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Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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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
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Originally posted by BrainLab on 20 Sep 2015, 14:18.
Last edited by BrainLab on 21 Sep 2015, 04:13, edited 1 time in total.



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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20 Sep 2015, 16:01
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 ?
a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216
Source: PerfectScores Good question. +1 For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things. Probability of any number from 1 to 6 = 1/6. Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not. 2 scenarios for a win: Scenario 1: 4 4 T N N , with T= 4 , N= not 4 Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots) Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible) Thus the final probability = 75/216 + 5/216 = 80/216 C is the correct answer.



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 01:38
to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2
to have any number except 4 is 5/6*1/6*1/6=5/216
5/216+1/2=5/216+108/216=113/216
what is wrong?



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 03:01
Temurkhon wrote: to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2
to have any number except 4 is 5/6*1/6*1/6=5/216
5/216+1/2=5/216+108/216=113/216
what is wrong? You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done. Look at my solution above. Hope this helps.



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 03:36
Hi Engr2012, Please help me with this question. I did the following way.. 4 4 X X X (X is the probability of obtaining another 4) Attempt 1: 1st X = 4. the probability is 1/6 (Game Won End of Game) Attempt 2: 2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won) Attempt 3 3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216 Total probability for 4= 91/ 216. The three XXX are a number (Say 5) probability = 1/216. I am getting 92/216. Where am I going wrong here. Please explain kindly. Engr2012 wrote: Temurkhon wrote: to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2
to have any number except 4 is 5/6*1/6*1/6=5/216
5/216+1/2=5/216+108/216=113/216
what is wrong? You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done. Look at my solution above. Hope this helps.
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Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 04:37
shriramvelamuri wrote: Hi Engr2012, Please help me with this question. I did the following way.. 4 4 X X X (X is the probability of obtaining another 4) Attempt 1: 1st X = 4. the probability is 1/6 (Game Won End of Game) Attempt 2: 2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won) Attempt 3 3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216 Total probability for 4= 91/ 216. The three XXX are a number (Say 5) probability = 1/216. I am getting 92/216. Where am I going wrong here. Please explain kindly. Engr2012 wrote: Temurkhon wrote: to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2
to have any number except 4 is 5/6*1/6*1/6=5/216
5/216+1/2=5/216+108/216=113/216
what is wrong? You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done. Look at my solution above. Hope this helps. Your assumption that the game "stops" once you get 3 of 4s is not correct. You still need to count for all 5 cases and then evaluate whether the person has won. Look at my solution above for scenario 1.



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 04:55
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 Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216 Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216 I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ?



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 04:58
kunal555 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 Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216 Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216 I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ? Nowhere is it mentioned that the game is stopped as soon he has won the game. Thus you still need to consider the other cases and not assume things that are not given.



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 05:29
Good question. Forgot to add the case where the next 3 could be identical and not 4s



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 07:51
Engr2012 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 ?
a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216
Source: PerfectScores Good question. +1 For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things. Probability of any number from 1 to 6 = 1/6. Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not. 2 scenarios for a win: Scenario 1: 4 4 T N N , with T= 4 , N= not 4 Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots) Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible) Thus the final probability = 75/216 + 5/216 = 80/216 C is the correct answer. Hi Engr2012, I've also set up 2 correct scenarios, but had problenbs calculating them  particularly the second scenario: 1] (1/6)*(5/6)*(5*6)*3!/2!=75/216 > Ok, here no questions Not 4*Not 4*4 (1/6) BUT 2] 4 4 N N N > 5/6*5/6*5/6 Not4*Not4*Not4, why do we have to choose 1/6 twice after 5/6 ?
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Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 08:02
BrainLab wrote: Engr2012 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 ?
a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216
Source: PerfectScores Good question. +1 For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things. Probability of any number from 1 to 6 = 1/6. Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not. 2 scenarios for a win: Scenario 1: 4 4 T N N , with T= 4 , N= not 4 Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots) Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible) Thus the final probability = 75/216 + 5/216 = 80/216 C is the correct answer. Hi Engr2012, I've also set up 2 correct scenarios, but had problenbs calculating them  particularly the second scenario: 1] (1/6)*(5/6)*(5*6)*3!/2!=75/216 > Ok, here no questions Not 4*Not 4*4 (1/6) BUT 2] 4 4 N N N > 5/6*5/6*5/6 Not4*Not4*Not4, why do we have to choose 1/6 twice after 5/6 ? That is because once you have 1 particular number out of (1,2,3,5,6), in order to "win" you MUST get that 1 particular number twice more in order to "win". Selecting 1 number out of 6 possible = 1/6. You can select 1 particular number of out of the remaining 5 numbers iwth probability of 5/6. The winning combinations can be 44555 or 44111 or 44222 or 44333 or 44666 Thus for 4 4 N N N , in which N \(\neq\)4 = (5/6)*(1/6)*(1/6) Hope this helps.



