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04 Jul 2025, 03:07
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A rental firm has only two distinct types of vehicles: minivans and sedans. Each vehicle runs on exactly one fuel type, either gasoline or electricity. If a vehicle is selected at random, is the probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline?
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
Let:
MG= Minivan on Gasoline
ME=Minivan on Electricity
SG= Sedan on Gasoline
SE= Sedan on Electricity
M= Minivan
S= Sedan
E= Electricity
G= Gasoline
Question: Prob (SE)> Prob (MG)
Statement 1
P(M or G)=2/5
P(M) + P(G)-P(MG)= 2/5
Since we don't know P(M) nor P(G) we cannot get P (MG).
Thus statement 1 is insufficient.
Statement 2
P(S or E)=5/8
P(S) or P(E) - P(SE) = 5/8
Likewise like statement 1 there are more than 2 unknowns in the expression which makes statement 2 insufficient
Jointly
We know from statement 1 that:
P(M) + P(G)-P(MG)= 2/5 ---- (1)
and from statement 2:
P(S) + P(E) - P(SE) = 5/8 ---- (2)
P(S) + P(E) = 5/8 +P(SE) ----(3)
we know that
P(M)+ P(S)=1 and
P(E) + P(G) =1
thus P(M)= 1-P(S)
P (G) =1- P(E)
We can express statement 1 in terms of statement 2
1-P(S) +1-P(E)-P(MG)= 2/5
2-P(S)-P(E)- P(MG)=2/5
2-(P(S)+P(E))=2/5+P(MG) -----(4)
substitute equation (3) into equation (4)
2 -(5/8+P(SE))= 2/5+P(MG)
2- (5/8 + P(SE) = 2/5+P(MG)
11/8- P(SE) = 2/5+P(MG)
P(SE) +P(MG) =11/8 - 2/5
P(SE) +P(MG) =39/40
This gives us the sum of the 2 probabilities but does not answer whether
Prob (SE)> Prob (MG)
Hence both statements are together insufficient. The answer is E
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
Let:
MG= Minivan on Gasoline
ME=Minivan on Electricity
SG= Sedan on Gasoline
SE= Sedan on Electricity
M= Minivan
S= Sedan
E= Electricity
G= Gasoline
Question: Prob (SE)> Prob (MG)
Statement 1
P(M or G)=2/5
P(M) + P(G)-P(MG)= 2/5
Since we don't know P(M) nor P(G) we cannot get P (MG).
Thus statement 1 is insufficient.
Statement 2
P(S or E)=5/8
P(S) or P(E) - P(SE) = 5/8
Likewise like statement 1 there are more than 2 unknowns in the expression which makes statement 2 insufficient
Jointly
We know from statement 1 that:
P(M) + P(G)-P(MG)= 2/5 ---- (1)
and from statement 2:
P(S) + P(E) - P(SE) = 5/8 ---- (2)
P(S) + P(E) = 5/8 +P(SE) ----(3)
we know that
P(M)+ P(S)=1 and
P(E) + P(G) =1
thus P(M)= 1-P(S)
P (G) =1- P(E)
We can express statement 1 in terms of statement 2
1-P(S) +1-P(E)-P(MG)= 2/5
2-P(S)-P(E)- P(MG)=2/5
2-(P(S)+P(E))=2/5+P(MG) -----(4)
substitute equation (3) into equation (4)
2 -(5/8+P(SE))= 2/5+P(MG)
2- (5/8 + P(SE) = 2/5+P(MG)
11/8- P(SE) = 2/5+P(MG)
P(SE) +P(MG) =11/8 - 2/5
P(SE) +P(MG) =39/40
This gives us the sum of the 2 probabilities but does not answer whether
Prob (SE)> Prob (MG)
Hence both statements are together insufficient. The answer is E
04 Jul 2025, 03:26
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Statement 1: insufficient
The chance that a vehicle is either a minivan or runs on gasoline is 2/5. This includes overlap but doesn’t separate how many are both or how many are only one.
Statement 2 : insufficient
The chance that a vehicle is either a sedan or runs on electricity is 5/8. Again, we don’t know how many are both or only one.
Combining statements 1 and 2, we know the total percentages for these two broad groups, but we still can’t isolate how many vehicles are sedans on electricity or minivans on gasoline.
Since we can't directly compare the two specific groups we need,
Answer: (E) Neither statement alone nor both statements together are sufficient.
The chance that a vehicle is either a minivan or runs on gasoline is 2/5. This includes overlap but doesn’t separate how many are both or how many are only one.
Statement 2 : insufficient
The chance that a vehicle is either a sedan or runs on electricity is 5/8. Again, we don’t know how many are both or only one.
Combining statements 1 and 2, we know the total percentages for these two broad groups, but we still can’t isolate how many vehicles are sedans on electricity or minivans on gasoline.
Since we can't directly compare the two specific groups we need,
Answer: (E) Neither statement alone nor both statements together are sufficient.
04 Jul 2025, 03:52
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Correct Answer: Yes the probability that it is a sedan running on electricity greater than the probability that it is a minivan running on gasoline.
Hence we can prove this statement by using both the statements but not any of the statement alone.
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
This statement alone cannot answer the question because we don't know anything about sedan vehicle and also electric vehicle too.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
Similarly this statement alone cannot answer the question because we don't know about minivan vehicle and gasoline type
If we have both statements together we have the information for both minivan and sedan vehicle, also we have information for gasoline as well as electric vehicle as well and we can compare and solve.
Hence we can prove this statement by using both the statements but not any of the statement alone.
(1) The probability that a randomly selected vehicle is either a minivan or runs on gasoline is 2/5.
This statement alone cannot answer the question because we don't know anything about sedan vehicle and also electric vehicle too.
(2) The probability that a randomly selected vehicle is either a sedan or runs on electricity is 5/8.
Similarly this statement alone cannot answer the question because we don't know about minivan vehicle and gasoline type
If we have both statements together we have the information for both minivan and sedan vehicle, also we have information for gasoline as well as electric vehicle as well and we can compare and solve.
