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Two thirds of the roads from A to B are at least 5 miles

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Two thirds of the roads from A to B are at least 5 miles  [#permalink]

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New post Updated on: 01 Feb 2012, 14:05
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A
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Two thirds of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
(A) 1/6
(B) 1/4
(C) 2/3
(D) 3/4
(E) 11/12

Originally posted by Madelaine88 on 26 Feb 2011, 12:36.
Last edited by Bunuel on 01 Feb 2012, 14:05, edited 1 time in total.
Edited the question and added the OA
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Re: Probability - roads  [#permalink]

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New post 26 Feb 2011, 12:48
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Madelaine88 wrote:
2/3 of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. If you randomly pick a road from B to C, what is the probability that at least one of the you pick is at least 5 miles long?


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Original question is:

Two thirds of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
(A) 1/6
(B) 1/4
(C) 2/3
(D) 3/4
(E) 11/12

Find the probability of the event that none of the roads you pick will be at least 5 miles long and subtract from 1 to get the probability that at least one of the roads you pick will be at least 5 miles long: P=1-1/3*3/4=3/4.

Answer: D.
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Re: Probability - roads  [#permalink]

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New post 28 Feb 2011, 03:00
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I'm just writing this for myself so as to understand fully :

we calculate 1-(none of the roads is at leat 5 miles) because it is easier than calculating all the cases where at least one road is at least 5 miles, ie :

AB >= 5miles & BC < 5miles : P = 2/3*3/4 = 6/12
AB >= 5miles & BC >= 5miles : P = 2/3*1/4 = 2/12
AB < 5miles & BC >= 5miles : P = 1/3*1/4 = 1/12

Sum all probabilities (because these events are independent) : P = 9/12 = 3/4


Is this right ???
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Re: Probability - roads  [#permalink]

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New post 28 Feb 2011, 03:14
chouky wrote:
I'm just writing this for myself so as to understand fully :

we calculate 1-(none of the roads is at leat 5 miles) because it is easier than calculating all the cases where at least one road is at least 5 miles, ie :

AB >= 5miles & BC < 5miles : P = 2/3*3/4 = 6/12
AB >= 5miles & BC >= 5miles : P = 2/3*1/4 = 2/12
AB < 5miles & BC >= 5miles : P = 1/3*1/4 = 1/12

Sum all probabilities (because these events are independent) : P = 9/12 = 3/4


Is this right ???
Thanks !


Yes, that's correct: the probability that at least one of the roads you pick is at least 5 miles long is the probability that either one or both of the roads are at least 5 miles long.
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Re: Probability - roads  [#permalink]

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New post 28 Feb 2011, 05:54
One suggestion, you might want to rephrase as :

Sum all probabilities (because these events are independent as mutually exclusive).

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Re: Probability - roads  [#permalink]

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New post 28 Feb 2011, 05:58
ok thanks bunuel and subhashghosh !
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Re: Two thirds of the roads from A to B are at least 5 miles  [#permalink]

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New post 29 Jan 2017, 12:18
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Sol 1: A to B –> 2/3 (5 m long) then 1/3 (not 5 m long)
Similarly B to C –> 1/4 (5 m long) then 3/4 (not 5 m long)
So probability that 1 of the road you pick is at least 5 miles =
( (A to B is 5 m long ) & (B to C is not 5 m long) + (A to B is not 5 m long ) & (B to C is 5 m long) + (A to B is 5 m long ) & (B to C is 5 m long) )
Now solving the above equation:
(A to B is 5 m long ) & (B to C is not 5 m long) => 2/3 * 3/4
(A to B is not 5 m long ) & (B to C is 5 m long) => 1/3 * 1/4
(A to B is 5 m long ) & (B to C is 5 m long) => 2/3 * 1/4
Then putting the value in equation – (2/3 * 3/4) + (1/3 * 1/4) + (2/3 * 1/4)
= ¾ Ans

Sol 2: using the 1 – p method
Equation will be => (1 – probability of none of the road is 5 m long)
So (probability of none of the road is 5 m long) = (A to B is not 5 m long) & (B to C is not 5 m long)
= (1/3) * (3/4) => (1/4)
Then
(1 – probability of none of the road is 5 m long) = (1 - (1/4)) => 3/4 Ans.
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Re: Two thirds of the roads from A to B are at least 5 miles  [#permalink]

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New post 29 Jan 2017, 13:48
devctg wrote:
Sol 1: A to B –> 2/3 (5 m long) then 1/3 (not 5 m long)
Similarly B to C –> 1/4 (5 m long) then 3/4 (not 5 m long)
So probability that 1 of the road you pick is at least 5 miles =
( (A to B is 5 m long ) & (B to C is not 5 m long) + (A to B is not 5 m long ) & (B to C is 5 m long) + (A to B is 5 m long ) & (B to C is 5 m long) )
Now solving the above equation:
(A to B is 5 m long ) & (B to C is not 5 m long) => 2/3 * 3/4
(A to B is not 5 m long ) & (B to C is 5 m long) => 1/3 * 1/4
(A to B is 5 m long ) & (B to C is 5 m long) => 2/3 * 1/4
Then putting the value in equation – (2/3 * 3/4) + (1/3 * 1/4) + (2/3 * 1/4)
= ¾ Ans

Sol 2: using the 1 – p method
Equation will be => (1 – probability of none of the road is 5 m long)
So (probability of none of the road is 5 m long) = (A to B is not 5 m long) & (B to C is not 5 m long)
= (1/3) * (3/4) => (1/4)
Then
(1 – probability of none of the road is 5 m long) = (1 - (1/4)) => 3/4 Ans.


