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Daniel is playing a dice game in which a dice is rolled 5 times

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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 02 Oct 2015, 04:49
Engr2012 wrote:
longfellow wrote:
Bunuel/ Karishma/ all:

Can you please provide an explanation for the question here, as my answer is not even present as a choice!

Explanation: game gets over after getting a unique number 3 times:

1. 444: 1/6 (can be written as 1/6*6/6*6/6 for calculations= 36/216)
2. 44N4: 5/6*1/6 (can be written as 5/6*1/6*6/6= 30/216)
3. 44NN4: 5/6*5/6*1/6 (25/216)
4. 44NNN:5/6*1/6*1/6 (5/216)

Sum equals: 96/216

Whats wrong with this approach?


The mistake you are doing is to assume that the game stops as soon as you get 3 is a row or even when you get a number 3 times out of 4. There will be 5 throws of the dice no matter what. Look at my solution above.




Got it! moral of the story: Do not assume!!

Thanks Engr2012!

And great turnaround time! Really appreciate it. :)
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 02 Oct 2015, 04:56
Quote:
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216


The answer I calculated is not in the answer options.

Case 1

He wins when number 4 comes on dice. It can happens in 3 ways. 4 on the third throw, 4 on the fourth throw, 4 on the fifth throw

Calculating the probability of all these cases and adding them :

1/6 + (5/6)(1/6) + (5/6)(5/6)(1/6) = 91/216

Case 2

He wins with throw of some other number in next 3 throws. Only one case i.e same number on next three throws.

Probability : (5/6)(1/6)(1/6) = 5/216

The final answer will be Case 1 + Case 2 = 97/216.

What is wrong with approach ?
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 02 Oct 2015, 05:13
longfellow wrote:
Engr2012 wrote:
longfellow wrote:
Bunuel/ Karishma/ all:

Can you please provide an explanation for the question here, as my answer is not even present as a choice!

Explanation: game gets over after getting a unique number 3 times:

1. 444: 1/6 (can be written as 1/6*6/6*6/6 for calculations= 36/216)
2. 44N4: 5/6*1/6 (can be written as 5/6*1/6*6/6= 30/216)
3. 44NN4: 5/6*5/6*1/6 (25/216)
4. 44NNN:5/6*1/6*1/6 (5/216)

Sum equals: 96/216

Whats wrong with this approach?


The mistake you are doing is to assume that the game stops as soon as you get 3 is a row or even when you get a number 3 times out of 4. There will be 5 throws of the dice no matter what. Look at my solution above.




Got it! moral of the story: Do not assume!!

Thanks Engr2012!

And great turnaround time! Really appreciate it. :)


It is not an assumption as the question clearly states that "dice is rolled 5 times". So be careful with what is mentioned in the question.
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 02 Oct 2015, 05:15
abhishekchaudhary wrote:
Quote:
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216


The answer I calculated is not in the answer options.

Case 1

He wins when number 4 comes on dice. It can happens in 3 ways. 4 on the third throw, 4 on the fourth throw, 4 on the fifth throw

Calculating the probability of all these cases and adding them :

1/6 + (5/6)(1/6) + (5/6)(5/6)(1/6) = 91/216

Case 2

He wins with throw of some other number in next 3 throws. Only one case i.e same number on next three throws.

Probability : (5/6)(1/6)(1/6) = 5/216

The final answer will be Case 1 + Case 2 = 97/216.

What is wrong with approach ?


