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# Two-thirds of the roads from A to B are at least 5 miles long, and of

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

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31 Oct 2017, 09:18
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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
[Reveal] Spoiler: OA

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

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31 Oct 2017, 10:04
Bunuel 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

let the number of roads from A to B are 3 and that from B to C are 4
number of roads of at least 5 miles long : A to B ---> 2/3 of 3 = 2
number of roads of at least 5 miles long : B to C ---> 1/4 of 4 = 1

total number of outcomes : 3 * 4 = 12
favorable outcomes : 2*1 = 2

probability = 2/12 = 1/6
IMO A
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Two-thirds of the roads from A to B are at least 5 miles long, and of [#permalink]

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01 Nov 2017, 16:39
Bunuel 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

I think D.

Probability of "at least one of the roads'" being at least 5 miles long is the complement of the event never occurring; that is, find probability of none for each event, then multiply. Subtract from one. (Because P(A) + P(not A) = 1.)

Probability of "none" =
For A to B: (1 - 2/3) = 1/3
For B to C: (1 - 1/4) = 3/4

P(none) = $$(\frac{1}{3}*\frac{3}{4})=\frac{3}{12}=\frac{1}{4}$$

1 - P(none) = (1 - $$\frac{1}{4})=\frac{3}{4}$$

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

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02 Nov 2017, 02:39
Bunuel 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

Prob of A to B who do not have 5 miles long = [1][/3]
Prob of B to C who do not have 5 miles long = [3][/4]
Prob that atleast one of the roads is 5 miles long = 1 - (1/3 * 3/4) = 3/4 (D)

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

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02 Nov 2017, 16:03
Bunuel 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 can use the following formula:

P(at least one of the roads you pick is at least 5 miles long) = 1 - P(none of the roads you pick is at least 5 miles long)

P(none of the roads you pick is at least 5 miles long) = 1/3 x 3/4 = 1/4

Thus, P(at least one of the roads you pick is at least 5 miles long) = 1 - 1/4 = 3/4.

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Re: Two-thirds of the roads from A to B are at least 5 miles long, and of   [#permalink] 02 Nov 2017, 16:03
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