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John and Amanda stand at opposite ends of a straight road and start [#permalink]
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Bunuel wrote:
John and Amanda stand at opposite ends of a straight road and start running towards each other at the same moment. Their rates are randomly selected in advance so that John runs at a constant rate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant rate of 4, 5, 6, or 7 miles per hour. What is the probability that John has traveled farther than Amanda by the time they meet?

(A) 3/16
(B) 5/16
(C) 3/8
(D) 1/2
(E) 13/16


John can run at 4 different speeds {3, 4, 5, 6}
and Amenda can run at 4 different speeds {4, 5, 6, 7}

Hence, Total Combination of the speeds maintained by the two runners can be given by = 4 x 4 = 16

Now Let's make FAVORABLE and UNFAVORABLE cases

Case 1: If John Runs are speed of 3 miles per hour

Amenda can run at speeds of {4, 5, 6, 7}

in every case Amenda runs at greater speed than the speed of John hence will cover more distance than John in each case

i.e. Favorable case = 0
i.e. Unfavorable case = 4

Case 2: If John Runs are speed of 4 miles per hour

If Amenda Runs at 4, The distance covered by them both will be same due to same speed of each runner

If Amenda Runs at 5, The distance covered by Amenda will be more than the distance covered by John till they meet due to higher speeed of Amenda than john's speed

If Amenda Runs at 6, The distance covered by Amenda will be more than the distance covered by John till they meet due to higher speeed of Amenda than john's speed

If Amenda Runs at 7, The distance covered by Amenda will be more than the distance covered by John till they meet due to higher speeed of Amenda than john's speed

i.e. Favorable case = 0
i.e. Unfavorable case = 4

Case 3: If John Runs are speed of 5 miles per hour

If Amenda Runs at 4, The distance covered by John will be more than the distance covered by Amenda till they meet due to higher speeed of John than Amenda's speed

If Amenda Runs at 5, The distance covered by them both will be same due to same speed of each runner

If Amenda Runs at 6, The distance covered by Amenda will be more than the distance covered by John till they meet due to higher speeed of Amenda than john's speed

If Amenda Runs at 7, The distance covered by Amenda will be more than the distance covered by John till they meet due to higher speeed of Amenda than john's speed

i.e. Favorable case = 1
i.e. Unfavorable case = 3

Case 4: If John Runs are speed of 6 miles per hour

If Amenda Runs at 4, The distance covered by John will be more than the distance covered by Amenda till they meet due to higher speeed of John than Amenda's speed

If Amenda Runs at 5, The distance covered by John will be more than the distance covered by Amenda till they meet due to higher speeed of John than Amenda's speed

If Amenda Runs at 6, The distance covered by them both will be same due to same speed of each runner

If Amenda Runs at 7, The distance covered by Amenda will be more than the distance covered by John till they meet due to higher speeed of Amenda than john's speed

i.e. Favorable case = 2
i.e. Unfavorable case = 2

Total Favorable cases when John covers more distance than distance covered by Amenda = 0+0+1+2 = 3
Total Cases = 16

Probability = Favorable cases / Total cases

i.e. Required Probability = 3/16

Answer: Option
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
I got this question wrong but here is my train of thought.
I would really like some pointers on how to identify when i should add or multiply the combinations and probabilities.
This confuses me a lot.

Anyways

I started by thinking that J > 4 to be able to run faster than A. A < 5.
The probability of J > 4 = 1/2 * probability of A < 5 = 1/4 = 1/8

Then the case of J > 5, A < 6.
The probability of J > 5 = 1/4 * probability of A < 6 = 1/2 = 1/8
Then add the 1/8 and 1/8 = 2/8

Which is wrong because it is not amongst the answer choices.
If someone could help this would be great, because I need to be able to answer this question with a lot of ease.
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tjerkrintjema wrote:
I got this question wrong but here is my train of thought.
I would really like some pointers on how to identify when i should add or multiply the combinations and probabilities.
This confuses me a lot.

Anyways

I started by thinking that J > 4 to be able to run faster than A. A < 5.
The probability of J > 4 = 1/2 * probability of A < 5 = 1/4 = 1/8

Then the case of J > 5, A < 6.
The probability of J > 5 = 1/4 * probability of A < 6 = 1/2 = 1/8
Then add the 1/8 and 1/8 = 2/8

Which is wrong because it is not amongst the answer choices.
If someone could help this would be great, because I need to be able to answer this question with a lot of ease.


