Ten balls are thrown randomly into three buckets.
$begingroup$
how would you solve this problem?
Ten balls are thrown randomly into three buckets. Compute the probability that each bucket contains at least one ball.
probability probability-distributions
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
$endgroup$
add a comment |
$begingroup$
how would you solve this problem?
Ten balls are thrown randomly into three buckets. Compute the probability that each bucket contains at least one ball.
probability probability-distributions
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
$endgroup$
add a comment |
$begingroup$
how would you solve this problem?
Ten balls are thrown randomly into three buckets. Compute the probability that each bucket contains at least one ball.
probability probability-distributions
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
$endgroup$
how would you solve this problem?
Ten balls are thrown randomly into three buckets. Compute the probability that each bucket contains at least one ball.
probability probability-distributions
probability probability-distributions
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
asked Jan 22 at 23:07
The Poor JewThe Poor Jew
476
476
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The answer is assuming that the balls can only land in the three buckets and each bucket is equally likely as a target.
The easier way to do it would be thinking go it in the other direction: What is the probability that at least one bucket has no balls?
Let the aforementioned probability be: P
Therefore, the answer you require is: 1 - P.
To calculate P:
$P = $ probability that ball lands only in any 2 buckets($=P_1$) $+$ probability that ball lands in only one bucket($=P_2$)
$$
P_1 = {3choose 2} (frac{2}{3})^{10}
$$
as, you choose any two buckets in which are the balls are supposed to fall ${3choose 2}$ and the probability that the ball falls only in those two is $frac{2}{3}$.
Since you are throwing 10 balls, simply multiple $frac{2}{3}$ 10 times to give $P_1$
$$
P_2 = {3choose 1} (frac{1}{3})^{10}
$$
as, you choose only buckets in which are the balls are supposed to fall ${3choose 1}$ and the probability that the ball falls only in that bucket is $frac{1}{3}$.
Since you are throwing 10 balls, simply multiple $frac{1}{3}$ 10 times to give $P_1$
Therefore, your answer is:
$$
1 - P_1 - P_2 = 1 - {3choose 2} (frac{2}{3})^{10} - {3choose 1} (frac{1}{3})^{10}
$$
answered Jan 23 at 4:03
HitenHiten
1477
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is assuming that the balls can only land in the three buckets and each bucket is equally likely as a target.
The easier way to do it would be thinking go it in the other direction: What is the probability that at least one bucket has no balls?
Let the aforementioned probability be: P
Therefore, the answer you require is: 1 - P.
To calculate P:
$P = $ probability that ball lands only in any 2 buckets($=P_1$) $+$ probability that ball lands in only one bucket($=P_2$)
$$
P_1 = {3choose 2} (frac{2}{3})^{10}
$$
as, you choose any two buckets in which are the balls are supposed to fall ${3choose 2}$ and the probability that the ball falls only in those two is $frac{2}{3}$.
Since you are throwing 10 balls, simply multiple $frac{2}{3}$ 10 times to give $P_1$
$$
P_2 = {3choose 1} (frac{1}{3})^{10}
$$
as, you choose only buckets in which are the balls are supposed to fall ${3choose 1}$ and the probability that the ball falls only in that bucket is $frac{1}{3}$.
Since you are throwing 10 balls, simply multiple $frac{1}{3}$ 10 times to give $P_1$
Therefore, your answer is:
$$
1 - P_1 - P_2 = 1 - {3choose 2} (frac{2}{3})^{10} - {3choose 1} (frac{1}{3})^{10}
$$
answered Jan 23 at 4:03
HitenHiten
1477
$endgroup$
add a comment |
$begingroup$
The answer is assuming that the balls can only land in the three buckets and each bucket is equally likely as a target.
The easier way to do it would be thinking go it in the other direction: What is the probability that at least one bucket has no balls?
Let the aforementioned probability be: P
Therefore, the answer you require is: 1 - P.
