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RenB
Joined: 13 Jul 2022
Last visit: 18 Nov 2025
Posts: 391
Given Kudos: 303
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Kellogg '26 (II) Booth '26 (D) Yale '26 (S) CBS '26 (II) Darden '26 (S) HBS '26 (D) Ross '26 (II) LBS '26 (D) Tuck '26 (II)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Kellogg '26 (II) Booth '26 (D) Yale '26 (S) CBS '26 (II) Darden '26 (S) HBS '26 (D) Ross '26 (II) LBS '26 (D) Tuck '26 (II)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 391
The probability that the sales representative will encounter road construction on the trip from City D to City E iS 0.7.: Yes
The probability that the sales representative will not encounter road construction on the trip from City F to City G is 0.7.: No
Rounded to the nearest hundredth, the probability that the sales representative will not encounter road construction during her journey is 0.03: Yes
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
48% (03:18) correct
52%
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wrong
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A sales representative based in City A is about to begin a journey, by car, to Cities B, C, D, E, F, and G, in this order. The table shows her estimates for the probability of encountering road construction for each of several routes between consecutive cities, with the routes numbered in the table from least to greatest distance. For example, there are 7 routes between City A and City B, and the probability of encountering road construction on the 5th longest of these routes is 0.9.
Suppose that the sales representative minimizes the probability, according to her estimates, of encountering road construction during the course of her journey from City A to City G. For each of the following statements, select Yes if the statement is therefore accurate, based on the representative's estimates. Otherwise, select No.
GMAT-Club-Forum-3jx71t8s.png [ 20.05 KiB | Viewed 3093 times ]
| Route | A to B | B ot C | C to D | D to E | E to F | F to G |
|---|---|---|---|---|---|---|
| 1 | 0.1 | 0.7 | 0.9 | 0.8 | 0.1 | 0.8 |
| 2 | 0 | 0.8 | 1 | 0.8 | 0.1 | 0.8 |
| 3 | 0.1 | 0.8 | 1 | 0.9 | 0.2 | 0.7 |
| 4 | 0.5 | 1 | 0.7 | 0.7 | 0.2 | 0.8 |
| 5 | 0.9 | 0.7 | 0.9 | 0.8 | 0.2 | 0.9 |
| 6 | 0.4 | 0 | 0.7 | 0.9 | 0.1 | 0.7 |
| 7 | 0.3 | 0.8 | 0.7 | 0.7 | 0.2 | 0.8 |
| 8 | - | 0.7 | 1 | 0.8 | 0.1 | 0.8 |
| 9 | - | 0.7 | - | 0.9 | 0 | 0.7 |
| 10 | - | - | - | 0.8 | - | 0.9 |
| 11 | - | - | - | 0.8 | - | - |
Suppose that the sales representative minimizes the probability, according to her estimates, of encountering road construction during the course of her journey from City A to City G. For each of the following statements, select Yes if the statement is therefore accurate, based on the representative's estimates. Otherwise, select No.
ID: 700195
Attachment:
GMAT-Club-Forum-3jx71t8s.png [ 20.05 KiB | Viewed 3093 times ]
| Yes | No | |
| The probability that the sales representative will encounter road construction on the trip from City D to City E iS 0.7. | ||
| The probability that the sales representative will not encounter road construction on the trip from City F to City G is 0.7. | ||
| Rounded to the nearest hundredth, the probability that the sales representative will not encounter road construction during her journey is 0.03 |
ShowHide Answer
Official Answer
The probability that the sales representative will encounter road construction on the trip from City D to City E iS 0.7.: Yes
The probability that the sales representative will not encounter road construction on the trip from City F to City G is 0.7.: No
Rounded to the nearest hundredth, the probability that the sales representative will not encounter road construction during her journey is 0.03: Yes
11 May 2024, 22:08
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RenB
You have not added the answers.
It should be Yes, No, Yes. Let me know if it is different.
Solution:
You have to realize many times the para may have info that would not matter, so scan the para and then the questions before spending and trying to understand the para more than what is required.
Just dont think if something i sgiven, it is a must for solving the question.
We are looking for least problematic path, that is where the probability of encountering disruptions is the least, and it is irrespective of the length of the route.
Let us write down the least problematic path by sorting each column.
A to B: 0
B to C: 0
C to D: 0.7
D to E: 0.7
E to F: 0
F to G: 0.7
1. The probability that the sales representative will encounter road construction on the trip from City D to City E iS 0.7.
Sort the table by column D-E, and the least is the answer, or check from above. So, 0.7 is correct.
2. The probability that the sales representative will not encounter road construction on the trip from City F to City G is 0.7.
Sort the table by column F-G, and the least is the probability of encountering construction, that is 0.7. TRAP, as you are looking for the opposite, that is NOT encountering.
Thus answer is 1-0.7 or 0.3, and you have to mark No.
3. Rounded to the nearest hundredth, the probability that the sales representative will not encounter road construction during her journey is 0.03
Again, it is NOT encountering the disruptions.
NOT in all means we will have to multiply probability of not encountering in each stretch.
Sort table each stretch wise, so that you do not make unforced error or copy from route marked above.
\(P = (1-0)*(1-0)*(1-0.7)*(1-0.7)*(1-0)*(1-0.7) = 1*1*0.3*0.3*1*0.3 = 0.027\) ~ 0.03, when rounded to nearest hundredth.
Mark Yes.
Answer: Yes, No, Yes
Attachment:
GMAT-Club-Forum-jlhgjmca.png [ 20.05 KiB | Viewed 3078 times ]
18 Aug 2024, 07:37
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abcd1234!!
The negation of the question is the probability to encounter constructions in AT LEAST ONE of the segments of the chosen path. You can't simply multiply the probability of encountering construction on the different segments of the road, as the value would represent the probability of encountering constructions in ALL the three segments.
Instead, the probability you need to compute includes three cases:
- The probablity to encounter constructions in only one of the segments = 3*(0.7*0.3*0.3) (3 represents the number of possible choices of segments where constructions are encountered 1 of 3)
- The probablity to encounter constructions in only two of the segments = 3*(0.7*0.7*0.3) (3 represents the number of possible choices of segments where constructions are encountered 2 of 3 )
- The probablity to encounter constructions in only three of the segments = 1*(0.7*0.7*0.7)
Therefore the probability to encounter constructions in AT LEAST ONE of the segments of the chosen path is:
3*(0.7*0.3*0.3)+3*(0.7*0.7*0.3)+1*(0.7*0.7*0.7) = 0.973
and therefore the probability to not encounter any of the constructions is:
1-0.973=0.027≈0.03
