frac{\frac{99*98}{2}}{\frac{100*99*98}{6}} = 0.03$

Why should the denominator be multiplied with 3?

P(Ist Place)  = $\frac{\binom{99}{2} \times 1}{\binom{100}{3}}$

P(2nd Place)  = $\frac{\binom{99}{2} \times 1}{\binom{100}{3}}$

P(3rd Place)  = $\frac{\binom{99}{2} \times 1}{\binom{100}{3}}$

so, totally,

$\frac{\binom{99}{2} \times 3}{\binom{100}{3}}$

right?

Where am I wrong?
Because, when you do 100C3, then you only choose 3 people, now after choosing, you can order them in 3 ways, so total number of ways would be 3*100C3.
After choosing 3(100c3) people, we can order them in 6 ways right?

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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probability of winning is not equally likely how can u assume it to be same for all the three
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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