To calculate P:
$P = $ probability that ball lands only in any 2 buckets($=P_1$) $+$ probability that ball lands in only one bucket($=P_2$)
$$
P_1 = {3choose 2} (frac{2}{3})^{10}
$$
as, you choose any two buckets in which are the balls are supposed to fall ${3choose 2}$ and the probability that the ball falls only in those two is $frac{2}{3}$.
Since you are throwing 10 balls, simply multiple $frac{2}{3}$ 10 times to give $P_1$
$$
P_2 = {3choose 1} (frac{1}{3})^{10}
$$
as, you choose only buckets in which are the balls are supposed to fall ${3choose 1}$ and the probability that the ball falls only in that bucket is $frac{1}{3}$.
Since you are throwing 10 balls, simply multiple $frac{1}{3}$ 10 times to give $P_1$
Therefore, your answer is:
$$
1 - P_1 - P_2 = 1 - {3choose 2} (frac{2}{3})^{10} - {3choose 1} (frac{1}{3})^{10}
$$
answered Jan 23 at 4:03
HitenHiten
1477
$endgroup$
add a comment |
$begingroup$
The answer is assuming that the balls can only land in the three buckets and each bucket is equally likely as a target.
The easier way to do it would be thinking go it in the other direction: What is the probability that at least one bucket has no balls?
Let the aforementioned probability be: P
Therefore, the answer you require is: 1 - P.
To calculate P:
$P = $ probability that ball lands only in any 2 buckets($=P_1$) $+$ probability that ball lands in only one bucket($=P_2$)
$$
P_1 = {3choose 2} (frac{2}{3})^{10}
$$
as, you choose any two buckets in which are the balls are supposed to fall ${3choose 2}$ and the probability that the ball falls only in those two is $frac{2}{3}$.
Since you are throwing 10 balls, simply multiple $frac{2}{3}$ 10 times to give $P_1$
$$
P_2 = {3choose 1} (frac{1}{3})^{10}
$$
as, you choose only buckets in which are the balls are supposed to fall ${3choose 1}$ and the probability that the ball falls only in that bucket is $frac{1}{3}$.
Since you are throwing 10 balls, simply multiple $frac{1}{3}$ 10 times to give $P_1$
Therefore, your answer is:
$$
1 - P_1 - P_2 = 1 - {3choose 2} (frac{2}{3})^{10} - {3choose 1} (frac{1}{3})^{10}
$$
answered Jan 23 at 4:03
HitenHiten
1477
$endgroup$
The answer is assuming that the balls can only land in the three buckets and each bucket is equally likely as a target.
The easier way to do it would be thinking go it in the other direction: What is the probability that at least one bucket has no balls?
Let the aforementioned probability be: P
Therefore, the answer you require is: 1 - P.
To calculate P:
$P = $ probability that ball lands only in any 2 buckets($=P_1$) $+$ probability that ball lands in only one bucket($=P_2$)
$$
P_1 = {3choose 2} (frac{2}{3})^{10}
$$
as, you choose any two buckets in which are the balls are supposed to fall ${3choose 2}$ and the probability that the ball falls only in those two is $frac{2}{3}$.
Since you are throwing 10 balls, simply multiple $frac{2}{3}$ 10 times to give $P_1$
$$
P_2 = {3choose 1} (frac{1}{3})^{10}
$$
as, you choose only buckets in which are the balls are supposed to fall ${3choose 1}$ and the probability that the ball falls only in that bucket is $frac{1}{3}$.
Since you are throwing 10 balls, simply multiple $frac{1}{3}$ 10 times to give $P_1$
Therefore, your answer is:
$$
1 - P_1 - P_2 = 1 - {3choose 2} (frac{2}{3})^{10} - {3choose 1} (frac{1}{3})^{10}
$$
answered Jan 23 at 4:03
HitenHiten
1477
answered Jan 23 at 4:03
HitenHiten
1477
answered Jan 23 at 4:03
HitenHiten
1477
answered Jan 23 at 4:03
HitenHiten
1477
1477
add a comment |
add a comment |
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