100 numbers chosen on unit interval
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100 numbers are chosen from the interval [0, 1], independently of each other. What is the probability that the 2nd largest # is <1/2? My reasoning was a) all 100 chosen are less then 1/2
b) 99 are less then 1/2 and one is more
This would give (1/2)^100 OR the probability of 99 less then 1/2 and one greater . Would i not need binomial theorem since im also factoring the number if ways to CHOOSE the 99 numbers ??
probability
asked Dec 31 '18 at 9:02
RandinRandin
347116
$endgroup$
add a comment |
$begingroup$
100 numbers are chosen from the interval [0, 1], independently of each other. What is the probability that the 2nd largest # is <1/2? My reasoning was a) all 100 chosen are less then 1/2
b) 99 are less then 1/2 and one is more
This would give (1/2)^100 OR the probability of 99 less then 1/2 and one greater . Would i not need binomial theorem since im also factoring the number if ways to CHOOSE the 99 numbers ??
probability
asked Dec 31 '18 at 9:02
RandinRandin
347116
$endgroup$
2
$begingroup$
If the second largest number is less than $frac{1}{2}$ then you are looking for the probability that at most one number is greater than or equal to $frac{1}{2}$. This is equivalent to finding the probability that you toss at most $1$ head in $100$ tosses of a fair coin so yes, the binomial theorem is a good way to go.
$endgroup$
– John Douma
Dec 31 '18 at 9:09
add a comment |
$begingroup$
100 numbers are chosen from the interval [0, 1], independently of each other. What is the probability that the 2nd largest # is <1/2? My reasoning was a) all 100 chosen are less then 1/2
b) 99 are less then 1/2 and one is more
This would give (1/2)^100 OR the probability of 99 less then 1/2 and one greater . Would i not need binomial theorem since im also factoring the number if ways to CHOOSE the 99 numbers ??
probability
asked Dec 31 '18 at 9:02
RandinRandin
347116
$endgroup$
100 numbers are chosen from the interval [0, 1], independently of each other. What is the probability that the 2nd largest # is <1/2? My reasoning was a) all 100 chosen are less then 1/2
b) 99 are less then 1/2 and one is more
This would give (1/2)^100 OR the probability of 99 less then 1/2 and one greater . Would i not need binomial theorem since im also factoring the number if ways to CHOOSE the 99 numbers ??
probability
probability
asked Dec 31 '18 at 9:02
RandinRandin
347116
asked Dec 31 '18 at 9:02
RandinRandin
347116
asked Dec 31 '18 at 9:02
RandinRandin
347116
asked Dec 31 '18 at 9:02
RandinRandin
347116
asked Dec 31 '18 at 9:02
RandinRandin
347116
347116
2
$begingroup$
If the second largest number is less than $frac{1}{2}$ then you are looking for the probability that at most one number is greater than or equal to $frac{1}{2}$. This is equivalent to finding the probability that you toss at most $1$ head in $100$ tosses of a fair coin so yes, the binomial theorem is a good way to go.
$endgroup$
– John Douma
Dec 31 '18 at 9:09
add a comment |
2
$begingroup$
If the second largest number is less than $frac{1}{2}$ then you are looking for the probability that at most one number is greater than or equal to $frac{1}{2}$. This is equivalent to finding the probability that you toss at most $1$ head in $100$ tosses of a fair coin so yes, the binomial theorem is a good way to go.
$endgroup$
– John Douma
Dec 31 '18 at 9:09
2
2
$begingroup$
If the second largest number is less than $frac{1}{2}$ then you are looking for the probability that at most one number is greater than or equal to $frac{1}{2}$. This is equivalent to finding the probability that you toss at most $1$ head in $100$ tosses of a fair coin so yes, the binomial theorem is a good way to go.
$endgroup$
– John Douma
Dec 31 '18 at 9:09
$begingroup$
If the second largest number is less than $frac{1}{2}$ then you are looking for the probability that at most one number is greater than or equal to $frac{1}{2}$. This is equivalent to finding the probability that you toss at most $1$ head in $100$ tosses of a fair coin so yes, the binomial theorem is a good way to go.
$endgroup$
– John Douma
Dec 31 '18 at 9:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would go about it using order statistics. Let the 100 independent random variables are $X_1,,X_2,,cdots,,X_{100}$ where all are uniformly distributed in $[0,,1]$. Arrange them as $$X_{(1)}leq X_{(2)}leqcdotsleq X_{(99)}leq X_{(100)}$$ Then you need to find $$Prleft[X_{(99)}<frac{1}{2}right]=sum_{i=99}^{100}{100choose i}F_X^i(0.5)left[1-F_X(0.5)right]^{100-i}$$ where $F_X(x)$ is the CDF of the random variables.
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
$endgroup$
$begingroup$
@greedoid What part?
$endgroup$
– BlackMath
Jan 8 at 0:08
add a comment |
$begingroup$
Indeed binomial distribution with parameters $n=100$ and $p=0.5$.
Let's say that there is a success if the chosen number does not exceed $0.5$.
The event that the second largest does not exceed $0.5$ is the same as the event that there are at least $99$ successes.