Look at the solution daniel-is-playing-a-dice-game-in-which-a-dice-is-rolled-5-times-205827.html#p1576074
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 29 Aug 2016, 03:28
4XY (5c2) =6*10
4XX = 5*3
XXX (5)
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 03 Sep 2016, 05:19
According to my calculation there are 3 scenarios possible for a win
Scenario 1: combinations 4xx,x4x,xx4 where x can be any number between 1 to 6 except 4 total ways will be 3*1*5*5 = 75
Scenario 2 :Combinations 44x,4x4,x44 total ways 3*1*1*5 = 15
Scenario 3 : 444 total ways = 1
total possible outcomes = 6*6*6 =216

So probability of winning the game = (75+15+1)/216 = 91/216
Please correct me if am wrong
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Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 03 Sep 2016, 05:54
gthrb22 wrote:
According to my calculation there are 3 scenarios possible for a win
Scenario 1: combinations 4xx,x4x,xx4 where x can be any number between 1 to 6 except 4 total ways will be 3*1*5*5 = 75
Scenario 2 :Combinations 44x,4x4,x44 total ways 3*1*1*5 = 15
Scenario 3 : 444 total ways = 1
total possible outcomes = 6*6*6 =216

So probability of winning the game = (75+15+1)/216 = 91/216
Please correct me if am wrong


You have missed the line "If a number turns up exactly 3 times then the game is won."

Notice the word EXACTLY. We are already given two 4s , so we MUST have only one four. So, Scenario 2 above is incorrect.

Also, notice Scenario 3 must be only for the numbers that are NOT 4.

Or It should be something like :

Scenario 1: combinations 4xx,x4x,xx4 where x can be any number between 1 to 6 except 4 total ways will be 3*1*5*5 = 75
Scenario 2 :Combinations 44x,4x4,x44 total ways 3*1*1*5 = 15
Scenario 3 : 444

Scenario 3 : XXX(X <> 4) total ways = 1 5.

total possible outcomes = 6*6*6 =216

So, probability = 80/216
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 09 Apr 2017, 07:29
BrainLab wrote:
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores


Please help...spent 2 hours on it...till wondering where did i go wriong....i didnt get any of the exlanation below.
My approach....to win he should get "at least" one more 4 in a total of three throws....
so the question becomes what is the probability that he get at least one 4 in three throws oif the dice...
First throw no chances of 4 = 5/6
second throw no chances of 4 = 5/6
fourth throw no chnces of 4 = 5/6
so in all the thrree throws chances of never getting a 4 = 75/216
so chances of getting at least one 4 = 1-75/216 = 141/216

what is wrong in my approach...help!!!!!!!!
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 27 Jun 2017, 19:35
saurabhsavant wrote:
BrainLab wrote:
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores


Please help...spent 2 hours on it...till wondering where did i go wriong....i didnt get any of the exlanation below.
My approach....to win he should get "at least" one more 4 in a total of three throws....
so the question becomes what is the probability that he get at least one 4 in three throws oif the dice...
First throw no chances of 4 = 5/6
second throw no chances of 4 = 5/6
fourth throw no chnces of 4 = 5/6
so in all the thrree throws chances of never getting a 4 = 75/216
so chances of getting at least one 4 = 1-75/216 = 141/216

what is wrong in my approach...help!!!!!!!!


Read the question again. The question says that he wins when he gets EXACTLY 3 of the same number and thus you only need 1 more 4 to win and not "atleast" one more as you are incorrectly assuming.
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 27 Jun 2017, 21:34
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Very good and tricky question
The game is won when 1 number show up Exactly 3 times
We are given first two throw result in 4 so we need one more 4 to win , this can be done in 1*1*5/6*5/6*1/6*3
where 5/6 is probability of not getting 4 and they can be arranged in 3 ways.
Now we can also win the game if we get rest of numbers 3 times i.e 1,2,3,5,6
so probability will be =5*1/6*1/6*1/6
The probability of winning =5/216+75/216=80/216
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Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 27 Jun 2017, 22:02
1. Daniel will win if he throws 4 once or any of the other numbers thrice in the next three throws.
2. Number of possibilities 4 can occur in the third throw out of 5 throws is 5^2=25. Similarly for the fourth and fifth throws for a total of 75
3. Number of possibilities that the other numbers are thrown thrice is 1 + 1+1+1+1=5
4. Probability is , (75+5)/216=80/216
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 11 Sep 2017, 09:37
Ans is 80/216
44 4xx => 25/216
44 x4x=> 25/216
44 xx4=>25/216
44 111 /222/333/555/666=> 5 x 1/216 = 5 /216
total = 80/216
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Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 17 May 2018, 05:54
Why is it assumed that after getting a 4 the other 2 numbers would be the same and this leads to divide by 2!. The 2 numbers can be 2 different numbers other than 4.
Pls help.

