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The probability of independent events.
Generalizing the total probability of simultaneous occurrences for independent eventsProbability in a series of events, during the series.Probability of expected value of independent eventsHow to find final probability if I know the probability of the individual events leading to it.Probability of multiple events occurring concurrentlyBayes Theorem to find nth timeLucky Six scratch tickets all have an independent probability of $0.05$Probability of getting pregnantDiscrete Probability: Determine if the events are Independent or notDoes probability make sense when the prior is unknown?
$begingroup$
- Suppose that there is a 1 in 50 chance of injury on a single skydiving attempt.
(a) (3 points) If we assume that the outcomes of different jumps are independent, what is the probability
that a skydiver is injured if she jumps twice?
The solution of this question was that 1-(the probability that she is not injured).
But I think it is not reasonable.
Let P(A)=the probability that she is not injured during two jumps.
1-P(A) contains the probability of (injured, safe), (injured, injured), (safe, injured). However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
probability
asked Mar 30 at 4:21
주혜민주혜민
133
$endgroup$
add a comment |
$begingroup$
- Suppose that there is a 1 in 50 chance of injury on a single skydiving attempt.
(a) (3 points) If we assume that the outcomes of different jumps are independent, what is the probability
that a skydiver is injured if she jumps twice?
The solution of this question was that 1-(the probability that she is not injured).
But I think it is not reasonable.
Let P(A)=the probability that she is not injured during two jumps.
1-P(A) contains the probability of (injured, safe), (injured, injured), (safe, injured). However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
probability
asked Mar 30 at 4:21
주혜민주혜민
133
$endgroup$
add a comment |
$begingroup$
- Suppose that there is a 1 in 50 chance of injury on a single skydiving attempt.
(a) (3 points) If we assume that the outcomes of different jumps are independent, what is the probability
that a skydiver is injured if she jumps twice?
The solution of this question was that 1-(the probability that she is not injured).
But I think it is not reasonable.
Let P(A)=the probability that she is not injured during two jumps.
1-P(A) contains the probability of (injured, safe), (injured, injured), (safe, injured). However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
probability
asked Mar 30 at 4:21
주혜민주혜민
133
$endgroup$
- Suppose that there is a 1 in 50 chance of injury on a single skydiving attempt.
(a) (3 points) If we assume that the outcomes of different jumps are independent, what is the probability
that a skydiver is injured if she jumps twice?
The solution of this question was that 1-(the probability that she is not injured).
But I think it is not reasonable.
Let P(A)=the probability that she is not injured during two jumps.
1-P(A) contains the probability of (injured, safe), (injured, injured), (safe, injured). However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
probability
probability
asked Mar 30 at 4:21
주혜민주혜민
133
asked Mar 30 at 4:21
주혜민주혜민
133
asked Mar 30 at 4:21
주혜민주혜민
133
asked Mar 30 at 4:21
주혜민주혜민
133
asked Mar 30 at 4:21
주혜민주혜민
133
133
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
If that were the case, then the events would not be independent.
You are asked to find the probability while assuming that the events are independent.
Therefore you should not consider injury to prevent (or otherwise influence) the second jump. If you like, think of it as having sufficient time between jumps for recovery or whatever.
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
Oh thank you . I didn't consider the meaning of "independent". :-D
$endgroup$
– 주혜민
Mar 30 at 6:02
add a comment |
$begingroup$
The complement event of “the skydiver is injured” is “the skydiver is safe in both jumps.” The skydiver has a chance of $1/50$ to be injured in an individual jump, so the chance that the skydiver is safe in an individual jump is
$$1 -frac150 = frac4950. $$
Since the two jumps are independent, the probability that she is safe in both jumps is:
$$biggl(frac4950biggr)^2 = frac24012500. $$
Thus, the probability that the skydiver is injured in either jump is
$$1 - frac24012500 = frac992500 = 3.96,%. $$
answered Mar 30 at 4:34
L. F.L. F.
20511
$endgroup$
add a comment |
$begingroup$
To answer your question more directly, if she's injured the first jump then the desired condition is already satisfied, and it doesn't matter whether she jumps again or not. (And if she jumps again, it doesn't matter if she's injured or safe the second time.)
So if you wish, you can consider there to be $3$ events instead of $4$:
With prob $p = 1/50$ she is injured the first jump. (The desired condition is satisfied, and it doesn't matter whether she jumps again or not. This combines your (injured, injured) and (injured, safe) scenarios into one. Note that I am making a mathematical point. I am not talking about typical human behavior at all.)
