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3 Answers

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Best answer
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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5 Comments

Here, michelle has 3 ways to be in first three places right? Then shouldn't we multiply with 3 ?
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Yes you are right, but then we have to multiply denominator also by 3, and then 3 is canceled from numerator and denominator, resulting to same result 0.03.

But yes, we should multiply by 3 to consider all cases, although that doesn't change answer here.
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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?
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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.
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After choosing 3(100c3) people, we can order them in 6 ways right?
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2 votes
2 votes

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
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1 vote
1 vote
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