So to be found is: $$P(Xgeq99)=P(X=99)+P(X=100)=0.5^{100}left(binom{100}{99}+binom{100}{100}right)=0.5^{100}cdot101$$
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
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add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
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$begingroup$
I would go about it using order statistics. Let the 100 independent random variables are $X_1,,X_2,,cdots,,X_{100}$ where all are uniformly distributed in $[0,,1]$. Arrange them as $$X_{(1)}leq X_{(2)}leqcdotsleq X_{(99)}leq X_{(100)}$$ Then you need to find $$Prleft[X_{(99)}<frac{1}{2}right]=sum_{i=99}^{100}{100choose i}F_X^i(0.5)left[1-F_X(0.5)right]^{100-i}$$ where $F_X(x)$ is the CDF of the random variables.
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
$endgroup$
$begingroup$
@greedoid What part?
$endgroup$
– BlackMath
Jan 8 at 0:08
add a comment |
$begingroup$
I would go about it using order statistics. Let the 100 independent random variables are $X_1,,X_2,,cdots,,X_{100}$ where all are uniformly distributed in $[0,,1]$. Arrange them as $$X_{(1)}leq X_{(2)}leqcdotsleq X_{(99)}leq X_{(100)}$$ Then you need to find $$Prleft[X_{(99)}<frac{1}{2}right]=sum_{i=99}^{100}{100choose i}F_X^i(0.5)left[1-F_X(0.5)right]^{100-i}$$ where $F_X(x)$ is the CDF of the random variables.
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
$endgroup$
$begingroup$
@greedoid What part?
$endgroup$
– BlackMath
Jan 8 at 0:08
add a comment |
$begingroup$
I would go about it using order statistics. Let the 100 independent random variables are $X_1,,X_2,,cdots,,X_{100}$ where all are uniformly distributed in $[0,,1]$. Arrange them as $$X_{(1)}leq X_{(2)}leqcdotsleq X_{(99)}leq X_{(100)}$$ Then you need to find $$Prleft[X_{(99)}<frac{1}{2}right]=sum_{i=99}^{100}{100choose i}F_X^i(0.5)left[1-F_X(0.5)right]^{100-i}$$ where $F_X(x)$ is the CDF of the random variables.
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
$endgroup$
I would go about it using order statistics. Let the 100 independent random variables are $X_1,,X_2,,cdots,,X_{100}$ where all are uniformly distributed in $[0,,1]$. Arrange them as $$X_{(1)}leq X_{(2)}leqcdotsleq X_{(99)}leq X_{(100)}$$ Then you need to find $$Prleft[X_{(99)}<frac{1}{2}right]=sum_{i=99}^{100}{100choose i}F_X^i(0.5)left[1-F_X(0.5)right]^{100-i}$$ where $F_X(x)$ is the CDF of the random variables.
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
answered Dec 31 '18 at 10:10
BlackMathBlackMath
31518
31518
$begingroup$
@greedoid What part?
$endgroup$
– BlackMath
Jan 8 at 0:08
add a comment |
$begingroup$
@greedoid What part?
$endgroup$
– BlackMath
Jan 8 at 0:08
$begingroup$
@greedoid What part?
$endgroup$
– BlackMath
Jan 8 at 0:08
$begingroup$
@greedoid What part?
$endgroup$
– BlackMath
Jan 8 at 0:08
add a comment |
$begingroup$
Indeed binomial distribution with parameters $n=100$ and $p=0.5$.
Let's say that there is a success if the chosen number does not exceed $0.5$.
The event that the second largest does not exceed $0.5$ is the same as the event that there are at least $99$ successes.
So to be found is: $$P(Xgeq99)=P(X=99)+P(X=100)=0.5^{100}left(binom{100}{99}+binom{100}{100}right)=0.5^{100}cdot101$$
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Indeed binomial distribution with parameters $n=100$ and $p=0.5$.
Let's say that there is a success if the chosen number does not exceed $0.5$.
The event that the second largest does not exceed $0.5$ is the same as the event that there are at least $99$ successes.
So to be found is: $$P(Xgeq99)=P(X=99)+P(X=100)=0.5^{100}left(binom{100}{99}+binom{100}{100}right)=0.5^{100}cdot101$$
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Indeed binomial distribution with parameters $n=100$ and $p=0.5$.
Let's say that there is a success if the chosen number does not exceed $0.5$.
The event that the second largest does not exceed $0.5$ is the same as the event that there are at least $99$ successes.
So to be found is: $$P(Xgeq99)=P(X=99)+P(X=100)=0.5^{100}left(binom{100}{99}+binom{100}{100}right)=0.5^{100}cdot101$$
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
$endgroup$
Indeed binomial distribution with parameters $n=100$ and $p=0.5$.
Let's say that there is a success if the chosen number does not exceed $0.5$.
The event that the second largest does not exceed $0.5$ is the same as the event that there are at least $99$ successes.
So to be found is: $$P(Xgeq99)=P(X=99)+P(X=100)=0.5^{100}left(binom{100}{99}+binom{100}{100}right)=0.5^{100}cdot101$$
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
answered Dec 31 '18 at 9:11
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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$begingroup$
If the second largest number is less than $frac{1}{2}$ then you are looking for the probability that at most one number is greater than or equal to $frac{1}{2}$. This is equivalent to finding the probability that you toss at most $1$ head in $100$ tosses of a fair coin so yes, the binomial theorem is a good way to go.
$endgroup$
– John Douma
Dec 31 '18 at 9:09