MGMAT Advanced problem

[umaa]

[spoiler]For the first question, it says we should multiply all the chances. ie, 2*2*2*2*2*2. Why didn't we take factorial? I'm pretty weak @ Prob, perm & combination. So, any valid explanation is welcome.[/spoiler]

[Brent@GMATPrepNow]

[spoiler]For the first question, it says we should multiply all the chances. ie, 2*2*2*2*2*2. Why didn't we take factorial? I'm pretty weak @ Prob, perm & combination. So, any valid explanation is welcome.[/spoiler]

If we wanted to, we could use factorials (in the form of combinations).
a) The total number of configurations could counted as follows:
(The total number of configurations with zero lights on) + (The total number of configurations with one light on) + (The total number of configurations with two lights on) + . . . and so on.
Since the order in which the lights are on doesn't matter (e.g., having lights 1, 2 and 5 on is the same as have lights 5, 1 and 2 on), we can use combinations.
So, we get: 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 which equals 1+6+15+20+15+6+1=64
Another way to calculate this is to break the counting task into stages and apply the Fundamental Counting Principle.
We get (the number of ways in which light 1 can be configured) x (the number of ways in which light 2 can be configured) x (the number of ways in which light 3 can be configured) x etc.
Since each light can be either on or off, each stage can be accomplished in 2 stages, so we get 2x2x2x2x2x2 = 64

b) The number of way in which 3 bulbs can be lit. We have 6 bulbs and we want to select 3 to be lit. Since order doesn't matter, this can be accomplished in 6C3 ways (20 ways)

c) probability = (# of outcomes favorable to having 3 lit bulbs)/(total # of outcomes) = 20/36 = 5/9

[umaa]

Thanks Brent.

By the way, for the third question isn't it 20/64?

[Brent@GMATPrepNow]

Thanks Brent.
By the way, for the third question isn't it 20/64?

Oops, yes it's 20/64 (5/16)
I had a hard time simply combining answers from parts a) and b)

[Stuart@KaplanGMAT]

Hi,

this is what we affectionately term a "pseud coin-flip question".

Any time you see "an equal chance of event A or B", think coin flips!

Let's rephrase the set up:

Kevin flips a fair coin 6 times.

Now the questions:

1) how many different results are possible?

For coin flip questions, there are always 2^n possible results. 2^6 = 64.

2) how many different ways can we get exactly 3 heads?

We want 3 results out of a total of 6 flips, so we use the combinations formula:

6C3 = 6!/3!3! = 6*5*4/3*2 = 20

3) what's the probability of getting exactly 3 heads?

Prob(k results out of n flips) = nCk/2^n = 6C3/2^6 = 20/64 = 5/16

(Note this this is just a specific application of the general probability formula:

probability = # of desired outcomes / total # of possibilities.)