Expected cost of eight successes where we can 'purchase' more attempts
$begingroup$
So we want to achieve $8$ successes where each trial has a probability of succeeding of $0.3$,
$$P(S) = 0.3.$$
We have $8$ "free" trials, and assuming we don't get lucky and get $8$ successes from our free trials, we can pay $1.00$ dollar for a $10%$ chance to regain one trial. So say we get $3$ successes from the free trials, then we would need to pay at least $5$ dollars [but expected value of $50$ dollars because $50times0.1 = 5$ successful $10%$ chances] to regain the $5$ unsuccessful attempts. What is our expected cost to obtain $8$ successes?
In my attempt I start with the expected number of successes from the $8$ free trials ($n times p = 8 times 0.3 = 2.4$) and then start a cycle; we have $5,6$ trials to be regained that we missed on the first trials, which would take an expected $$56.00$ (or fifty-six $10%$ chances) to regain.
Then from that we get an expected ($5.6 times 0.3 =$) $1.68$ successes and have $4.08$ successes total. Wash and repeat until we hit $8$ successes. I'm getting around $$185.00$ as an answer but I'm not sure of my answer and also curious if there's a better or less tedious way of doing this problem.
Any thoughts are appreciated. Thanks!
probability statistics
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
$endgroup$
add a comment |
$begingroup$
So we want to achieve $8$ successes where each trial has a probability of succeeding of $0.3$,
$$P(S) = 0.3.$$
We have $8$ "free" trials, and assuming we don't get lucky and get $8$ successes from our free trials, we can pay $1.00$ dollar for a $10%$ chance to regain one trial. So say we get $3$ successes from the free trials, then we would need to pay at least $5$ dollars [but expected value of $50$ dollars because $50times0.1 = 5$ successful $10%$ chances] to regain the $5$ unsuccessful attempts. What is our expected cost to obtain $8$ successes?
In my attempt I start with the expected number of successes from the $8$ free trials ($n times p = 8 times 0.3 = 2.4$) and then start a cycle; we have $5,6$ trials to be regained that we missed on the first trials, which would take an expected $$56.00$ (or fifty-six $10%$ chances) to regain.
Then from that we get an expected ($5.6 times 0.3 =$) $1.68$ successes and have $4.08$ successes total. Wash and repeat until we hit $8$ successes. I'm getting around $$185.00$ as an answer but I'm not sure of my answer and also curious if there's a better or less tedious way of doing this problem.
Any thoughts are appreciated. Thanks!
probability statistics
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
$endgroup$
add a comment |
$begingroup$
So we want to achieve $8$ successes where each trial has a probability of succeeding of $0.3$,
$$P(S) = 0.3.$$
We have $8$ "free" trials, and assuming we don't get lucky and get $8$ successes from our free trials, we can pay $1.00$ dollar for a $10%$ chance to regain one trial. So say we get $3$ successes from the free trials, then we would need to pay at least $5$ dollars [but expected value of $50$ dollars because $50times0.1 = 5$ successful $10%$ chances] to regain the $5$ unsuccessful attempts. What is our expected cost to obtain $8$ successes?
In my attempt I start with the expected number of successes from the $8$ free trials ($n times p = 8 times 0.3 = 2.4$) and then start a cycle; we have $5,6$ trials to be regained that we missed on the first trials, which would take an expected $$56.00$ (or fifty-six $10%$ chances) to regain.
Then from that we get an expected ($5.6 times 0.3 =$) $1.68$ successes and have $4.08$ successes total. Wash and repeat until we hit $8$ successes. I'm getting around $$185.00$ as an answer but I'm not sure of my answer and also curious if there's a better or less tedious way of doing this problem.
Any thoughts are appreciated. Thanks!
probability statistics
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
$endgroup$
So we want to achieve $8$ successes where each trial has a probability of succeeding of $0.3$,
$$P(S) = 0.3.$$
We have $8$ "free" trials, and assuming we don't get lucky and get $8$ successes from our free trials, we can pay $1.00$ dollar for a $10%$ chance to regain one trial. So say we get $3$ successes from the free trials, then we would need to pay at least $5$ dollars [but expected value of $50$ dollars because $50times0.1 = 5$ successful $10%$ chances] to regain the $5$ unsuccessful attempts. What is our expected cost to obtain $8$ successes?
