Combinatorics and Probability Question

[Houman]

I'm having a difficult time comprehending this solution and if someone can explain it to me I would very much appreciate it!

(this is not a OG question but a practice problem from MGMAT)

Question: 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?

Solution:
Step 1) (1/6)^3(5/6) + (1/6)^3(5/6)+(1/6)^3(5/6)+(1/6)^3(5/6) = 4(1/6)^3(5/6) --> probability of rolling a two exactly 3 times out of 4 attempts.
Step 2) ADD the probability of rolling a two exactly 4 times (1/6)^4

Final answer: 4(1/6)^3(5/6)+(1/6)^4

What I don't get is why do we have to add the probability of rolling two exactly 4 times to step 1? Doesn't step 1 answer the question which is the probability of rolling a two exactly 3 times out of 4? What am I missing here? :roll:

Thanks in advance for your help.

[Brent@GMATPrepNow]

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

P(roll 2 at least 3 times) = P(roll 2 three times OR roll 2 four times)
= P(roll 2 three times) + P(roll 2 four times)

P(roll 2 three times)
There are 4 different ways we can roll a 2 three times:
a) 2, 2, 2, non-2
b) 2, 2, non-2, 2
c) 2, non-2, 2, 2
d) non-2, 2, 2, 2
The fortunate thing is that the probability of each occurrence is equal. Here's what I mean:
P(2, 2, 2, non-2) = (1/6)(1/6)(1/6)(5/6) = [(1/6)^3](5/6)
Similarly, P(non-2, 2, 2, 2) = (5/6)(1/6)(1/6)(1/6) = [(1/6)^3](5/6)
etc...
So, P((roll 2 three times) equals the sum of the four probabilities = 4[(1/6)^3](5/6)


P(roll 2 four times)
P(roll 2 four times) = P(2, 2, 2, 2)
= (1/6)(1/6)(1/6)(1/6)
= (1/6)^4


So, P(roll 2 at least 3 times) = 4[(1/6)^3](5/6) + (1/6)^4
= D

Cheers,
Brent

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Hi azizi2011,

Brent has explained the calculation, so I won't double up on that. For future reference, in probability questions, it's fairly common that there will be "more than one way to do something". You'll have to keep track of "what you want to have happen" and all the various ways to make it happen. While probability math almost always involves multiplication of some type, it also frequently involves addition.

Here's a simple example:

How many ways are there to roll a sum of 7 when rolling 2 fair 6-sided dice?

Total possibilities: 6 x 6 = 36

Total ways to sum to 7:
1 + 6
6 + 1
2 + 5
5 + 2
3 + 4
4 + 3

Add up all of the ways = 6

Total Probability = 6/36 = 1/6

As an aside, if you're ever in a casino, then you can use this type of math to figure out all the bets on a craps table (and why they all work against you).

GMAT assassins aren't born, they're made,
Rich

[[email protected]]

Hi azizi2011,

Brent has explained the calculation, so I won't double up on that. For future reference, in probability questions, it's fairly common that there will be "more than one way to do something". You'll have to keep track of "what you want to have happen" and all the various ways to make it happen. While probability math almost always involves multiplication of some type, it also frequently involves addition.

Here's a simple example:

How many ways are there to roll a sum of 7 when rolling 2 fair 6-sided dice?

Total possibilities: 6 x 6 = 36

Total ways to sum to 7:
1 + 6
6 + 1
2 + 5
5 + 2
3 + 4
4 + 3

Add up all of the ways = 6

Total Probability = 6/36 = 1/6

As an aside, if you're ever in a casino, then you can use this type of math to figure out all the bets on a craps table (and why they all work against you).

GMAT assassins aren't born, they're made,
Rich

[Houman]

Thank you both!