With prob $(1-p)p$ she is safe the first time and injured the second time.
With orb $(1-p)^2$ she is safe both times.
You can find your desired prob either as $1 - (1-p)^2$ (as in the answer by @L.F.) or as $p + (1-p)p$. They both equal $2p - p^2$.
answered Mar 30 at 4:59
antkamantkam
2,737312
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
If that were the case, then the events would not be independent.
You are asked to find the probability while assuming that the events are independent.
Therefore you should not consider injury to prevent (or otherwise influence) the second jump. If you like, think of it as having sufficient time between jumps for recovery or whatever.
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
Oh thank you . I didn't consider the meaning of "independent". :-D
$endgroup$
– 주혜민
Mar 30 at 6:02
add a comment |
$begingroup$
However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
If that were the case, then the events would not be independent.
You are asked to find the probability while assuming that the events are independent.
Therefore you should not consider injury to prevent (or otherwise influence) the second jump. If you like, think of it as having sufficient time between jumps for recovery or whatever.
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
$begingroup$
Oh thank you . I didn't consider the meaning of "independent". :-D
$endgroup$
– 주혜민
Mar 30 at 6:02
add a comment |
$begingroup$
However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
If that were the case, then the events would not be independent.
You are asked to find the probability while assuming that the events are independent.
Therefore you should not consider injury to prevent (or otherwise influence) the second jump. If you like, think of it as having sufficient time between jumps for recovery or whatever.
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
However, in general, if she gets injured at first jump, then she can't jump anymore. So, I think (injured, safe) case is impossible.
If that were the case, then the events would not be independent.
You are asked to find the probability while assuming that the events are independent.
Therefore you should not consider injury to prevent (or otherwise influence) the second jump. If you like, think of it as having sufficient time between jumps for recovery or whatever.
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
answered Mar 30 at 5:18
Graham KempGraham Kemp
87.8k43578
87.8k43578
$begingroup$
Oh thank you . I didn't consider the meaning of "independent". :-D
$endgroup$
– 주혜민
Mar 30 at 6:02
add a comment |
$begingroup$
Oh thank you . I didn't consider the meaning of "independent". :-D
$endgroup$
– 주혜민
Mar 30 at 6:02
$begingroup$
Oh thank you . I didn't consider the meaning of "independent". :-D
$endgroup$
– 주혜민
Mar 30 at 6:02
$begingroup$
Oh thank you . I didn't consider the meaning of "independent". :-D
$endgroup$
– 주혜민
Mar 30 at 6:02
add a comment |
$begingroup$
The complement event of “the skydiver is injured” is “the skydiver is safe in both jumps.” The skydiver has a chance of $1/50$ to be injured in an individual jump, so the chance that the skydiver is safe in an individual jump is
$$1 -frac150 = frac4950. $$
Since the two jumps are independent, the probability that she is safe in both jumps is:
$$biggl(frac4950biggr)^2 = frac24012500. $$
Thus, the probability that the skydiver is injured in either jump is
$$1 - frac24012500 = frac992500 = 3.96,%. $$
answered Mar 30 at 4:34
L. F.L. F.
20511
$endgroup$
add a comment |
$begingroup$
The complement event of “the skydiver is injured” is “the skydiver is safe in both jumps.” The skydiver has a chance of $1/50$ to be injured in an individual jump, so the chance that the skydiver is safe in an individual jump is
$$1 -frac150 = frac4950. $$
Since the two jumps are independent, the probability that she is safe in both jumps is:
$$biggl(frac4950biggr)^2 = frac24012500. $$
Thus, the probability that the skydiver is injured in either jump is
$$1 - frac24012500 = frac992500 = 3.96,%. $$
answered Mar 30 at 4:34
L. F.L. F.
20511
$endgroup$
add a comment |
$begingroup$
The complement event of “the skydiver is injured” is “the skydiver is safe in both jumps.” The skydiver has a chance of $1/50$ to be injured in an individual jump, so the chance that the skydiver is safe in an individual jump is
$$1 -frac150 = frac4950. $$
Since the two jumps are independent, the probability that she is safe in both jumps is:
$$biggl(frac4950biggr)^2 = frac24012500. $$
Thus, the probability that the skydiver is injured in either jump is
$$1 - frac24012500 = frac992500 = 3.96,%. $$
answered Mar 30 at 4:34
L. F.L. F.
20511
$endgroup$
The complement event of “the skydiver is injured” is “the skydiver is safe in both jumps.” The skydiver has a chance of $1/50$ to be injured in an individual jump, so the chance that the skydiver is safe in an individual jump is
$$1 -frac150 = frac4950. $$
Since the two jumps are independent, the probability that she is safe in both jumps is:
$$biggl(frac4950biggr)^2 = frac24012500. $$
Thus, the probability that the skydiver is injured in either jump is
$$1 - frac24012500 = frac992500 = 3.96,%. $$
answered Mar 30 at 4:34
L. F.L. F.
20511
answered Mar 30 at 4:34
L. F.L. F.
20511
answered Mar 30 at 4:34
L. F.L. F.
20511
answered Mar 30 at 4:34
L. F.L. F.
20511
20511
add a comment |
add a comment |
$begingroup$
To answer your question more directly, if she's injured the first jump then the desired condition is already satisfied, and it doesn't matter whether she jumps again or not. (And if she jumps again, it doesn't matter if she's injured or safe the second time.)
So if you wish, you can consider there to be $3$ events instead of $4$:
With prob $p = 1/50$ she is injured the first jump. (The desired condition is satisfied, and it doesn't matter whether she jumps again or not. This combines your (injured, injured) and (injured, safe) scenarios into one. Note that I am making a mathematical point. I am not talking about typical human behavior at all.)
With prob $(1-p)p$ she is safe the first time and injured the second time.
With orb $(1-p)^2$ she is safe both times.
You can find your desired prob either as $1 - (1-p)^2$ (as in the answer by @L.F.) or as $p + (1-p)p$. They both equal $2p - p^2$.
answered Mar 30 at 4:59
antkamantkam
2,737312
$endgroup$
add a comment |
$begingroup$
To answer your question more directly, if she's injured the first jump then the desired condition is already satisfied, and it doesn't matter whether she jumps again or not. (And if she jumps again, it doesn't matter if she's injured or safe the second time.)
So if you wish, you can consider there to be $3$ events instead of $4$:
With prob $p = 1/50$ she is injured the first jump. (The desired condition is satisfied, and it doesn't matter whether she jumps again or not. This combines your (injured, injured) and (injured, safe) scenarios into one. Note that I am making a mathematical point. I am not talking about typical human behavior at all.)
With prob $(1-p)p$ she is safe the first time and injured the second time.
With orb $(1-p)^2$ she is safe both times.
You can find your desired prob either as $1 - (1-p)^2$ (as in the answer by @L.F.) or as $p + (1-p)p$. They both equal $2p - p^2$.
answered Mar 30 at 4:59
antkamantkam
2,737312
$endgroup$
add a comment |
$begingroup$
To answer your question more directly, if she's injured the first jump then the desired condition is already satisfied, and it doesn't matter whether she jumps again or not. (And if she jumps again, it doesn't matter if she's injured or safe the second time.)
So if you wish, you can consider there to be $3$ events instead of $4$:
With prob $p = 1/50$ she is injured the first jump. (The desired condition is satisfied, and it doesn't matter whether she jumps again or not. This combines your (injured, injured) and (injured, safe) scenarios into one. Note that I am making a mathematical point. I am not talking about typical human behavior at all.)
With prob $(1-p)p$ she is safe the first time and injured the second time.
With orb $(1-p)^2$ she is safe both times.
You can find your desired prob either as $1 - (1-p)^2$ (as in the answer by @L.F.) or as $p + (1-p)p$. They both equal $2p - p^2$.
answered Mar 30 at 4:59
antkamantkam
2,737312
$endgroup$
To answer your question more directly, if she's injured the first jump then the desired condition is already satisfied, and it doesn't matter whether she jumps again or not. (And if she jumps again, it doesn't matter if she's injured or safe the second time.)
So if you wish, you can consider there to be $3$ events instead of $4$:
With prob $p = 1/50$ she is injured the first jump. (The desired condition is satisfied, and it doesn't matter whether she jumps again or not. This combines your (injured, injured) and (injured, safe) scenarios into one. Note that I am making a mathematical point. I am not talking about typical human behavior at all.)
With prob $(1-p)p$ she is safe the first time and injured the second time.
With orb $(1-p)^2$ she is safe both times.
You can find your desired prob either as $1 - (1-p)^2$ (as in the answer by @L.F.) or as $p + (1-p)p$. They both equal $2p - p^2$.
answered Mar 30 at 4:59
antkamantkam
2,737312
answered Mar 30 at 4:59
antkamantkam
2,737312
answered Mar 30 at 4:59
antkamantkam
2,737312
answered Mar 30 at 4:59
antkamantkam
2,737312
2,737312
add a comment |
add a comment |
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