Probability question: calculating probability of attendance in training camp.
Assume there are $400$ athletes in a training camp, who are required to attend the morning drill starting at $4$ am. The attendance in morning drills is $70%$, i.e. on an average, $280$ athletes are present. Fifty new athletes are admitted in this batch.
What is the probability of attendance being at least $70%$ among the new
athletes, thus ensuring the overall attendance does not fall below $70%$?The training coach thinks that this probability will increase if the new batch size is $40$ instead of $50$. Is he assuming right?
I tried solving the first part by following method.
Let New athlete = N.A, Old athlete = O.A.
probability of N.A = $(50 / 400 ) = 1 / 8$.
probability of O.A = $(350 / 400 ) = 7 / 8$.
Then according to the question the ans should be $0.7*(0.125) + 0.8*(0.125) + 0.9*(0.125) + 1*(0.125)$ which is $0.425$.
But I am not sure of this answer.
probability binomial-theorem
asked Nov 26 at 10:53
Aditya Jain
12
|
show 1 more comment
Assume there are $400$ athletes in a training camp, who are required to attend the morning drill starting at $4$ am. The attendance in morning drills is $70%$, i.e. on an average, $280$ athletes are present. Fifty new athletes are admitted in this batch.
What is the probability of attendance being at least $70%$ among the new
athletes, thus ensuring the overall attendance does not fall below $70%$?The training coach thinks that this probability will increase if the new batch size is $40$ instead of $50$. Is he assuming right?
I tried solving the first part by following method.
Let New athlete = N.A, Old athlete = O.A.
probability of N.A = $(50 / 400 ) = 1 / 8$.
probability of O.A = $(350 / 400 ) = 7 / 8$.
Then according to the question the ans should be $0.7*(0.125) + 0.8*(0.125) + 0.9*(0.125) + 1*(0.125)$ which is $0.425$.
But I am not sure of this answer.
probability binomial-theorem
asked Nov 26 at 10:53
Aditya Jain
12
1
This is not clear. We have no information at all regarding the new athletes. We might guess that they follow the same statistics as the old, but even if we do that wouldn't "ensure" any attendance. rate.
– lulu
Nov 26 at 10:57
I guess it is quite clear. Among 400 athletes, 50 are the new ones.
– Aditya Jain
Nov 26 at 11:00
Sorry, that doesn't clarify anything. Actually, it contradicts what you wrote before...as your post says that the $50$ are new...that is, they aren't among the original $400$.
– lulu
Nov 26 at 11:02
@lulu , Considering 50 as a part of total 400 athletes may give us some solution
– Aditya Jain
Nov 26 at 11:04
Please edit your post to ask a coherent and complete question. As it stands, nothing is clear.
– lulu
Nov 26 at 11:05
|
show 1 more comment
Assume there are $400$ athletes in a training camp, who are required to attend the morning drill starting at $4$ am. The attendance in morning drills is $70%$, i.e. on an average, $280$ athletes are present. Fifty new athletes are admitted in this batch.
What is the probability of attendance being at least $70%$ among the new
athletes, thus ensuring the overall attendance does not fall below $70%$?The training coach thinks that this probability will increase if the new batch size is $40$ instead of $50$. Is he assuming right?
I tried solving the first part by following method.
Let New athlete = N.A, Old athlete = O.A.
probability of N.A = $(50 / 400 ) = 1 / 8$.
probability of O.A = $(350 / 400 ) = 7 / 8$.
Then according to the question the ans should be $0.7*(0.125) + 0.8*(0.125) + 0.9*(0.125) + 1*(0.125)$ which is $0.425$.
But I am not sure of this answer.
probability binomial-theorem
asked Nov 26 at 10:53
Aditya Jain
12
Assume there are $400$ athletes in a training camp, who are required to attend the morning drill starting at $4$ am. The attendance in morning drills is $70%$, i.e. on an average, $280$ athletes are present. Fifty new athletes are admitted in this batch.
What is the probability of attendance being at least $70%$ among the new
athletes, thus ensuring the overall attendance does not fall below $70%$?The training coach thinks that this probability will increase if the new batch size is $40$ instead of $50$. Is he assuming right?
I tried solving the first part by following method.
Let New athlete = N.A, Old athlete = O.A.
probability of N.A = $(50 / 400 ) = 1 / 8$.
probability of O.A = $(350 / 400 ) = 7 / 8$.