In my attempt I start with the expected number of successes from the $8$ free trials ($n times p = 8 times 0.3 = 2.4$) and then start a cycle; we have $5,6$ trials to be regained that we missed on the first trials, which would take an expected $$56.00$ (or fifty-six $10%$ chances) to regain.
Then from that we get an expected ($5.6 times 0.3 =$) $1.68$ successes and have $4.08$ successes total. Wash and repeat until we hit $8$ successes. I'm getting around $$185.00$ as an answer but I'm not sure of my answer and also curious if there's a better or less tedious way of doing this problem.
Any thoughts are appreciated. Thanks!
probability statistics
probability statistics
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
edited Dec 17 '18 at 4:04
Tianlalu
3,09421138
3,09421138
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
asked Dec 17 '18 at 3:02
Nance GaronNance Garon
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes there is a simpler way. $frac{8}{0.3} = 26.667$ which rounds up to $27$ trials to get an expected $8$ successes. So you have to pay for another $19$ trials for which the expected number of $$1$ payments would $frac{19}{0.1} = 190$ hence the expected cost would be $$190$.
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
$endgroup$
1
$begingroup$
If this is an expected value, then there should be no issue I think with the expected number of required trials being fractional.
$endgroup$
– Riley
Dec 17 '18 at 3:37
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Yes there is a simpler way. $frac{8}{0.3} = 26.667$ which rounds up to $27$ trials to get an expected $8$ successes. So you have to pay for another $19$ trials for which the expected number of $$1$ payments would $frac{19}{0.1} = 190$ hence the expected cost would be $$190$.
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
$endgroup$
1
$begingroup$
If this is an expected value, then there should be no issue I think with the expected number of required trials being fractional.
$endgroup$
– Riley
Dec 17 '18 at 3:37
add a comment |
$begingroup$
Yes there is a simpler way. $frac{8}{0.3} = 26.667$ which rounds up to $27$ trials to get an expected $8$ successes. So you have to pay for another $19$ trials for which the expected number of $$1$ payments would $frac{19}{0.1} = 190$ hence the expected cost would be $$190$.
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
$endgroup$
1
$begingroup$
If this is an expected value, then there should be no issue I think with the expected number of required trials being fractional.
$endgroup$
– Riley
Dec 17 '18 at 3:37
add a comment |
$begingroup$
Yes there is a simpler way. $frac{8}{0.3} = 26.667$ which rounds up to $27$ trials to get an expected $8$ successes. So you have to pay for another $19$ trials for which the expected number of $$1$ payments would $frac{19}{0.1} = 190$ hence the expected cost would be $$190$.
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
$endgroup$
Yes there is a simpler way. $frac{8}{0.3} = 26.667$ which rounds up to $27$ trials to get an expected $8$ successes. So you have to pay for another $19$ trials for which the expected number of $$1$ payments would $frac{19}{0.1} = 190$ hence the expected cost would be $$190$.
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
answered Dec 17 '18 at 3:33
Phil HPhil H
4,2582312
4,2582312
1
$begingroup$
If this is an expected value, then there should be no issue I think with the expected number of required trials being fractional.
$endgroup$
– Riley
Dec 17 '18 at 3:37
add a comment |
1
$begingroup$
If this is an expected value, then there should be no issue I think with the expected number of required trials being fractional.
$endgroup$
– Riley
Dec 17 '18 at 3:37
1
1
$begingroup$
If this is an expected value, then there should be no issue I think with the expected number of required trials being fractional.
$endgroup$
– Riley
Dec 17 '18 at 3:37
$begingroup$
If this is an expected value, then there should be no issue I think with the expected number of required trials being fractional.
$endgroup$
– Riley
Dec 17 '18 at 3:37
add a comment |
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