Hi. I'm happy to contribute my 2¢ on this.
First, I will say: I completely agree with
vishal.pathak and the answer he got. I am writing in response to
amit.trivedi, who wrote: "
Could anybody help to understand this question as according to me the answer should be either C or E. A alone does not give me the answer..."
The prompt: "
A team of workers including Tom and Dick work in the same office according to a schedule that ensures that exactly two team members will be present at a given time, and that in the course of the week all the team members work an equal number of hours. What is the probability that a visitor to the office who doesn't know the schedule arrives to find both Tom and Dick in the office?"
So, when the visitor arrives, essentially he will find two randomly chosen members of the team. The question really boils down to: what is the probability that, if we select two team members at random, the two selected happen to be Tom & Dick.
Statement #1: The team has three members.
If the team has 3 members, n = 3, then the total number of pairs possible is given by the general formula for combinations. When you have a pool of n, and you are going to select from this pool a set of r, the total number of combinations = nCr = [n!]/[(r!)*((n-r)!)], where the "!" is the factorial symbol, and "n!" means the product of n and every integer below n, down to one. Thus, 5! = 5*4*3*2*1 = 120.
Here, the number of combinations is 3C2 = (3!)/[(2!)*(1!)] = 3. There are three possible pairs to be chosen from a group of three. Only one of those is the pair Tom & Dick, so the probability that the visitor arrives and find Tom & Dick there is 1/3.
I want to underscore:
as a general rule, you do not need to calculate all the way to an answer on the DS questions. I was only showing the entire calculation so that
amit.trivedi would be clear. I'll add: if combinations are something unfamiliar, it would be worthwhile studying them in whatever review materials you have.
So, with Statement #1, we calculated an exact value, so Statement #1 is
sufficient.
Statement #2: Tom and Dick worked together for the whole of the previous day.
This is a variant of the old probability question --- if I flip 10 heads in a row, what is the probability that the next flip is also heads? This is a deep idea about probability: the past doesn't matter. Things like coins have no memory, so no matter what happened on previous flips, the probability of heads on the next flip will always be 1/2.
Here, it's obviously not as pure a situation as a coin flip, but since we have no further information about the nature of the schedule, we can't draw any conclusions about what working one day would mean for the probability of working the next day. Not enough information. Statement #2 is
insufficient.
Thus, as vishal.pathak said, answer =
A.
Here's a free probability question, just for practice:
https://gmat.magoosh.com/questions/1035
Does all that make sense? Please let me know if you have questions on any of this.
Mike