Also do we need to learn conditional probability and Bayes theorem etc for GMAT?
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 23 May 2018, 10:35
Probability of Daniel not winning the game = 5/6 x 5/6 x 5/6 = 125/216
Probability of win = 1- 125/216 = 91/216.

Do not understand why the solution has to be so complicated. Please explain if it's wrong.
Thanks in advance.
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 23 May 2018, 10:59
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nidhiprasad wrote:
Why is it assumed that after getting a 4 the other 2 numbers would be the same and this leads to divide by 2!. The 2 numbers can be 2 different numbers other than 4.
Pls help.

Also do we need to learn conditional probability and Bayes theorem etc for GMAT?


Hey nidhiprasad ,

We are no where assuming that case. When I am saying 5 possibilities for each of 444XX, I can say it could be anything(same or different). Look at the solution posted by me here and let me know in case of further concerns.

Also, there is no need to learn probability and Bayes theorem etc for GMAT.

duttarupam2344 wrote:
Probability of Daniel not winning the game = 5/6 x 5/6 x 5/6 = 125/216
Probability of win = 1- 125/216 = 91/216.

Do not understand why the solution has to be so complicated. Please explain if it's wrong.
Thanks in advance.


Hey duttarupam2344 ,

You have missed a scenario when your last three numbers are infact the same. In that case also, the player will win. Hence, you cannot directly talk 5/6 for all the three scenarios.

Does that make sense?
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 21 Jun 2018, 09:48
abhishekchaudhary wrote:
Quote:
Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216


The answer I calculated is not in the answer options.

Case 1

He wins when number 4 comes on dice. It can happens in 3 ways. 4 on the third throw, 4 on the fourth throw, 4 on the fifth throw

Calculating the probability of all these cases and adding them :

1/6 + (5/6)(1/6) + (5/6)(5/6)(1/6) = 91/216

Case 2

He wins with throw of some other number in next 3 throws. Only one case i.e same number on next three throws.

Probability : (5/6)(1/6)(1/6) = 5/216

The final answer will be Case 1 + Case 2 = 97/216.

What is wrong with approach ?


Haha... I also did exactly same mistake. But understood after reading other's answer.
Please read question again. It clearly says that exactly 3 times. Not less and not more.
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 21 Jun 2018, 09:52
nidhiprasad wrote:
Why is it assumed that after getting a 4 the other 2 numbers would be the same and this leads to divide by 2!. The 2 numbers can be 2 different numbers other than 4.
Pls help.

Also do we need to learn conditional probability and Bayes theorem etc for GMAT?


It is not assumed that after getting a 4 the other 2 numbers would be the same. Basically it does not make any difference whether two numbers are same or not.
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Re: Daniel is playing a dice game in which a dice is rolled 5 times  [#permalink]

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New post 21 Jun 2018, 09:54
duttarupam2344 wrote:
Probability of Daniel not winning the game = 5/6 x 5/6 x 5/6 = 125/216
Probability of win = 1- 125/216 = 91/216.

Do not understand why the solution has to be so complicated. Please explain if it's wrong.
Thanks in advance.


You say that Probability of Daniel not winning the game = 5/6 x 5/6 x 5/6. But it is incorrect.
What if he gets all three same number, Then he will win. So this is not the correct approach to reach answer.
You need to solve by taking cases.
Re: Daniel is playing a dice game in which a dice is rolled 5 times &nbs [#permalink] 21 Jun 2018, 09:54

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