$P(A_k ∪ B_n−k)$, for $k = 0, 1, . . . , n$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Distribution of # balls in each container (balls dropped randomly)Proof: Probability using InductionSampling from weighted sum distributionFind Combined Probability of one Die and two Coins TossedDo probability using combinationHow to account for revealed cards in a model of a poorly shuffled deck?Urn with an infinite number of balls of finite, uniformly distributed coloursExpected Balls in Bins of Unequal CapacitiesProbability that sum of three digits is the same as sum of other three digitsAn algorithm randomly generates a sequence $(c_1, c_2,… c_n)$
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$P(A_k ∪ B_n−k)$, for $k = 0, 1, . . . , n$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Distribution of # balls in each container (balls dropped randomly)Proof: Probability using InductionSampling from weighted sum distributionFind Combined Probability of one Die and two Coins TossedDo probability using combinationHow to account for revealed cards in a model of a poorly shuffled deck?Urn with an infinite number of balls of finite, uniformly distributed coloursExpected Balls in Bins of Unequal CapacitiesProbability that sum of three digits is the same as sum of other three digitsAn algorithm randomly generates a sequence $(c_1, c_2,… c_n)$
$begingroup$
An algorithm randomly generates a sequence $(c_1, c_2,... c_n)$, where each $c_i$ can assume the values $0$, $1$ or $2$. Given $A_k =$ $k$ values of the sequence are equal to $0$ and $B_j$ = $j$ values of the sequence are equal to $1$.
How to calculate $P (A_k)$, for $k = 0, 1,. . . , n$?
I did this:
$$P(A_k)=binomnkp^kq^n-k.$$
How to calculate $P(A_k ∪ B_n−k)$, for $k = 0, 1, . . . , n$?
I don't know how to calculate this.
probability statistics
asked Mar 27 at 20:02
user658422user658422
63
$endgroup$
add a comment |
$begingroup$
An algorithm randomly generates a sequence $(c_1, c_2,... c_n)$, where each $c_i$ can assume the values $0$, $1$ or $2$. Given $A_k =$ $k$ values of the sequence are equal to $0$ and $B_j$ = $j$ values of the sequence are equal to $1$.
How to calculate $P (A_k)$, for $k = 0, 1,. . . , n$?
I did this:
$$P(A_k)=binomnkp^kq^n-k.$$
How to calculate $P(A_k ∪ B_n−k)$, for $k = 0, 1, . . . , n$?
I don't know how to calculate this.
probability statistics
asked Mar 27 at 20:02
user658422user658422
63
$endgroup$
$begingroup$
What does $p$ and $q$ mean? Is $p=P(c_i=0)$?
$endgroup$
– Milten
Mar 27 at 20:18
$begingroup$
$p$ is the probability of success and it's 1/3, q is the probability of unsuccess and it's 2/3
$endgroup$
– user658422
Mar 27 at 20:30
$begingroup$
Well, $A_k=B_n-k$ Is not it?
$endgroup$
– Phicar
Mar 27 at 20:31
1
$begingroup$
@Phicar No. That would be true if the $c_i$ could only be 0 or 1.
$endgroup$
– Milten
Mar 27 at 20:56
$begingroup$
@Milten Thanks, i didn't read that.
$endgroup$
– Phicar
Mar 27 at 21:10
add a comment |
$begingroup$
An algorithm randomly generates a sequence $(c_1, c_2,... c_n)$, where each $c_i$ can assume the values $0$, $1$ or $2$. Given $A_k =$ $k$ values of the sequence are equal to $0$ and $B_j$ = $j$ values of the sequence are equal to $1$.
How to calculate $P (A_k)$, for $k = 0, 1,. . . , n$?
I did this:
$$P(A_k)=binomnkp^kq^n-k.$$
How to calculate $P(A_k ∪ B_n−k)$, for $k = 0, 1, . . . , n$?
I don't know how to calculate this.
probability statistics
asked Mar 27 at 20:02
user658422user658422
63
$endgroup$
An algorithm randomly generates a sequence $(c_1, c_2,... c_n)$, where each $c_i$ can assume the values $0$, $1$ or $2$. Given $A_k =$ $k$ values of the sequence are equal to $0$ and $B_j$ = $j$ values of the sequence are equal to $1$.
How to calculate $P (A_k)$, for $k = 0, 1,. . . , n$?
I did this:
$$P(A_k)=binomnkp^kq^n-k.$$
How to calculate $P(A_k ∪ B_n−k)$, for $k = 0, 1, . . . , n$?
I don't know how to calculate this.
probability statistics
probability statistics
asked Mar 27 at 20:02
user658422user658422
63
asked Mar 27 at 20:02
user658422user658422
63
asked Mar 27 at 20:02
user658422user658422
63
asked Mar 27 at 20:02
user658422user658422
63
asked Mar 27 at 20:02
user658422user658422
63
63
$begingroup$
What does $p$ and $q$ mean? Is $p=P(c_i=0)$?