Hi tjerkrintjema,

In the simplest Language

We Multiple when we use the word [b]"AND" between two events e.g If Probability of x to happen is a and probability of y to happen is y then Probability of both events x ANDy to happen simultaneously is axb[/b]

We Add the two probability when we use the word "OR" between two events e.g If Probability of x to happen is a and probability of y to happen is y then Probability of Either events x happens OR event y to happens is a+b

Here, Let's use the same concept

Case I: Probability of the speed of John J > 4 and probability of A < 5. will be (2/4)x(1/4) = 1/8

Case II: Probability of the speed of John J > 5 and probability of A < 6. will be (1/4)x(2/4) = 1/8

But the problem here is that we have taken One case common between Cases I and II J= 6 and A=4 which has the probability (1/4)x(1/4)= 1/16

So the probability will be (1/8)+(1/8)-(1/16) = 3/16

Hence either we have to take such cases out or the best way is to just make cases
e.g.

J = 3 No favorable Case

J= 4 No favorable case as A can't be less than J

J = 5, A = 4

J = 6, A = 4

J = 6, A = 5


Only 3 favorable cases out of total 4 x 4 = 16 cases

Hence Probability = 3/16

Answer: Option
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
GMATinsight wrote:
tjerkrintjema wrote:
I got this question wrong but here is my train of thought.
I would really like some pointers on how to identify when i should add or multiply the combinations and probabilities.
This confuses me a lot.

Anyways

I started by thinking that J > 4 to be able to run faster than A. A < 5.
The probability of J > 4 = 1/2 * probability of A < 5 = 1/4 = 1/8

Then the case of J > 5, A < 6.
The probability of J > 5 = 1/4 * probability of A < 6 = 1/2 = 1/8
Then add the 1/8 and 1/8 = 2/8

Which is wrong because it is not amongst the answer choices.
If someone could help this would be great, because I need to be able to answer this question with a lot of ease.


Hi tjerkrintjema,

In the simplest Language

We Multiple when we use the word [b]"AND" between two events e.g If Probability of x to happen is a and probability of y to happen is y then Probability of both events x ANDy to happen simultaneously is axb[/b]

We Add the two probability when we use the word "OR" between two events e.g If Probability of x to happen is a and probability of y to happen is y then Probability of Either events x happens OR event y to happens is a+b

Here, Let's use the same concept

Case I: Probability of the speed of John J > 4 and probability of A < 5. will be (2/4)x(1/4) = 1/8

Case II: Probability of the speed of John J > 5 and probability of A < 6. will be (1/4)x(2/4) = 1/8

But the problem here is that we have taken One case common between Cases I and II J= 6 and A=4 which has the probability (1/4)x(1/4)= 1/16

So the probability will be (1/8)+(1/8)-(1/16) = 3/16

Hence either we have to take such cases out or the best way is to just make cases
e.g.

J = 3 No favorable Case

J= 4 No favorable case as A can't be less than J

J = 5, A = 4

J = 6, A = 4

J = 6, A = 5


Only 3 favorable cases out of total 4 x 4 = 16 cases

Hence Probability = 3/16

Answer: Option


Wow thanks, this really helps. I'm also kind of glad that i was on the right track with my method as well, in the sense that addition was right, but that i just forgot to subtract the 1/16 like you mentioned. Although I believe the other ways mentioned here are way better, at least it makes me feel less stupid :D
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John and Amanda stand at opposite ends of a straight road and start [#permalink]
Yep, I also did this:

J: 3 - 4 - 5 - 6
A: 4- 5 - 6 - 7

There are 16 possible combinations of rates between them.

Looking at J, if we runs at 3 or 4 he will be slower or equally fast as A. We want him to be quicker.
If he runs at 5, he has one chance of being faster (A runs at 3).
If he runs at 6, he has two chances of being faster (A runs at 4 or 5).
In total, he has three chances of being faster out of the 16 possible combinations.

So, 3/16 ANS A.
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
If John and Amanda run at the same rate, they will meet each other exactly in the middle. John will only run farther than Amanda if John's rate is greater than Amanda's. In math terms, distance = rate × time, and since John and Amanda run for the same time, their relative distances depend solely on their relative rates. We can rephrase the question to “what is the probability that John ran faster than Amanda?” There are four possible rates for John (3, 4, 5, and 6) and four for Amanda (4, 5, 6, and 7). In total, there are (4)(4) = 16 possible rate scenarios.
Of John's four possible rates, only two (5 and 6) are greater than some of Amanda's possible rates (4and 5). We can easily list the 3 rate scenarios that result in a faster speed (greater distance) for John:
John ran 5 mph, and Amanda ran 4 mph.
John ran 6 mph, and Amanda ran 4 mph.
John ran 6 mph, and Amanda ran 5 mph.
Since there are 16 possible combinations of rates, the probability that John ran farther than Amanda
is 3/16.
The correct answer is A.
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
I looked at when John's speed would be greater than Amanda's: John at 5 and John at 6.

