The probability of Bus A arriving before Bus BExpected time of last bus leftProbability at least one of two buses arrive on timeBus stop independent events expected valueWhat is the expected time you have to wait until the first bus comes?Probabilty of 2 buses or more arriving at a bus s top at the same timeContinuous Probability - Bus ArrivingFred-to-bus and bus-to-bus average timesBus arrival probability…Average time waiting for busBus arrival times and minimum of exponential random variables
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The probability of Bus A arriving before Bus B
Expected time of last bus leftProbability at least one of two buses arrive on timeBus stop independent events expected valueWhat is the expected time you have to wait until the first bus comes?Probabilty of 2 buses or more arriving at a bus s top at the same timeContinuous Probability - Bus ArrivingFred-to-bus and bus-to-bus average timesBus arrival probability…Average time waiting for busBus arrival times and minimum of exponential random variables
$begingroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
New contributor
$endgroup$
add a comment |
$begingroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
New contributor
$endgroup$
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
37 mins ago
add a comment |
$begingroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
New contributor
$endgroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
probability
New contributor
New contributor
edited 1 hour ago
IrinaS
New contributor
asked 1 hour ago
IrinaSIrinaS
62
62
New contributor
New contributor
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
37 mins ago
add a comment |
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
37 mins ago
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
37 mins ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
37 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
$endgroup$
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
8 mins ago
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
add a comment |
Your Answer
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3 Answers
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3 Answers
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oldest
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$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
add a comment |
$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
add a comment |
$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
answered 1 hour ago
leonbloyleonbloy
41.8k647108
41.8k647108
add a comment |
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
$endgroup$
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
8 mins ago
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
$endgroup$
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
8 mins ago
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
$endgroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
edited 43 mins ago
answered 58 mins ago
farruhotafarruhota
21.4k2841
21.4k2841
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
8 mins ago
add a comment |
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
8 mins ago
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
8 mins ago
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
8 mins ago
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
answered 56 mins ago
Robert ShoreRobert Shore
3,410323
3,410323
add a comment |
add a comment |
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
37 mins ago