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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
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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: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
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One suggestion, you might want to rephrase as :

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

Regards,
Subhash
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
ok thanks bunuel and subhashghosh !
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
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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 long, and 1/4 [#permalink]
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.


Great thank you! Kudos ;)
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
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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 long, and 1/4 [#permalink]
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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 long, and 1/4 [#permalink]
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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...

Final Answer:

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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
A-B: P(5+ miles) = 2/3
B-C: P(5+ miles) = 1/4

a) A-B YES B-C NO : 2/3 x 3/4 = 6/12
b) A-B NO B-C YES : 1/3 x 1/4 = 1/12
c) A-B YES B-C YES : 2/3 x 1/4 = 2/12

6/12 + 1/12 + 2/12 = 9/12 = 3/4

Answer is D.
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
EMPOWERgmatRichC wrote:
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...


Love the logical approach EMPOWERgmatRichC - often overlooked for probability questions!
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
Why can't the answer be 2/3 x 1/4 ???
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
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vinsin03 wrote:
Why can't the answer be 2/3 x 1/4 ???


Hi vinsin03,

When dealing with any GMAT question, you have to pay careful attention to the question that is ASKED. This question asks for the probability of AT LEAST one of the two roads being at least 5 miles long. This means that there are 3 'successful outcomes' that you have to consider:

1) Road A-to-B is at least 5 miles long, but Road B-to-C is not.
2) Road B-to-C is at least 5 miles long, but Road A-to-B is not.
3) Both roads are at least 5 miles long.

The calculation (2/3)(1/4) is the probability that BOTH roads are at least 5 miles long, but that's not what the question asks for. In simple terms, this calculation is only part of the overall solution.

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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
"Hi vinsin03,

When dealing with any GMAT question, you have to pay careful attention to the question that is ASKED. This question asks for the probability of AT LEAST one of the two roads being at least 5 miles long. This means that there are 3 'successful outcomes' that you have to consider:

1) Road A-to-B is at least 5 miles long, but Road B-to-C is not.
2) Road B-to-C is at least 5 miles long, but Road A-to-B is not.
3) Both roads are at least 5 miles long.

The calculation (2/3)(1/4) is the probability that BOTH roads are at least 5 miles long, but that's not what the question asks for. In simple terms, this calculation is only part of the overall solution"



:) Thanks a lot for the clarification :)
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
If you don't subtract the probability of both out then you will have accounted for the probability of both twice in your calculations. This is why you need to subtract out both.
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Re: Two thirds of the roads from A to B are at least 5 miles long, and 1/4 [#permalink]
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