So:
Probability of John running at 6 = 1/4 AND Amanda running at 5 or 4 = 1/2 --> (1/4)(1/2) = 1/8
PLUS
Probability of John running at 5 = 1/4 AND Amanda running at 4 = 1/4 --> (1/4)(1/4) = 1/16
1/8 + 1/16 = 3/16.
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
Only 2 rates of John is faster than Amanda's
i.e. 5 and 6
Probability of J to run at 5=1/4
Probability of J to run at 6=1/4
Total probability of J =1/16
If J runs at 5, then the probability of A being slower than J is 1/4
If J runs at 6, then the probability of A being slower than J is 2/4=1/2
Total probability of A=1/8
probability of J being faster than A = (1/16)+(1/8)=3/16

Option A
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
Bunuel wrote:
John and Amanda stand at opposite ends of a straight road and start running towards each other at the same moment. Their rates are randomly selected in advance so that John runs at a constant rate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant rate of 4, 5, 6, or 7 miles per hour. What is the probability that John has traveled farther than Amanda by the time they meet?

(A) 3/16
(B) 5/16
(C) 3/8
(D) 1/2
(E) 13/16


if John runs at 3 mph. possible ways = 0
if john runs at 4 mph, Possible ways = 0
if john runs at 5 mph, Possible ways is 4 mph for amanda = 1
if john runs at 6 mph, Possible ways is 4mph and 5 mph = 2

Adding, we get total ways = 3

Possible ways = 4C1 x 4C1 = 16

probability=3/16

Option A
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
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Bunuel wrote:
John and Amanda stand at opposite ends of a straight road and start running towards each other at the same moment. Their rates are randomly selected in advance so that John runs at a constant rate of 3, 4, 5, or 6 miles per hour and Amanda runs at a constant rate of 4, 5, 6, or 7 miles per hour. What is the probability that John has traveled farther than Amanda by the time they meet?

(A) 3/16
(B) 5/16
(C) 3/8
(D) 1/2
(E) 13/16



John will travel farther than Amanda if John is traveling faster than Amanda. That is, John’s rate must be greater than Amanda’s rate. Thus, John must travel either 5 or 6 miles per hour in order to have a chance to be faster than Amanda.

If John travels 5 mph (there is a 1/4 chance he would do that), then Amanda has to travel 4 mph (there is a 1/4 chance she would do that). Thus, the probability is 1/4 x 1/4 = 1/16.

If John travels 6 mph (there is a 1/4 chance he would do that), then Amanda has to travel either 4 or 5 mph (there is a 2/4 = 1/2 chance she would do that). Thus, the probability is 1/4 x 1/2 = 1/8.
Therefore, the overall probability that John will travel farther than Amanda is 1/16 + 1/8 = 3/16.

Alternate Solution:

Since there are 4 possibilities for the rates of John and Amanda each, there are 4 x 4 = 16 possible ways to assign rates for them.

Of these 16 possibilities, the only way John travels further than Amanda is if John = 6 mph and Amanda = 5 mph; John = 6 mph and Amanda = 4 mph; or if John = 5 mph and Amanda = 4 mph. Since there are 3 favorable outcomes in a total of 16 outcomes, the probability is 3/16.

Answer: A
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Re: John and Amanda stand at opposite ends of a straight road and start [#permalink]
UJs wrote:
Note: this one is like a Data Sufficiency question, you don't have to solve all of it to get the answer. :)
its a 2 min question,key here is -> think your approach (before you jump into solving)

    John speed options = 3, 4, 5, or 6 miles per hour.
    Amanda speed options = 4, 5, 6, or 7 miles per hour.

now when can John travel farther than Amanda ?
only in cases where (speed of John) > (Speed of Amanda) , that's possible when John travels at (5,6) and Amanda less than that, lets solve it now :

    \(Probability = \frac{Favorable-outcomes}{Total-possible-outcomes}\)

    Total-possible-outcomes = John (4 speeds to pick from) * Amanda (4 speeds to pick from) \(= 4 * 4 = 16\)

    when John speed 5 , Amanda can be 4 = 1 outcome
    when John speed 6 , Amanda can be 4,5 = 2 outcome
    Favorable-outcomes\(= 1 + 2 = 3\)

    \(Probability = \frac{3}{16}\)
Ans : A


Very good explanation
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