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28 Oct 2021, 04:00
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (02:47) correct
66%
(03:17)
wrong
based on 137
sessions
History
Date
Time
Result
Not Attempted Yet
One week a certain vehicle rental outlet had a total of 40 cars, 12 trucks, 28 vans, and 20 SUVs available. Andre and Barbara went to the vehicle rental outlet and chose 2 vehicles at random, with the condition that Andre and Barbara would not select two of the same type of vehicle (in other words, if one of them has an SUV, the other won't take an SUV, so the second person doesn't even consider the SUVs). What is the probability that, of the two vehicles, one of them is a car or a van?
A. 63/1100
B. 133/275
C. 991/1100
D. 1037/1100
E. 427/442
A. 63/1100
B. 133/275
C. 991/1100
D. 1037/1100
E. 427/442
MBhavya
They pick one vehicle each.
P(a car or a van) =
= 1 - P(Not a car or a van) =
= 1 - P(a truck and SUV) =
= 1 - (P(Andre = truck and Barbara = SUV) + P(Andre = SUV and Barbara = truck )) =
= 1 - (12/100 * 20/88 + 20/100 * 12/80) =
= 1,037/1,100
= 1 - P(Not a car or a van) =
= 1 - P(a truck and SUV) =
= 1 - (P(Andre = truck and Barbara = SUV) + P(Andre = SUV and Barbara = truck )) =
= 1 - (12/100 * 20/88 + 20/100 * 12/80) =
= 1,037/1,100
Answer: D.
General Discussion
28 Oct 2021, 09:29
Kudos
Bookmarks
Question wants to know the the chances that a single one of the vehicles is either a car or a van or both. The easiest method will be to work in reverse and figure out the probability that neither of the picks is a car or van and subtract that from 1 (Ie probability of a truck AND a van being picked).
First pick a truck and the second pick is an SUV:
Probability that the first vehicle chosen is a truck: \(\frac{12}{100}\) or simplified \(\frac{3}{25}\)
Probability that the second vehicle chosen is an SUV after a truck was chosen: \(\frac{20}{88}\) (over 88 due to their agreement that they will never choose the same type of vehicle) or simplified \(\frac{5}{22}\)
Probability that this event occurs: \(\frac{3}{25} * \frac{5}{22} \)
\(=\frac{3}{5} * \frac{1}{22} \)
\(\frac{3}{110}\)
First pick an SUV and the second pick is a truck:
Probability that the first vehicle chosen is an SUV: \(\frac{20}{100}\) or simplified \(\frac{1}{5}\)
Probability that the second vehicle chosen is a truck after an SUV was chosen: \(\frac{12}{80}\) (over 80 as no truck can be chosen) or simplified \(\frac{3}{20}\)
Probability that this event occurs: \(\frac{1}{5} * \frac{3}{20}\)
\(\frac{3}{100}\)
Probability that either of these events occurs:
\(\frac{3}{110}+\frac{3}{100}\)
\(\frac{(300+330)}{11000}\)
\(\frac{630}{11000}\)
\(\frac{63}{1100}\)
Probability that a car or van or both are chosen:
\(1 - \frac{63}{1100}\)
\(\frac{1037}{1100}\)
Answer D
First pick a truck and the second pick is an SUV:
Probability that the first vehicle chosen is a truck: \(\frac{12}{100}\) or simplified \(\frac{3}{25}\)
Probability that the second vehicle chosen is an SUV after a truck was chosen: \(\frac{20}{88}\) (over 88 due to their agreement that they will never choose the same type of vehicle) or simplified \(\frac{5}{22}\)
Probability that this event occurs: \(\frac{3}{25} * \frac{5}{22} \)
\(=\frac{3}{5} * \frac{1}{22} \)
\(\frac{3}{110}\)
First pick an SUV and the second pick is a truck:
Probability that the first vehicle chosen is an SUV: \(\frac{20}{100}\) or simplified \(\frac{1}{5}\)
Probability that the second vehicle chosen is a truck after an SUV was chosen: \(\frac{12}{80}\) (over 80 as no truck can be chosen) or simplified \(\frac{3}{20}\)
Probability that this event occurs: \(\frac{1}{5} * \frac{3}{20}\)
\(\frac{3}{100}\)
Probability that either of these events occurs:
\(\frac{3}{110}+\frac{3}{100}\)
\(\frac{(300+330)}{11000}\)
\(\frac{630}{11000}\)
\(\frac{63}{1100}\)
Probability that a car or van or both are chosen:
\(1 - \frac{63}{1100}\)
\(\frac{1037}{1100}\)
Answer D
