Mixing
Probability of No Collision
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Four Athletes are standing at the vertices of a square park.When a whistle is blown, each of the athletes starts running along the side of the park at a speed of 20 Km/hr.The probability that they do no collide with each other is?
I am completely blank at this.
Please provide a clue to solve this.
probability
asked Jan 5 at 9:12
user3767495user3767495
3818
$endgroup$
add a comment |
$begingroup$
Four Athletes are standing at the vertices of a square park.When a whistle is blown, each of the athletes starts running along the side of the park at a speed of 20 Km/hr.The probability that they do no collide with each other is?
I am completely blank at this.
Please provide a clue to solve this.
probability
asked Jan 5 at 9:12
user3767495user3767495
3818
$endgroup$
$begingroup$
If they all run clockwise or counterclockwise, they'll never collide.
$endgroup$
– Raskolnikov
Jan 5 at 9:18
$begingroup$
And Each one can do so independently with a probability of $frac{1}{2}$ right?
$endgroup$
– user3767495
Jan 5 at 9:20
$begingroup$
It's not explicitly stated in the question, but I suppose that is indeed what you have to assume.
$endgroup$
– Raskolnikov
Jan 5 at 9:20
$begingroup$
I think $frac{1}{2}$ is suitable because only two directions are possible and an athlete can choose only one at a time right?
$endgroup$
– user3767495
Jan 5 at 9:24
1
$begingroup$
The athlete could decide to prefer one direction over another, or he could decide to choose a direction with probability $pneqfrac{1}{2}$.
$endgroup$
– Raskolnikov
Jan 5 at 9:25
add a comment |
$begingroup$
Four Athletes are standing at the vertices of a square park.When a whistle is blown, each of the athletes starts running along the side of the park at a speed of 20 Km/hr.The probability that they do no collide with each other is?
I am completely blank at this.
Please provide a clue to solve this.
probability
asked Jan 5 at 9:12
user3767495user3767495
3818
$endgroup$
Four Athletes are standing at the vertices of a square park.When a whistle is blown, each of the athletes starts running along the side of the park at a speed of 20 Km/hr.The probability that they do no collide with each other is?
I am completely blank at this.
Please provide a clue to solve this.
probability
probability
asked Jan 5 at 9:12
user3767495user3767495
3818
asked Jan 5 at 9:12
user3767495user3767495
3818
asked Jan 5 at 9:12
user3767495user3767495
3818
asked Jan 5 at 9:12
user3767495user3767495
3818
asked Jan 5 at 9:12
user3767495user3767495
3818
3818
$begingroup$
If they all run clockwise or counterclockwise, they'll never collide.
$endgroup$
– Raskolnikov
Jan 5 at 9:18
$begingroup$
And Each one can do so independently with a probability of $frac{1}{2}$ right?
$endgroup$
– user3767495
Jan 5 at 9:20
$begingroup$
It's not explicitly stated in the question, but I suppose that is indeed what you have to assume.
$endgroup$
– Raskolnikov
Jan 5 at 9:20
$begingroup$
I think $frac{1}{2}$ is suitable because only two directions are possible and an athlete can choose only one at a time right?
$endgroup$
– user3767495
Jan 5 at 9:24
1
$begingroup$
The athlete could decide to prefer one direction over another, or he could decide to choose a direction with probability $pneqfrac{1}{2}$.
$endgroup$
– Raskolnikov
Jan 5 at 9:25
add a comment |
$begingroup$
If they all run clockwise or counterclockwise, they'll never collide.
$endgroup$
– Raskolnikov
Jan 5 at 9:18
$begingroup$
And Each one can do so independently with a probability of $frac{1}{2}$ right?
$endgroup$
– user3767495
Jan 5 at 9:20
$begingroup$
It's not explicitly stated in the question, but I suppose that is indeed what you have to assume.
$endgroup$
– Raskolnikov
Jan 5 at 9:20
$begingroup$
I think $frac{1}{2}$ is suitable because only two directions are possible and an athlete can choose only one at a time right?
$endgroup$
– user3767495
Jan 5 at 9:24
1
$begingroup$
The athlete could decide to prefer one direction over another, or he could decide to choose a direction with probability $pneqfrac{1}{2}$.
$endgroup$
– Raskolnikov
Jan 5 at 9:25
$begingroup$
If they all run clockwise or counterclockwise, they'll never collide.
$endgroup$
– Raskolnikov
Jan 5 at 9:18
$begingroup$
If they all run clockwise or counterclockwise, they'll never collide.
$endgroup$
– Raskolnikov
Jan 5 at 9:18
$begingroup$
And Each one can do so independently with a probability of $frac{1}{2}$ right?
$endgroup$
– user3767495
Jan 5 at 9:20
$begingroup$
And Each one can do so independently with a probability of $frac{1}{2}$ right?
$endgroup$
– user3767495
Jan 5 at 9:20
$begingroup$
It's not explicitly stated in the question, but I suppose that is indeed what you have to assume.
$endgroup$
– Raskolnikov
Jan 5 at 9:20
$begingroup$
It's not explicitly stated in the question, but I suppose that is indeed what you have to assume.
$endgroup$
– Raskolnikov
Jan 5 at 9:20
$begingroup$
I think $frac{1}{2}$ is suitable because only two directions are possible and an athlete can choose only one at a time right?
$endgroup$
– user3767495
Jan 5 at 9:24
$begingroup$
I think $frac{1}{2}$ is suitable because only two directions are possible and an athlete can choose only one at a time right?
$endgroup$
– user3767495
Jan 5 at 9:24
1
1
$begingroup$
The athlete could decide to prefer one direction over another, or he could decide to choose a direction with probability $pneqfrac{1}{2}$.
