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

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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.

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

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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.

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

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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.

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

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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.

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

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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

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

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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

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

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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

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

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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.

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

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27 Jun 2017, 21:34

1

This post received KUDOS

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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We are more often frightened than hurt; and we suffer more from imagination than from reality

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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