A family with 5 members (Father, Mother, two daughters and one son)

then what is meant by only one of the parents can drive?

There are 2 cases possible
1. when father is driving-
2 sisters can sit in 4 ways ( when 1 in the front, other in the either of the 3 back seats and when both are in the window seats in the back), now both sister can be arranged in 2! ways and mother and son can be sit in 2! ways on remaining 2 seats
total number of ways possible in this case= 4*2!*2!=16
2. When mother is driving
total number of ways possible in this case= 4*2!*2!=16(we can find in the similar manner as the 1st case)
Total number of ways with the given restrictions=16+16=32
Probability= 32/120= 4/15

nick1816 Saurabhminocha mbaprep2016 GMATinsight
But only one parent can drive -- therefore there's only one case, mother or father drives, and the total is 16/120 = 2/15. The OA is incorrect.

choose driver as mother = 1
rest of the car = 4! = 24 arrangements
bad outcomes have backseat = DDX or XDD, where daughters can swap, so 4 bad orders and 2 choices of F or S for the position X (where person not chosen gets shotgun) = 4*2 = 8 arrangements

M F/S
D(1/2) D(2/1) S/F

M F/S
S/F D(1/2) D(2/1)

good outcomes = total of 24 - bad 8 = 16

total number of ways =5!

Acceptable arrangement can be in the following way
either mother of father can drive the car (2 ways)
the seat beside the driver can be filled in 4 ways(mother/father,daughter1,daughter2,son)
and the rear 3 seats can be filled in 6 ways
therefore the total number of arrangement is 48.
But we require the daughters to not seat adjacent to each other. the only ways wherein they're next to one another are if either of the daughter occupies the middle rear seat and the other daughter sits on any of the remaining rear seat . The number of ways in which such an arrangement is possible is 16. (fix the positions if the daughters on the rear seats and assign values to the 3 vacant seats appropriately)
so the total acceptable ways are 48-16=32

And the probability asked in this question is = 32/120 ; 120=5!
=4/15

Give kudos if this was helpful.
thank you.

"only one of the parents can drive"

but parents = mother or father
so the number of parents is 2.
you can assign two individuals as the driver .

It doesn't say "only one can drive at a time," since that's obvious (how cars work). It's says only one can drive, meaning only one is capable.
If the question really intends for you to interpret it that way, why include the restriction at all? It should say "only a parent can drive."

since we dont know who is capable of driving we have to consider all the possible cases.
in case 1 the driver can be the mother
and in the second case the father can be the driver.

since the question explicitly doesn't mention whether who the driver is , its safe to assume two cases as stated above and proceed with the solution.
In either case only one parent is capable of driving the car and it satisfies the condition given in the question.

"only one of the parents can drive"
you read this as: only one of the parents can drive == either parent
I read this as: only one (of the parents) can drive == either mom or dad == there's only one case

Super unclear.

I tried the following way:

1- P(both girls sit together)

Only 1 parent can drive: so driver seat can be filled in 2 ways
both girls can sit together only in 2 of the 3 seats at the back in 2x2 = 4 ways
the remaining 2 seats can be occupied in 2 ways

so total ways: girls can sit together: 2 x 4 x 2 = 16 ways

Total ways of arranging 5 people: 5!

1 - (16/5!) = (2/15)

The OA however, is (4/15). What am I missing here?

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