$endgroup$
– Milten
Mar 27 at 20:18
$begingroup$
$p$ is the probability of success and it's 1/3, q is the probability of unsuccess and it's 2/3
$endgroup$
– user658422
Mar 27 at 20:30
$begingroup$
Well, $A_k=B_n-k$ Is not it?
$endgroup$
– Phicar
Mar 27 at 20:31
1
$begingroup$
@Phicar No. That would be true if the $c_i$ could only be 0 or 1.
$endgroup$
– Milten
Mar 27 at 20:56
$begingroup$
@Milten Thanks, i didn't read that.
$endgroup$
– Phicar
Mar 27 at 21:10
add a comment |
$begingroup$
What does $p$ and $q$ mean? Is $p=P(c_i=0)$?
$endgroup$
– Milten
Mar 27 at 20:18
$begingroup$
$p$ is the probability of success and it's 1/3, q is the probability of unsuccess and it's 2/3
$endgroup$
– user658422
Mar 27 at 20:30
$begingroup$
Well, $A_k=B_n-k$ Is not it?
$endgroup$
– Phicar
Mar 27 at 20:31
1
$begingroup$
@Phicar No. That would be true if the $c_i$ could only be 0 or 1.
$endgroup$
– Milten
Mar 27 at 20:56
$begingroup$
@Milten Thanks, i didn't read that.
$endgroup$
– Phicar
Mar 27 at 21:10
$begingroup$
What does $p$ and $q$ mean? Is $p=P(c_i=0)$?
$endgroup$
– Milten
Mar 27 at 20:18
$begingroup$
What does $p$ and $q$ mean? Is $p=P(c_i=0)$?
$endgroup$
– Milten
Mar 27 at 20:18
$begingroup$
$p$ is the probability of success and it's 1/3, q is the probability of unsuccess and it's 2/3
$endgroup$
– user658422
Mar 27 at 20:30
$begingroup$
$p$ is the probability of success and it's 1/3, q is the probability of unsuccess and it's 2/3
$endgroup$
– user658422
Mar 27 at 20:30
$begingroup$
Well, $A_k=B_n-k$ Is not it?
$endgroup$
– Phicar
Mar 27 at 20:31
$begingroup$
Well, $A_k=B_n-k$ Is not it?
$endgroup$
– Phicar
Mar 27 at 20:31
1
1
$begingroup$
@Phicar No. That would be true if the $c_i$ could only be 0 or 1.
$endgroup$
– Milten
Mar 27 at 20:56
$begingroup$
@Phicar No. That would be true if the $c_i$ could only be 0 or 1.
$endgroup$
– Milten
Mar 27 at 20:56
$begingroup$
@Milten Thanks, i didn't read that.
$endgroup$
– Phicar
Mar 27 at 21:10
$begingroup$
@Milten Thanks, i didn't read that.
$endgroup$
– Phicar
Mar 27 at 21:10
add a comment |
1 Answer
1
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oldest
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$begingroup$
HINT: Use the identity $P(A_kcup B_n-k) = P(A_k) + P(B_n-k) - P(A_kcap B_n-k)$. Can you figure out the last probability?
answered Mar 27 at 20:52
MiltenMilten
3746
$endgroup$
add a comment |
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$begingroup$
HINT: Use the identity $P(A_kcup B_n-k) = P(A_k) + P(B_n-k) - P(A_kcap B_n-k)$. Can you figure out the last probability?
answered Mar 27 at 20:52
MiltenMilten
3746
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add a comment |
$begingroup$
HINT: Use the identity $P(A_kcup B_n-k) = P(A_k) + P(B_n-k) - P(A_kcap B_n-k)$. Can you figure out the last probability?
answered Mar 27 at 20:52
MiltenMilten
3746
$endgroup$
add a comment |
$begingroup$
HINT: Use the identity $P(A_kcup B_n-k) = P(A_k) + P(B_n-k) - P(A_kcap B_n-k)$. Can you figure out the last probability?
answered Mar 27 at 20:52
MiltenMilten
3746
$endgroup$
HINT: Use the identity $P(A_kcup B_n-k) = P(A_k) + P(B_n-k) - P(A_kcap B_n-k)$. Can you figure out the last probability?
answered Mar 27 at 20:52
MiltenMilten
3746
answered Mar 27 at 20:52
MiltenMilten
3746
answered Mar 27 at 20:52
MiltenMilten
3746
answered Mar 27 at 20:52
MiltenMilten
3746
3746
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$begingroup$
What does $p$ and $q$ mean? Is $p=P(c_i=0)$?
$endgroup$
– Milten
Mar 27 at 20:18
$begingroup$
$p$ is the probability of success and it's 1/3, q is the probability of unsuccess and it's 2/3
$endgroup$
– user658422
Mar 27 at 20:30
$begingroup$
Well, $A_k=B_n-k$ Is not it?
$endgroup$
– Phicar
Mar 27 at 20:31
1
$begingroup$
@Phicar No. That would be true if the $c_i$ could only be 0 or 1.
$endgroup$
– Milten
Mar 27 at 20:56
$begingroup$
@Milten Thanks, i didn't read that.
$endgroup$
– Phicar
Mar 27 at 21:10