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 08:08
Thanks engr2012, yes this information helped me. I realised my mistake in the moment I've written my question.
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Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 11:20
Hi Bunuel , please solve this question . Please did not understand the above solution .
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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 13:03
kunal555 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 Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216 Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216 I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ? Hi Kunal555, I was also having the same question as you have. But look at the details of question. It says exactly 3 times. If it is not exactly 3 times, he will lose the game. Hence, the game should go on for total 5 throws, then only we can decide whether he won the game or not. If we stop the game at the third throw when he gets 4, then we are eliminating his chance of losing the game because we don't know the result of 4th and 5th throw. If he gets 4 in the fourth throw also, then it means he will lose the game because now, the total number of 4s are not exactly 3. Please award kudos if you agree with my explanation.
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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 13:37
abhigoyal1989 wrote: kunal555 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 Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216 Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216 I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ? Hi Kunal555, I was also having the same question as you have. But look at the details of question. It says exactly 3 times. If it is not exactly 3 times, he will lose the game. Hence, the game should go on for total 5 throws, then only we can decide whether he won the game or not. If we stop the game at the third throw when he gets 4, then we are eliminating his chance of losing the game because we don't know the result of 4th and 5th throw. If he gets 4 in the fourth throw also, then it means he will lose the game because now, the total number of 4s are not exactly 3. Please award kudos if you agree with my explanation. You are right. Also, first statement says "Daniel is playing a dice game in which a dice is rolled 5 times." IMO it is simple and clear that he plays a game in which a dice HAS TO BE ROLLED 5 TIMES. And if in 5 trials, a number shows up more than 3 times, he will loose.



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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21 Sep 2015, 16:13
shriramvelamuri wrote: Hi Engr2012, Please help me with this question. I did the following way.. 4 4 X X X (X is the probability of obtaining another 4) Attempt 1: 1st X = 4. the probability is 1/6 (Game Won End of Game) Attempt 2: 2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won) Attempt 3 3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216 Total probability for 4= 91/ 216. The three XXX are a number (Say 5) probability = 1/216. I am getting 92/216. Where am I going wrong here. Please explain kindly. Engr2012 wrote: Temurkhon wrote: to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2
to have any number except 4 is 5/6*1/6*1/6=5/216
5/216+1/2=5/216+108/216=113/216
what is wrong? You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done. Look at my solution above. Hope this helps. Hi Shriram, You are missing one crucial information that he has to throw dice 5 times. and he has to get same number exactly 3 times. If you are stopping at "attempt 1" where you are saying game will end and he will won because he has thrown same number 3 times, you are missing "exactly" term in the question. Thus to ensure he won the game, he has to throw a number other than 4. That is why we multiply 5/6 two times (4th throw and 5th throw). Your logic would have been fine in case the question says "he will win as soon as he gets three times a same number" Please award kudos if you like my explanation. Thanks
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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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25 Sep 2015, 08:07
Engr2012 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 ?
a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216
Source: PerfectScores Good question. +1 For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things. Probability of any number from 1 to 6 = 1/6. Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not. 2 scenarios for a win: Scenario 1: 4 4 T N N , with T= 4 , N= not 4 Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots) Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible) Thus the final probability = 75/216 + 5/216 = 80/216 C is the correct answer. Excellent explanation! But just out of curiosity, can this problem also be solved using combinatorics approach? Thanks



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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]
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25 Sep 2015, 12:18
Rk91 wrote: Engr2012 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 ?
a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216
Source: PerfectScores Good question. +1 For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things. Probability of any number from 1 to 6 = 1/6. Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not. 2 scenarios for a win: Scenario 1: 4 4 T N N , with T= 4 , N= not 4 Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots) Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible) Thus the final probability = 75/216 + 5/216 = 80/216 C is the correct answer. Excellent explanation! But just out of curiosity, can this problem also be solved using combinatorics approach? Thanks The solution above does use combinatorics with counting principle used to come up with the final probability. I dont think pure combinatorics method will be time effective strategy for such questions.



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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, 04:23
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?



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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, 04:30
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.




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