$endgroup$
– Raskolnikov
Jan 5 at 9:25
$begingroup$
The athlete could decide to prefer one direction over another, or he could decide to choose a direction with probability $pneqfrac{1}{2}$.
$endgroup$
– Raskolnikov
Jan 5 at 9:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To calculate the probability, we need to make an assumption about which direction each athlete runs. Let's assume, as you suggest, that
Each one can [run clockwise or counterclockwise] independently with a probability of $frac12$.
Now that this assumption is added to the problem, there are 16 possbilities: each of the four athletes can run Clockwise or Counterclockwise, and all $16$ possibilities are equally likely.
If one athlete runs Clockwise and another runs Counterclockwise then the two athletes will collide (convince yourself of this if it isn't clear).
So the only possibilities where the athletes do not collide are two:
Clockwise, Clockwise, Clockwise, Clockwise
Counterclockwise, Counterclockwise, Counterclockwise, Counterclockwise
That's $2$ out of $16$ equally-likely possibilities; therefore, the probability is $frac{2}{16} = frac{1}{8}$.
answered Jan 5 at 9:41
60056005
36.1k751125
$endgroup$
add a comment |
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1 Answer
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$begingroup$
To calculate the probability, we need to make an assumption about which direction each athlete runs. Let's assume, as you suggest, that
Each one can [run clockwise or counterclockwise] independently with a probability of $frac12$.
Now that this assumption is added to the problem, there are 16 possbilities: each of the four athletes can run Clockwise or Counterclockwise, and all $16$ possibilities are equally likely.
If one athlete runs Clockwise and another runs Counterclockwise then the two athletes will collide (convince yourself of this if it isn't clear).
So the only possibilities where the athletes do not collide are two:
Clockwise, Clockwise, Clockwise, Clockwise
Counterclockwise, Counterclockwise, Counterclockwise, Counterclockwise
That's $2$ out of $16$ equally-likely possibilities; therefore, the probability is $frac{2}{16} = frac{1}{8}$.
answered Jan 5 at 9:41
60056005
36.1k751125
$endgroup$
add a comment |
$begingroup$
To calculate the probability, we need to make an assumption about which direction each athlete runs. Let's assume, as you suggest, that
Each one can [run clockwise or counterclockwise] independently with a probability of $frac12$.
Now that this assumption is added to the problem, there are 16 possbilities: each of the four athletes can run Clockwise or Counterclockwise, and all $16$ possibilities are equally likely.
If one athlete runs Clockwise and another runs Counterclockwise then the two athletes will collide (convince yourself of this if it isn't clear).
So the only possibilities where the athletes do not collide are two:
Clockwise, Clockwise, Clockwise, Clockwise
Counterclockwise, Counterclockwise, Counterclockwise, Counterclockwise
That's $2$ out of $16$ equally-likely possibilities; therefore, the probability is $frac{2}{16} = frac{1}{8}$.
answered Jan 5 at 9:41
60056005
36.1k751125
$endgroup$
add a comment |
$begingroup$
To calculate the probability, we need to make an assumption about which direction each athlete runs. Let's assume, as you suggest, that
Each one can [run clockwise or counterclockwise] independently with a probability of $frac12$.
Now that this assumption is added to the problem, there are 16 possbilities: each of the four athletes can run Clockwise or Counterclockwise, and all $16$ possibilities are equally likely.
If one athlete runs Clockwise and another runs Counterclockwise then the two athletes will collide (convince yourself of this if it isn't clear).
So the only possibilities where the athletes do not collide are two:
Clockwise, Clockwise, Clockwise, Clockwise
Counterclockwise, Counterclockwise, Counterclockwise, Counterclockwise
That's $2$ out of $16$ equally-likely possibilities; therefore, the probability is $frac{2}{16} = frac{1}{8}$.
answered Jan 5 at 9:41
60056005
36.1k751125
$endgroup$
To calculate the probability, we need to make an assumption about which direction each athlete runs. Let's assume, as you suggest, that
Each one can [run clockwise or counterclockwise] independently with a probability of $frac12$.
Now that this assumption is added to the problem, there are 16 possbilities: each of the four athletes can run Clockwise or Counterclockwise, and all $16$ possibilities are equally likely.
If one athlete runs Clockwise and another runs Counterclockwise then the two athletes will collide (convince yourself of this if it isn't clear).
So the only possibilities where the athletes do not collide are two:
Clockwise, Clockwise, Clockwise, Clockwise
Counterclockwise, Counterclockwise, Counterclockwise, Counterclockwise
That's $2$ out of $16$ equally-likely possibilities; therefore, the probability is $frac{2}{16} = frac{1}{8}$.
answered Jan 5 at 9:41
60056005
36.1k751125
answered Jan 5 at 9:41
60056005
36.1k751125
answered Jan 5 at 9:41
60056005
36.1k751125
answered Jan 5 at 9:41
60056005
36.1k751125
36.1k751125
add a comment |
add a comment |
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$begingroup$
If they all run clockwise or counterclockwise, they'll never collide.
$endgroup$
– Raskolnikov
Jan 5 at 9:18
$begingroup$
And Each one can do so independently with a probability of $frac{1}{2}$ right?
$endgroup$
– user3767495
Jan 5 at 9:20
$begingroup$
It's not explicitly stated in the question, but I suppose that is indeed what you have to assume.
$endgroup$
– Raskolnikov
Jan 5 at 9:20
$begingroup$
I think $frac{1}{2}$ is suitable because only two directions are possible and an athlete can choose only one at a time right?
$endgroup$
– user3767495
Jan 5 at 9:24
1
$begingroup$
The athlete could decide to prefer one direction over another, or he could decide to choose a direction with probability $pneqfrac{1}{2}$.
$endgroup$
– Raskolnikov
Jan 5 at 9:25