Probability of mutually Independent events The 2019 Stack Overflow Developer Survey Results Are InProbability of 2 EventsMutually exclusive eventsMutually exclusive and independent probabilityEasy Probability Question (Independent Events)If A and B are independent events but NOT mutually exclusive, find P(A and B)?Probability Question: Mutually Exclusive EventsPairwise independent events but not mutually independentProbability involving independence and mutually exclusive eventsUsing average to find probability of eventsGiven $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$
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Probability of mutually Independent events
The 2019 Stack Overflow Developer Survey Results Are InProbability of 2 EventsMutually exclusive eventsMutually exclusive and independent probabilityEasy Probability Question (Independent Events)If A and B are independent events but NOT mutually exclusive, find P(A and B)?Probability Question: Mutually Exclusive EventsPairwise independent events but not mutually independentProbability involving independence and mutually exclusive eventsUsing average to find probability of eventsGiven $A, B$, and $C$ are mutually independent events, find $ P(A cap B' cap C')$
$begingroup$
I'm a bit confused on how to solve this problem.
Suppose the probability of dying in a skydiving accident is $0.04%$ per jump, mutually independent with all other jumps.
What is the largest integer $k$ such that the probability of surviving $k$ jumps is at least $99%$?
So we know that probability of surviving is 99.96%. Since the jumps are mutually independent, would the answer be $.9996^kgeq .99$? Am I on the right track?
probability
edited Mar 24 at 7:04
MarianD
2,2611618
asked Mar 24 at 6:57
RobinRobin
625
$endgroup$
add a comment |
$begingroup$
I'm a bit confused on how to solve this problem.
Suppose the probability of dying in a skydiving accident is $0.04%$ per jump, mutually independent with all other jumps.
What is the largest integer $k$ such that the probability of surviving $k$ jumps is at least $99%$?
So we know that probability of surviving is 99.96%. Since the jumps are mutually independent, would the answer be $.9996^kgeq .99$? Am I on the right track?
probability
edited Mar 24 at 7:04
MarianD
2,2611618
asked Mar 24 at 6:57
RobinRobin
625
$endgroup$
1
$begingroup$
Yes you are right. Since events are independent, they simply multiply leading to this relation
$endgroup$
– Tojrah
Mar 24 at 7:06
add a comment |
$begingroup$
I'm a bit confused on how to solve this problem.
Suppose the probability of dying in a skydiving accident is $0.04%$ per jump, mutually independent with all other jumps.
What is the largest integer $k$ such that the probability of surviving $k$ jumps is at least $99%$?
So we know that probability of surviving is 99.96%. Since the jumps are mutually independent, would the answer be $.9996^kgeq .99$? Am I on the right track?
probability
edited Mar 24 at 7:04
MarianD
2,2611618
asked Mar 24 at 6:57
RobinRobin
625
$endgroup$
I'm a bit confused on how to solve this problem.
Suppose the probability of dying in a skydiving accident is $0.04%$ per jump, mutually independent with all other jumps.
What is the largest integer $k$ such that the probability of surviving $k$ jumps is at least $99%$?
So we know that probability of surviving is 99.96%. Since the jumps are mutually independent, would the answer be $.9996^kgeq .99$? Am I on the right track?
probability
probability
edited Mar 24 at 7:04
MarianD
2,2611618
asked Mar 24 at 6:57
RobinRobin
625
edited Mar 24 at 7:04
MarianD
2,2611618
asked Mar 24 at 6:57
RobinRobin
625
edited Mar 24 at 7:04
MarianD
2,2611618
edited Mar 24 at 7:04
MarianD
2,2611618
edited Mar 24 at 7:04
MarianD
2,2611618
2,2611618
asked Mar 24 at 6:57
RobinRobin
625
asked Mar 24 at 6:57
RobinRobin
625
asked Mar 24 at 6:57
RobinRobin
625
625
1
$begingroup$
Yes you are right. Since events are independent, they simply multiply leading to this relation
$endgroup$
– Tojrah
Mar 24 at 7:06
add a comment |
1
$begingroup$
Yes you are right. Since events are independent, they simply multiply leading to this relation
$endgroup$
– Tojrah
Mar 24 at 7:06
1
1
$begingroup$
Yes you are right. Since events are independent, they simply multiply leading to this relation
$endgroup$
– Tojrah
Mar 24 at 7:06
$begingroup$
Yes you are right. Since events are independent, they simply multiply leading to this relation
$endgroup$
– Tojrah
Mar 24 at 7:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes. You’re absolutely on the right track. Two events are independent if
$$P(A cap B) = P(A)P(B)$$. In other words, the probability of them both occurring is the probability of the first event occurring multiplied by the probability of the second occurring.
