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Probability + Permutation - MGMAT

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Probability + Permutation - MGMAT

by Lifetron » Tue Feb 12, 2013 9:49 pm
For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled
4 times, which of the following is the probability that the outcome will be a two at
least 3 times?

(A) (1/6)^4
(B) 2 * (1/6)^3 + (1/6)^4
(C) 3 * (1/6)^3 * (5/6) + (1/6)^4
(D) 4 * (1/6)^3 * (5/6) + (1/6)^4
(E) 6 * (1/6)^3 * (5/6) + (1/6)^4

OA D

Source : MGMAT

Somehow I got the answer, but I need a proper work-flow for this type of questions. They can be a bit confusing when the question gets more complex.

Thank you !
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by hemant_rajput » Tue Feb 12, 2013 11:09 pm
gughanbose wrote:For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled
4 times, which of the following is the probability that the outcome will be a two at
least 3 times?

(A) (1/6)^4
(B) 2 * (1/6)^3 + (1/6)^4
(C) 3 * (1/6)^3 * (5/6) + (1/6)^4
(D) 4 * (1/6)^3 * (5/6) + (1/6)^4
(E) 6 * (1/6)^3 * (5/6) + (1/6)^4

OA D

Source : MGMAT

Somehow I got the answer, but I need a proper work-flow for this type of questions. They can be a bit confusing when the question gets more complex.

Thank you !
this is my approach, it may not be the best but easy enough to deduce your own methodology.

you want to find the outcome will come 2, at least 3 times.

we have only two scenarios for outcome to come as two, every time we roll dice, 2 comes exactly 3 times or 2 comes exactly 4 times.

for 2 coming exactly 3 times:

1\6 , 1\6 , 1\6 and 5\6. Now in how many ways we can arrange this combination 4C3(for arranging 2 in 3 place out of 4 place) or 4C1(for arranging any number but 2, in 1 place out of 4 place).

this is equal to (1\6)^3 * (5\6) * 4

for 2 coming exactly 4 times:

(1/6)^4

so now we have (1\6)^3 * (5\6) * 4 + (1/6)^4

hope it helps.
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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by lunarpower » Wed Feb 13, 2013 9:24 am
gughanbose wrote:For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled
4 times, which of the following is the probability that the outcome will be a two at
least 3 times?
well, for this problem, there's no need to use combinatorial formulas at all. there are only five ways this outcome can happen, so you can just list out all five (remember, order matters here), find the probabilities, and add them together.

if "2" stands for rolling a 2, and "N" stands for rolling something that's not a 2, then you need one of the following 5 outcomes:
222N
22N2
2N22
N222
2222

the probability of each of the first 4 events is 1/6 x 1/6 x 1/6 x 5/6. (these products will originally appear in different orders -- e.g., the second one will appear as 1/6 x 1/6 x 5/6 x 1/6, and so forth -- but you can rearrange products however you want to.)

the probability of the last event is 1/6 x 1/6 x 1/6 x 1/6.

add them together and you get [4 x (1/6)^3 x 5/6] + [(1/6)^4].
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by ceilidh.erickson » Wed Feb 13, 2013 9:27 am
Hemant is exactly right. For a more general approach to solving these types of questions, think about this:
Probability = (desired outcome)/(total number of possible outcomes).

The first thing you should do is ask yourself - how many different desired outcomes are there? In this case, there are 2 outcomes that would work: all four 2's, or three 2's, and something else. If these two outcomes have different probabilities, you'll need to find them separately, then add them together. Right away, we can eliminate A, because it only takes into account the first possibility.

The second question to ask is - how many ways are there to get the desired outcome?

Desired outcome 1: all four 2's... there's only 1 way to get it: [2, 2, 2, 2]. So calculate the probabilities of each and multiply. (1/6)(1/6)(1/6)(1/6) = (1/6)^4

Desired outcome 2: three 2's and something else... there are 4 ways to get it: [2, 2, 2, x], [2, 2, x, 2], [2, x, 2, 2], [x, 2, 2, 2]. All 4 of these events will have the same probability, so we can just find the probability of one of them, and multiply it by 4.
[2, 2, 2, x]: (1/6)(1/6)(1/6)(5/6). Then simply multiply by 4.

If you want to save time, you don't even have to do this last step! As soon as you recognize that there are 4 ways to get it, you know it's going to be 4 times something. The only option for that is D!
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by lunarpower » Wed Feb 13, 2013 9:43 am
3 minutes apart. lol
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by ceilidh.erickson » Wed Feb 13, 2013 10:06 am
Great minds think alike! (At least as far as the GMAT is concerned. That's definitely not true in real life)
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by lunarpower » Wed Feb 13, 2013 10:15 am
ceilidh.erickson wrote:Great minds think alike! (At least as far as the GMAT is concerned. That's definitely not true in real life)
i'll leave that particular discussion to the great minds.

cheers!
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by SFGMATtrainer » Sat Feb 16, 2013 11:25 am
When the question uses "at least" or "or", the probabilities of each outcome need to added and/or subtracted from 1. Here, adding would be enough: (P of exactly three 2s) + (P of all four 2s)

I have a 5-step method for probability of multiple events that might help.
Probability of exactly three 2s:
1) Lay out the number of events (four): _ _ _ _
2) Label the events (underneath each slot) with one example of the desired outcome: 2 2 2 N2
3) Label the probability of each specific event and multiply: 1/6 x 1/6 x 1/6 x 5/6 = 5/6^4
4) Determine the number of possible of desired outcomes: 4 ways to roll exactly three 2s.
5) Same specific probabilities: multiply result of step 4) by product in step 3): 4 x (5/6^4)

Probability of all four 2s:
1) _ _ _ _
2) 2 2 2 2
3) 1/6 x 1/6 x 1/6 x 1/6 = 1/6^4
4) Only one possibility.
5) 1 x 1/6^4 = 1/6^4

So probability of at rolling at least three 2s = 4(5/6^4) + 1/6^4

I hope this helps.
Any "thanks" would be greatly appreciated!
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