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Re: Two thirds of the roads from A to B are at least 5 miles  [#permalink]

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New post 29 Jan 2017, 15:14
A useful probability pattern: When a probability question has the phrase at least in the phrasing of the actual question (as in, "what is the probability that at least one six is rolled?", or "what is the probability that "at least one heads is tossed?) it is often faster to find the probability of NOT getting what the question asks about and then subtracting from one, than it is to find the probability directly.

Here, we are asked what the probability that at least one of the roads is at least 5 miles long.

Well, there is only one way for that NOT to happen. The only way to not have at least one road that is 5 or more miles long is to have NEITHER road 5 or more miles long.

So the probability that NEITHER is 5 or more miles long:

Two things need to happen if we're never going to get a 5-mile road: The first road needs to be less than five miles AND the second road needs to be five miles.

The probability that the first road is less than 5 miles is 1/3 (if 2/3 of the roads are at least 5 miles long, the other one third are less than 5 miles)

The probability that the second road is less than 5 miles long is 3/4 (if 1/4 of the roads are at least 5 miles long, the other 3/4 are less than 5 miles)

To find the probability of one event AND another event occurring together, we multiply the individual probabilities.

Thus, the prob of getting a road less than 5 miles for the first leg AND for the second leg is (1/3)(3/4) = 1/4

So the prob that NEITHER road is 5 or more miles is 1/4. That means that the prob that at least one of the roads is 5 or more (i.e every other situation) miles long is 1 - 1/4 = 3/4. D.
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Re: Two thirds of the roads from A to B are at least 5 miles  [#permalink]

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New post 30 Jan 2017, 17:34
Madelaine88 wrote:
Two thirds of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
(A) 1/6
(B) 1/4
(C) 2/3
(D) 3/4
(E) 11/12



We are given that 2/3 of the roads from A to B are at least 5 miles long (which means that 1/3 of the roads from A to B are less than 5 miles long) and 1/4 of the roads from B to C are at least 5 miles long (which means that 3/4 of the roads from A to B are less than 5 miles long). We need to determine the probability that when picking a road from A to B and B to C, at least one of the roads is at least 5 miles long.

We can use the following formula:

P(selecting at least 1 road that is at least 5 miles long) + P(selecting no roads that are at least 5 miles long) = 1

P(selecting at least 1 road that is 5 miles long) = 1 - P(selecting no roads that are at least 5 miles long)

P(selecting at least 1 road that is 5 miles long) = 1 - P(selecting all roads that are less than 5 miles long)

Thus, if we can determine the probability of selecting all roads that are less than 5 miles long, we’ll quickly be able to calculate the probability of selecting at least 1 road that is at least 5 miles long.

The probability of selecting all roads that are less than 5 miles long is: 1/3 x 3/4 = 1/4

Thus, the probability of selecting at least 1 road that is at least 5 miles long is: 1 - 1/4 = 3/4

Answer: D
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Re: Two thirds of the roads from A to B are at least 5 miles  [#permalink]

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New post 01 Feb 2019, 16:39
Hi All,

We're told that 2/3 of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. We're asked to randomly pick a road from A to B and then randomly pick a road from B to C and determine the probability that AT LEAST ONE of the roads you pick is at least 5 miles long. This question can be approached in a number of different ways. In most cases, when a question asks for the probability of 'at least one' outcome occurring, the fastest approach is to determine that the outcome does NOT occur at all - and then subtract that probability from the number 1 (as swerve has shown). The answers to this question are sufficiently 'spread out' that you don't actually have to do any math to get the solution - a little logic is all that's required.

To start, since 2/3 of the roads from A to B are at least 5 miles long, the probability of randomly choosing such a road from A to B OR B to C (OR both) MUST be GREATER than 2/3. Eliminate Answers A, B and C.

With the two remaining answers, we have a reasonable answer (re: Answer D: 75% of the time) and an unreasonable one (Answer E: 11/12, which is over 90% of the time). Since the probability from B to C is relatively small (just 1/4 = 25%), there's no way that the overall probability would be so close to 100%. Thus, there's only one answer that makes sense...

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Re: Two thirds of the roads from A to B are at least 5 miles   [#permalink] 01 Feb 2019, 16:39
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