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"The die is cast" Problem

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"The die is cast" Problem

by XLogic » Fri Feb 18, 2011 11:32 am
Ok, I came across the following question in the Manhattan GMAT strategy guide -- Free publicity :-)

"If a fair coin is tossed 3 times, what is the probability that it will turn up heads exactly twice?"

To solve, we list combos: HHH HHT HTH THH HTT THT TTH TTT
> 8 total combinations, and only 3 out of 8 have 2 Heads, therefore the answer is 3/8

Question: How would you solve the following? (Please provide detailed steps)

1. "If a die is cast 3 times, what is the probability that it will turn up 6 exactly twice?"
2. "If a die is cast 4 times, what is the probability that it will turn up 6 on the first try, and 5 on the third try exactly?"

Thanks.

PS. Provide shortcuts too.
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by Night reader » Fri Feb 18, 2011 12:07 pm
XLogic wrote:Ok, I came across the following question in the Manhattan GMAT strategy guide -- Free publicity :-)

"If a fair coin is tossed 3 times, what is the probability that it will turn up heads exactly twice?"

To solve, we list combos: HHH HHT HTH THH HTT THT TTH TTT
> 8 total combinations, and only 3 out of 8 have 2 Heads, therefore the answer is 3/8

Question: How would you solve the following? (Please provide detailed steps)

1. "If a die is cast 3 times, what is the probability that it will turn up 6 exactly twice?"
2. "If a die is cast 4 times, what is the probability that it will turn up 6 on the first try, and 5 on the third try exactly?"

Thanks.

PS. Provide shortcuts too.
the events are independent, occurrence of one doesn't affect the occurrence of the other event.
1. luckily it says one die --> P(6, twice)=(1/6)^2 * 5/6 OR 5/216, total possibilities 3C2=3*2/2=3 (out of 3!). Hence (5/216)*3=[spoiler]5/72[/spoiler]
2. again got lucky --> P(6, 1st try)=1/6, P(5, third try)=1/6 --> (1/6)*((5/6)*2)*(1/6)*(5/6)=[spoiler]25/648[/spoiler]
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by XLogic » Fri Feb 18, 2011 12:59 pm
Nice.

What is 3C2? And can you solve for the two dice situation you mentioned.

Example:

If two die are rolled together 3 times, what is the probability that on exactly two rolls, 1 die number is odd and both die numbers together (i.e. placed side-by-side) is divisible by 11?

OR

If two die are rolled together 3 times, what is the probability that on exactly two rolls, 1 die number is odd and the sum of both die numbers is even? (not sure this is solvable)

I doubt something like this will show up on the GMAT. Just curious.
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by Pilot » Fri Feb 18, 2011 1:13 pm
Night reader wrote:
XLogic wrote:Ok, I came across the following question in the Manhattan GMAT strategy guide -- Free publicity :-)

"If a fair coin is tossed 3 times, what is the probability that it will turn up heads exactly twice?"

To solve, we list combos: HHH HHT HTH THH HTT THT TTH TTT
> 8 total combinations, and only 3 out of 8 have 2 Heads, therefore the answer is 3/8

Question: How would you solve the following? (Please provide detailed steps)

1. "If a die is cast 3 times, what is the probability that it will turn up 6 exactly twice?"
2. "If a die is cast 4 times, what is the probability that it will turn up 6 on the first try, and 5 on the third try exactly?"

Thanks.

PS. Provide shortcuts too.
the events are independent, occurrence of one doesn't affect the occurrence of the other event.
1. luckily it says one die --> P(6, twice)=(1/6)^2 * 5/6 OR 5/216, total possibilities 3C2=3*2/2=3 (out of 3!). Hence (5/216)*3=[spoiler]5/72[/spoiler]
2. again got lucky --> P(6, 1st try)=1/6, P(5, third try)=1/6 --> (1/6)*((5/6)*2)*(1/6)*(5/6)=[spoiler]25/648[/spoiler]
Do you suppose that 6 and 5 occure exectly one time on the second part of your solution?
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by Night reader » Fri Feb 18, 2011 1:32 pm
XLogic wrote:Nice.

What is 3C2? And can you solve for the two dice situation you mentioned.
actually, made extra assumption in the first solution - clued two exact probabilities - this one must be
the events are independent, occurrence of one doesn't affect the occurrence of the other event.
1. luckily it says one die --> P(6, twice)=(1/6)^2 * 5/6 OR 5/216, total possibilities 3C2=3*2/2=3 (out of 3!). Hence (5/216)*3!=5/36
2. again got lucky --> P(6, 1st try)=1/6, P(5, third try)=1/6 --> (1/6)*((5/6)*2)*(1/6)*(5/6)=25/648
instead of 3C2 which is C(3;2) should be 3P2 (P, permutation)

let me try further
XLogic wrote: Example:

If two die are rolled together 3 times, what is the probability that on exactly two rolls, 1 die number is odd and both die numbers together (i.e. placed side-by-side) is divisible by 11?
P(one die is odd)=1/2 and P(another die is not odd)=1/2, P(both dies divisible by 11)=1/18 --> (1/2)^2 *(1/2)^2 *(1/18)=1/288 ---> 1/288 *3!= 1/48 Ooops

XLogic wrote: OR

If two die are rolled together 3 times, what is the probability that on exactly two rolls, 1 die number is odd and the sum of both die numbers is even? (not sure this is solvable)

I doubt something like this will show up on the GMAT. Just curious.
die number is odd and the sum of both die numbers is even means one die is odd and the other is odd too, together their sum is even; the third pair of dies can not be odd, hence both even, BUT then we contradict to the condition on exactly two rolls ... the sum of both die numbers is even - we may have sums of the third pairs being even too and this makes more than two exactly rolls.

So IOM unsolvable
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by ruplun » Fri Apr 29, 2011 1:11 pm
1. luckily it says one die --> P(6, twice)=(1/6)^2 * 5/6 OR 5/216, total possibilities 3C2=3*2/2=3 (out of 3!). Hence (5/216)*3!=5/36
2. again got lucky --> P(6, 1st try)=1/6, P(5, third try)=1/6 --> (1/6)*((5/6)*2)*(1/6)*(5/6)=25/648

please can anyone explain step by step how this values are getting formed... why the square is reqd?
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by ruplun » Thu Jul 28, 2011 11:15 pm
can u explain in detail the logic behind solving the 3 problems ... am not able to get the sloution provided
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by XLogic » Fri Jul 29, 2011 12:25 am
ruplun wrote:can u explain in detail the logic behind solving the 3 problems ... am not able to get the sloution provided
It's been a while, I'll do the first one for now. Let me know if it makes sense.

1. If a fair coin is tossed 3 times, what is the probability that it will turn up heads exactly twice?

> We are looking for the probability of exactly 2 heads and 1 tail (on 3 tosses)
What are the possible outcomes?
a. First toss: Head, Second Toss: Head, Third toss: Tail
b. First toss: Head, Second toss: Tail, Third toss: Head
c. First toss: Tail, Second toss: Head, Third toss: Head

(there are no other possibilities)

(a) Probability (head and head and tail) = 1/2 * 1/2 * 1/2 = 1/8
(b) Probability (head and tail and head) = 1/2 * 1/2 * 1/2 = 1/8
(c) Probability (tail and head and head) = 1/2 * 1/2 * 1/2 = 1/8

Either outcome a, b, or c will guarantee exactly 2 heads and 1 tail
therefore, Probability (a or b or c) = P(a) + P(b) + P(c) = 3 * (1/8) = 3/8

Short Cut:

Notice that you need 2 heads and one tail.
Each head/tail probability is 1/2, therefore, 2 heads and one tail = (1/2)^3
How many ways can we arrange 2 heads and one tail? 3!/2!1! = 6/2 = 3
So, to get 2 heads and one tail --> 3 different ways = 3 * (1/8) = 3/8

I hope I got it all. What do you think?
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by XLogic » Fri Jul 29, 2011 1:05 am
Okay, let's try the others. Already ate my spinach :)

2. If a die is cast 3 times, what is the probability that it will turn up 6 exactly twice?

Possibilities:
6,6, not-6
6, not-6, 6
not-6, 6, 6

not-6 = P(1 or 2 or 3 or 4 or 5) = 5/6

P(Exactly two 6's in 3 tosses) = 1/6 * 1/6 * 5/6 = 5/216

5/216 --> 3 ways = 3 * 5/216 = 15/216
15/216 = 5/72

3. If a die is cast 4 times, what is the probability that it will turn up 6 on the first try, and 5 on the third try exactly?

<Hmm.. Not sure the original explanation for this one is correct. I'll post later..>

Note: I don't think questions like these will show up on the GMAT (I may be wrong). Either way, at the time, I thought it was good exercise :)
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by XLogic » Fri Jul 29, 2011 9:26 am
XLogic wrote: 3. If a die is cast 4 times, what is the probability that it will turn up 6 on the first try, and 5 on the third try exactly?
I think this question is incomplete, or at least, we'd have to make some assumptions to solve it properly. A more complete question would be:

If a die is cast 4 times, what is the probability that it will turn up "6" on the first try, and "5" on the third try, and neither 6 nor 5 show up any other times.

If so, we are looking for:
[6] [not-5,6] [5] [not-5,6]

Probability(not 5,6) = 1 or 2 or 3 or 4 = (4 * 1/6) = 2/3
Probability(6,X,5,X) = 1/6 * 2/3 * 1/6 * 2/3 = 1/81

I'll be interested in other takes on this. (Hope I didn't miss a step or two)
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