Because of this, we know that the probability of surviving and then surviving again ($P(Acap B)$) is just the product of their probabilities since the events are given to be mutually independent. So as you correctly observed, $.9996^2$. But then the probability of surviving again after this is the previous result multiplied by $.9996$. This continues for a total of $k$ steps until the result is less than $.99$.
answered Mar 24 at 8:58
rb612rb612
1,9131924
$endgroup$
add a comment |
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$begingroup$
Yes. You’re absolutely on the right track. Two events are independent if
$$P(A cap B) = P(A)P(B)$$. In other words, the probability of them both occurring is the probability of the first event occurring multiplied by the probability of the second occurring.
Because of this, we know that the probability of surviving and then surviving again ($P(Acap B)$) is just the product of their probabilities since the events are given to be mutually independent. So as you correctly observed, $.9996^2$. But then the probability of surviving again after this is the previous result multiplied by $.9996$. This continues for a total of $k$ steps until the result is less than $.99$.
answered Mar 24 at 8:58
rb612rb612
1,9131924
$endgroup$
add a comment |
$begingroup$
Yes. You’re absolutely on the right track. Two events are independent if
$$P(A cap B) = P(A)P(B)$$. In other words, the probability of them both occurring is the probability of the first event occurring multiplied by the probability of the second occurring.
Because of this, we know that the probability of surviving and then surviving again ($P(Acap B)$) is just the product of their probabilities since the events are given to be mutually independent. So as you correctly observed, $.9996^2$. But then the probability of surviving again after this is the previous result multiplied by $.9996$. This continues for a total of $k$ steps until the result is less than $.99$.
answered Mar 24 at 8:58
rb612rb612
1,9131924
$endgroup$
add a comment |
$begingroup$
Yes. You’re absolutely on the right track. Two events are independent if
$$P(A cap B) = P(A)P(B)$$. In other words, the probability of them both occurring is the probability of the first event occurring multiplied by the probability of the second occurring.
Because of this, we know that the probability of surviving and then surviving again ($P(Acap B)$) is just the product of their probabilities since the events are given to be mutually independent. So as you correctly observed, $.9996^2$. But then the probability of surviving again after this is the previous result multiplied by $.9996$. This continues for a total of $k$ steps until the result is less than $.99$.
answered Mar 24 at 8:58
rb612rb612
1,9131924
$endgroup$
Yes. You’re absolutely on the right track. Two events are independent if
$$P(A cap B) = P(A)P(B)$$. In other words, the probability of them both occurring is the probability of the first event occurring multiplied by the probability of the second occurring.
Because of this, we know that the probability of surviving and then surviving again ($P(Acap B)$) is just the product of their probabilities since the events are given to be mutually independent. So as you correctly observed, $.9996^2$. But then the probability of surviving again after this is the previous result multiplied by $.9996$. This continues for a total of $k$ steps until the result is less than $.99$.
answered Mar 24 at 8:58
rb612rb612
1,9131924
answered Mar 24 at 8:58
rb612rb612
1,9131924
answered Mar 24 at 8:58
rb612rb612
1,9131924
answered Mar 24 at 8:58
rb612rb612
1,9131924
1,9131924
add a comment |
add a comment |
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Yes you are right. Since events are independent, they simply multiply leading to this relation
$endgroup$
– Tojrah
Mar 24 at 7:06