After n balls what is the probability that there is a ball in a bucket
Answers
Answered by
3
I will assume that we are throwing balls sequentially towards the buckets, with at any stage each bucket equally likely to receive a ball, and independence between throws. Then the probability that bucket ii has no balls in it after KK balls have been thrown is equal to
(N−1N)K.
(N−1N)K.
Let Xi=1Xi=1 if the ii-th bucket ends up with least 11 ball, and let Xi=0Xi=0 otherwise. Then
P(Xi=1)=1−(N−1N)K.
P(Xi=1)=1−(N−1N)K.
Let YY be the number of buckets with at least 11 ball. Then
Y=∑i=1NXi.
Y=∑i=1NXi.
Now use the linearity of expectation. We can easily compute E(Xi)E(Xi).
Remark: The XiXi are not independent, but that makes no difference to the calculation. That's the beauty of the formula
E(a1X1+a2X2+⋯+aNXN)=a1E(X1)+a2E(X2)+⋯+anE(XN).
E(a1X1+a2X2+⋯+aNXN)=a1E(X1)+a2E(X2)+⋯+anE(XN).
We do not need to know the distribution of the random variable ∑aiXi∑aiXi to find its expectation.
(N−1N)K.
(N−1N)K.
Let Xi=1Xi=1 if the ii-th bucket ends up with least 11 ball, and let Xi=0Xi=0 otherwise. Then
P(Xi=1)=1−(N−1N)K.
P(Xi=1)=1−(N−1N)K.
Let YY be the number of buckets with at least 11 ball. Then
Y=∑i=1NXi.
Y=∑i=1NXi.
Now use the linearity of expectation. We can easily compute E(Xi)E(Xi).
Remark: The XiXi are not independent, but that makes no difference to the calculation. That's the beauty of the formula
E(a1X1+a2X2+⋯+aNXN)=a1E(X1)+a2E(X2)+⋯+anE(XN).
E(a1X1+a2X2+⋯+aNXN)=a1E(X1)+a2E(X2)+⋯+anE(XN).
We do not need to know the distribution of the random variable ∑aiXi∑aiXi to find its expectation.
Answered by
0
Answer:
that is a ball inside a bucket
Step-by-step explanation:
Similar questions