Then according to the question the ans should be $0.7*(0.125) + 0.8*(0.125) + 0.9*(0.125) + 1*(0.125)$ which is $0.425$.
But I am not sure of this answer.
probability binomial-theorem
probability binomial-theorem
asked Nov 26 at 10:53
Aditya Jain
12
asked Nov 26 at 10:53
Aditya Jain
12
edited Nov 26 at 11:08
asked Nov 26 at 10:53
Aditya Jain
12
asked Nov 26 at 10:53
Aditya Jain
12
asked Nov 26 at 10:53
Aditya Jain
12
12
1
This is not clear. We have no information at all regarding the new athletes. We might guess that they follow the same statistics as the old, but even if we do that wouldn't "ensure" any attendance. rate.
– lulu
Nov 26 at 10:57
I guess it is quite clear. Among 400 athletes, 50 are the new ones.
– Aditya Jain
Nov 26 at 11:00
Sorry, that doesn't clarify anything. Actually, it contradicts what you wrote before...as your post says that the $50$ are new...that is, they aren't among the original $400$.
– lulu
Nov 26 at 11:02
@lulu , Considering 50 as a part of total 400 athletes may give us some solution
– Aditya Jain
Nov 26 at 11:04
Please edit your post to ask a coherent and complete question. As it stands, nothing is clear.
– lulu
Nov 26 at 11:05
|
show 1 more comment
1
This is not clear. We have no information at all regarding the new athletes. We might guess that they follow the same statistics as the old, but even if we do that wouldn't "ensure" any attendance. rate.
– lulu
Nov 26 at 10:57
I guess it is quite clear. Among 400 athletes, 50 are the new ones.
– Aditya Jain
Nov 26 at 11:00
Sorry, that doesn't clarify anything. Actually, it contradicts what you wrote before...as your post says that the $50$ are new...that is, they aren't among the original $400$.
– lulu
Nov 26 at 11:02
@lulu , Considering 50 as a part of total 400 athletes may give us some solution
– Aditya Jain
Nov 26 at 11:04
Please edit your post to ask a coherent and complete question. As it stands, nothing is clear.
– lulu
Nov 26 at 11:05
1
1
This is not clear. We have no information at all regarding the new athletes. We might guess that they follow the same statistics as the old, but even if we do that wouldn't "ensure" any attendance. rate.
– lulu
Nov 26 at 10:57
This is not clear. We have no information at all regarding the new athletes. We might guess that they follow the same statistics as the old, but even if we do that wouldn't "ensure" any attendance. rate.
– lulu
Nov 26 at 10:57
I guess it is quite clear. Among 400 athletes, 50 are the new ones.
– Aditya Jain
Nov 26 at 11:00
I guess it is quite clear. Among 400 athletes, 50 are the new ones.
– Aditya Jain
Nov 26 at 11:00
Sorry, that doesn't clarify anything. Actually, it contradicts what you wrote before...as your post says that the $50$ are new...that is, they aren't among the original $400$.
– lulu
Nov 26 at 11:02
Sorry, that doesn't clarify anything. Actually, it contradicts what you wrote before...as your post says that the $50$ are new...that is, they aren't among the original $400$.
– lulu
Nov 26 at 11:02
@lulu , Considering 50 as a part of total 400 athletes may give us some solution
– Aditya Jain
Nov 26 at 11:04
@lulu , Considering 50 as a part of total 400 athletes may give us some solution
– Aditya Jain
Nov 26 at 11:04
Please edit your post to ask a coherent and complete question. As it stands, nothing is clear.
– lulu
Nov 26 at 11:05
Please edit your post to ask a coherent and complete question. As it stands, nothing is clear.
– lulu
Nov 26 at 11:05
|
show 1 more comment
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1
This is not clear. We have no information at all regarding the new athletes. We might guess that they follow the same statistics as the old, but even if we do that wouldn't "ensure" any attendance. rate.
– lulu
Nov 26 at 10:57
I guess it is quite clear. Among 400 athletes, 50 are the new ones.
– Aditya Jain
Nov 26 at 11:00
Sorry, that doesn't clarify anything. Actually, it contradicts what you wrote before...as your post says that the $50$ are new...that is, they aren't among the original $400$.
– lulu
Nov 26 at 11:02
@lulu , Considering 50 as a part of total 400 athletes may give us some solution
– Aditya Jain
Nov 26 at 11:04
Please edit your post to ask a coherent and complete question. As it stands, nothing is clear.
– lulu
Nov 26 at 11:05