Binomial situation

[PetriF258]

I am losing my mind! So what is the probability of rolling two 5's when you roll 10 dices?

So using binomial rule: 10C2 (1/6)^2 (5/6)^8

If I multiply this I get about 29%.

Now. Lets say I want to calculate this another method. There are 6^10 possibilities for the dice. How many of these possibilities represent two 5's? 10C2 = 45. So then the probability of rolling two 5's is 45 / 6^10 which is not 29%?

It works with the prob of getting 2 heads if you flip a coin 3 times:
Binomial rule: 3C2 (1/2)^2 (1/2)^1 = 3/8.

And 2^3 = 8 possibilities; 3C2 = 3, therfore 3/8.

Full disclosure: I have purchased both Manhattan Interact and Magoosh and this question comes from a Magoos lecture.

Thanks

[PetriF258]

I think I just got it. When I considered the number of possibilities where there are two 5's, I forgot about the other 8 dices. So number of times that two 5's could be rolled is 10C2 times 5^8. For each time I throw two 5's there is the actually 5^8 number of possibilities for those two 5's. For example: If 5 =x
(x)(x)(5)(5)(5)(5)(5)(5)(5)(5); 5 representing the possibilities for the other dices; ie: 1,2,3,4,6.

Therefor the probability is 10C2 5^8 / 6^10 = 29%

So why did this work with the coin? The number of possibilities for the other coin is only 1; (H)(H)(CAN ONLY BE TAILS:1)

Ok, I think I understand but now something that I struggle with: When should I go straight probability "(1/6)" and when should I go with "successes / total number of outcomes"?

[PetriF258]

I know, I am bumping my own thread :-)

One final question and I would love it if the MG staff could indicate whether 1) the answer is correct and 2) what level this might be as I thought of the question myself:

What is the probability of rolling a dice 10 times and getting no more than one double. In other words, 5 rolls must be one number, and the other 5 rolls must be the other 5 numbers?

Ok, pick any number, roll that number 4 more times, then roll each of the other 5 numbers once:

(6/6)(1/6)^4(5/6)(4/6)(3/6)(2/6)(1/6) this would give me the probability of rolling any number for the first 5 rolls and then each other number on the next 5 rolls. But, it does not necessarily have to be in this order, there are various combinations, 5C5, representing the various combinations in which we can roll the dice with the same number. Therefore, my final asnwer:

5C5 (6/6)(1/6)^4(5/6)(4/6)(3/6)(2/6)(1/6)

Thanks!

[RonPurewal]

I think I just got it. When I considered the number of possibilities where there are two 5's, I forgot about the other 8 dices. So number of times that two 5's could be rolled is 10C2 times 5^8. For each time I throw two 5's there is the actually 5^8 number of possibilities for those two 5's. For example: If 5 =x
(x)(x)(5)(5)(5)(5)(5)(5)(5)(5); 5 representing the possibilities for the other dices; ie: 1,2,3,4,6.

Therefor the probability is 10C2 5^8 / 6^10 = 29%

So why did this work with the coin? The number of possibilities for the other coin is only 1; (H)(H)(CAN ONLY BE TAILS:1)

All good.

Ok, I think I understand but now something that I struggle with: When should I go straight probability "(1/6)" and when should I go with "successes / total number of outcomes"?

You should learn both of them. Then, when you see a problem, just try whichever one comes to mind first. If you get stuck, then try the other one.

Do not burden yourself with a series of decisions to make BEFORE working a problem. That's bad news all around. Shoot first, ask questions later.

[RonPurewal]

I know, I am bumping my own thread :-)

One final question and I would love it if the MG staff could indicate whether 1) the answer is correct and 2) what level this might be as I thought of the question myself:

What is the probability of rolling a dice 10 times and getting no more than one double. In other words, 5 rolls must be one number, and the other 5 rolls must be the other 5 numbers?

Ok, pick any number, roll that number 4 more times, then roll each of the other 5 numbers once:

(6/6)(1/6)^4(5/6)(4/6)(3/6)(2/6)(1/6) this would give me the probability of rolling any number for the first 5 rolls and then each other number on the next 5 rolls. But, it does not necessarily have to be in this order, there are various combinations, 5C5, representing the various combinations in which we can roll the dice with the same number. Therefore, my final asnwer:

5C5 (6/6)(1/6)^4(5/6)(4/6)(3/6)(2/6)(1/6)

Thanks!

"5c5" is 1, so, old-fashioned common sense is enough to determine that "5c5" is wrong.

If you replace "5c5" with "10c5", I think this solution should work.

The chance that you'll see such a technically intense problem on the GMAT is nil. There has never been an official problem demanding anywhere close to this level of manipulation of "c" and/or "p" formulas.
In fact, the whole point of the GMAT is to contain problems that, though challenging, are NOT very technically demanding. (Take a look through some official problems, and note the degree of technical skill required to solve them. Quite modest.)

[PetriF258]

Hi Ron

My apologies for only replying now. Just want to say thanks.

Regards

[RonPurewal]

no worries. you're welcome.