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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.

Kudos [?]: 1776 [0], given: 794

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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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Re: Daniel is playing a dice game in which a dice is rolled 5 times   [#permalink] 11 Sep 2017, 09:37

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