Probability of being first, second, or third in a contest
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Question in the book:
Suppose that 100 people enter a contest and that different
winners are selected at random for first, second, and third
prizes. What is the probability that Michelle wins one of
these prizes if she is one of the contestants?
My work:
Being first: 1/100
Being second: 1/99
Being third: 1/98
So total probablity is 1/100 + 1/99 + 1/98
Book Answer:
But the book says 3/100, I understand why kind of, but I don't understand what I did wrong in my work?
probability combinatorics
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
$endgroup$
add a comment |
$begingroup$
Question in the book:
Suppose that 100 people enter a contest and that different
winners are selected at random for first, second, and third
prizes. What is the probability that Michelle wins one of
these prizes if she is one of the contestants?
My work:
Being first: 1/100
Being second: 1/99
Being third: 1/98
So total probablity is 1/100 + 1/99 + 1/98
Book Answer:
But the book says 3/100, I understand why kind of, but I don't understand what I did wrong in my work?
probability combinatorics
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
$endgroup$
$begingroup$
You can draw a parallel to this problem: You think of three numbers between 1 and 100. What is the probability that I can guess one of them right on the first guess? Maybe by looking at the problem like this, you can see what the right and wrong approaches are.
$endgroup$
– Matti P.
Dec 5 '18 at 13:49
$begingroup$
Why should the probability to become second in this (weird) contest be less than the probability to become first? What if there are e.g. only $3$ contestants? Try to grasp!
$endgroup$
– drhab
Dec 5 '18 at 13:50
add a comment |
$begingroup$
Question in the book:
Suppose that 100 people enter a contest and that different
winners are selected at random for first, second, and third
prizes. What is the probability that Michelle wins one of
these prizes if she is one of the contestants?
My work:
Being first: 1/100
Being second: 1/99
Being third: 1/98
So total probablity is 1/100 + 1/99 + 1/98
Book Answer:
But the book says 3/100, I understand why kind of, but I don't understand what I did wrong in my work?
probability combinatorics
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
$endgroup$
Question in the book:
Suppose that 100 people enter a contest and that different
winners are selected at random for first, second, and third
prizes. What is the probability that Michelle wins one of
these prizes if she is one of the contestants?
My work:
Being first: 1/100
Being second: 1/99
Being third: 1/98
So total probablity is 1/100 + 1/99 + 1/98
Book Answer:
But the book says 3/100, I understand why kind of, but I don't understand what I did wrong in my work?
probability combinatorics
probability combinatorics
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
asked Dec 5 '18 at 13:45
2000mroliver2000mroliver
132
132
$begingroup$
You can draw a parallel to this problem: You think of three numbers between 1 and 100. What is the probability that I can guess one of them right on the first guess? Maybe by looking at the problem like this, you can see what the right and wrong approaches are.
$endgroup$
– Matti P.
Dec 5 '18 at 13:49
$begingroup$
Why should the probability to become second in this (weird) contest be less than the probability to become first? What if there are e.g. only $3$ contestants? Try to grasp!
$endgroup$
– drhab
Dec 5 '18 at 13:50
add a comment |
$begingroup$
You can draw a parallel to this problem: You think of three numbers between 1 and 100. What is the probability that I can guess one of them right on the first guess? Maybe by looking at the problem like this, you can see what the right and wrong approaches are.
$endgroup$
– Matti P.
Dec 5 '18 at 13:49
$begingroup$
Why should the probability to become second in this (weird) contest be less than the probability to become first? What if there are e.g. only $3$ contestants? Try to grasp!
$endgroup$
– drhab
Dec 5 '18 at 13:50
$begingroup$
You can draw a parallel to this problem: You think of three numbers between 1 and 100. What is the probability that I can guess one of them right on the first guess? Maybe by looking at the problem like this, you can see what the right and wrong approaches are.
$endgroup$
– Matti P.
Dec 5 '18 at 13:49
$begingroup$
You can draw a parallel to this problem: You think of three numbers between 1 and 100. What is the probability that I can guess one of them right on the first guess? Maybe by looking at the problem like this, you can see what the right and wrong approaches are.
$endgroup$
– Matti P.
Dec 5 '18 at 13:49
$begingroup$
Why should the probability to become second in this (weird) contest be less than the probability to become first? What if there are e.g. only $3$ contestants? Try to grasp!
$endgroup$
– drhab
Dec 5 '18 at 13:50
$begingroup$
Why should the probability to become second in this (weird) contest be less than the probability to become first? What if there are e.g. only $3$ contestants? Try to grasp!
$endgroup$
– drhab
Dec 5 '18 at 13:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The probability of being second/third is not $frac 1{99}, frac 1{98}$. Then, is the probability of coming $99$th $frac{1}{100-99} = 1$? Clearly not,right?
Choosing a position for Michelle does not depend upon what position it is. The probability that she comes first, is equal to the probability that she comes second, is equal to the probability that she comes third, because in each case, we have to choose one place out of hundred : whether that is the first place, or second place, or third place, or seventy-seventh place, it is still one place out of hundred. Therefore, the probability of each of these events is $frac 1{100}$, and given that coming first,second and third are mutually exclusive events gives the answer $frac 3{100}$.
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The probability of being second/third is not $frac 1{99}, frac 1{98}$. Then, is the probability of coming $99$th $frac{1}{100-99} = 1$? Clearly not,right?
Choosing a position for Michelle does not depend upon what position it is. The probability that she comes first, is equal to the probability that she comes second, is equal to the probability that she comes third, because in each case, we have to choose one place out of hundred : whether that is the first place, or second place, or third place, or seventy-seventh place, it is still one place out of hundred. Therefore, the probability of each of these events is $frac 1{100}$, and given that coming first,second and third are mutually exclusive events gives the answer $frac 3{100}$.
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
$endgroup$
add a comment |
$begingroup$
The probability of being second/third is not $frac 1{99}, frac 1{98}$. Then, is the probability of coming $99$th $frac{1}{100-99} = 1$? Clearly not,right?
Choosing a position for Michelle does not depend upon what position it is. The probability that she comes first, is equal to the probability that she comes second, is equal to the probability that she comes third, because in each case, we have to choose one place out of hundred : whether that is the first place, or second place, or third place, or seventy-seventh place, it is still one place out of hundred. Therefore, the probability of each of these events is $frac 1{100}$, and given that coming first,second and third are mutually exclusive events gives the answer $frac 3{100}$.
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
$endgroup$
add a comment |
$begingroup$
The probability of being second/third is not $frac 1{99}, frac 1{98}$. Then, is the probability of coming $99$th $frac{1}{100-99} = 1$? Clearly not,right?
Choosing a position for Michelle does not depend upon what position it is. The probability that she comes first, is equal to the probability that she comes second, is equal to the probability that she comes third, because in each case, we have to choose one place out of hundred : whether that is the first place, or second place, or third place, or seventy-seventh place, it is still one place out of hundred. Therefore, the probability of each of these events is $frac 1{100}$, and given that coming first,second and third are mutually exclusive events gives the answer $frac 3{100}$.
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
$endgroup$
The probability of being second/third is not $frac 1{99}, frac 1{98}$. Then, is the probability of coming $99$th $frac{1}{100-99} = 1$? Clearly not,right?
Choosing a position for Michelle does not depend upon what position it is. The probability that she comes first, is equal to the probability that she comes second, is equal to the probability that she comes third, because in each case, we have to choose one place out of hundred : whether that is the first place, or second place, or third place, or seventy-seventh place, it is still one place out of hundred. Therefore, the probability of each of these events is $frac 1{100}$, and given that coming first,second and third are mutually exclusive events gives the answer $frac 3{100}$.
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
answered Dec 5 '18 at 13:52
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.8k33376
37.8k33376
add a comment |
add a comment |
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$begingroup$
You can draw a parallel to this problem: You think of three numbers between 1 and 100. What is the probability that I can guess one of them right on the first guess? Maybe by looking at the problem like this, you can see what the right and wrong approaches are.
$endgroup$
– Matti P.
Dec 5 '18 at 13:49
$begingroup$
Why should the probability to become second in this (weird) contest be less than the probability to become first? What if there are e.g. only $3$ contestants? Try to grasp!
$endgroup$
– drhab
Dec 5 '18 at 13:50