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If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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04 Nov 2015, 05:15
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If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled? A. 25/216 B. 50/216 C. 25/72 D. 25/36 E. 5/6
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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shyind wrote: If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled?
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6 Total ways in which a 6sided die can be rolled three times = 6*6*6 = 216 To get exactly one 3, there are three ways: A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways. The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75 Required Probability = 75/216 = 25/72 Answer (C)
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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04 Nov 2015, 21:22
shyind wrote: If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled?
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6 Probability if getting 3 on a die is: P(3)=1/6 Probability of not getting 3 on a die is nP(3)=5/6 Let the three dies be denoted by P1,P2 and P3 Probability of getting 3 on the first die and not getting 3 on the other two dies is given by: =P1(3)*nP(3)*nP(3) =1/6*5/6*5/6 =25/216 As there are 3 dies,so similarly for the rest of the two dies probability will be 25/216 Summing up all the 3 cases we get: =3*25/216 =25/72



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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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05 Nov 2015, 01:22
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled? A. 25/216 B. 50/216 C. 25/72 D. 25/36 E. 5/6 Let O be represented as 3 being rolled, and X be represented as no 3 being rolled. Then the case of exactly one 3 being rolled can be represented as (O, X, X), (X, O, X), (X, X, O). The probability of (O, X, X) is (1/6)*(5/6)*(5/6), that of (X, O, X) is (5/6)*(1/6)*(5/6) and that of (X, X, O) is (5/6)*(5/6)*(1/6). So the probability is (1/6)*(5/6)*(5/6) + (5/6)*(1/6)*(5/6) + (5/6)*(5/6)*(1/6) = 3*(1/6)*(5/6)*(5/6) = 25/72. The answer is (C).
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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08 Jul 2017, 12:50
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shyind wrote: If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled?
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6 Probability of getting 3 \(= \frac{1}{6}\) Probability of not getting 3 \(= \frac{5}{6}\) Probability of getting three on the first roll \(= \frac{1}{6} * \frac{5}{6} * \frac{5}{6}\) Probability of getting three on the first roll \(= \frac{25}{216}\) As we can get 3 on either second or the third time we will multiply the above probability by 3 \(= \frac{25}{216} * 3\) \(= \frac{25}{72}\) Hence, Answer is C
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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06 Jan 2018, 17:32
Here is the full answer, from the ground up. To understand the mathematics behind this question we need to understand three basic probability rules: (1) A probability is calculated by dividing the number of options when the event occurs by the total number of possibilities.(2) When calculating the probability of multiple events that must ALL occur to meet a certain condition, we must multiply the separate probabilities of each event together. Thus, the probability of all the events \(A\), \(B\), and \(C\) occurring would be: \(P_{combined} = (P_A)*(P_B)*(P_C)\)(3) When calculating the probability of a series of mutuallyexclusive events, where ANY event occurring would be sufficient, we must add the separate probabilities of each event together. Thus, the probability of any of the events \(A\), \(B\), and \(C\) occurring would be: \(P_{combined} = (P_A)+(P_B)+(P_C)\)Let's leverage these three basic rules to solve this problem. First, the probability of rolling a single value (in this case 3) on a fair 6sided die would be one out of six (or \(\frac{1}{6}\)). The probability of NOT rolling that number would be five out of six (or \(\frac{5}{6}\)). So, if you need to roll a single value on one die, while simultaneously not rolling that value on the other two dice, then we invoke rule #2 from above (i.e., we multiply the probabilities.) Thus, \(P=(\frac{1}{6})*(\frac{5}{6})*(\frac{5}{6})=\frac{25}{216}\) Notice that this is answer choice A. However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work. There are three possibilities: \(3, N, N\) \(N, 3, N\) \(N, N, 3\) The probability of each of these orders is the same (\(\frac{25}{216}\)), but each is separate and independent. We now need to invoke rule #3 from above and add the separate probabilities together: \(P_{Total} = \frac{25}{216} + \frac{25}{216} + \frac{25}{216} = \frac{3*25}{216}\) Recognizing common factors in the top and bottom of this fraction allow us to get to the answer without any messy math. \(216 = 6*6*6 = 2*3*6*6\). Thus, \(\frac{3*25}{216}=\frac{3*25}{2*3*6*6}=\frac{25}{2*6*6}=\frac{25}{72}\) The answer is C.
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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07 Feb 2018, 12:46
Hi All, You can approach the math in this question in a couple of different ways, so you have to think about what would be the easiest way for YOU to organize your work. The questions tells us to roll a 6sided dice 3 times. We're asked for the probability of rolling EXACTLY one '3' on those three rolls. Here's a way to break the calculation down into 3 smaller calculations: (first roll is 3)(second roll is NOT 3)(third roll is NOT 3) = (1/6)(5/6)(5/6) = 25/216 (first roll is NOT 3)(second roll is 3)(third roll is NOT 3) = (5/6)(1/6)(5/6) = 25/216 (first roll is NOT 3)(second roll is NOT 3)(third roll is 3) = (5/6)(5/6)(1/6) = 25/216 Total = 3(25/216) = 75/216 = 25/72 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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13 Apr 2018, 01:38
niel1989 wrote: shyind wrote: If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled?
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6 Probability if getting 3 on a die is: P(3)=1/6 Probability of not getting 3 on a die is nP(3)=5/6 Let the three dies be denoted by P1,P2 and P3 Probability of getting 3 on the first die and not getting 3 on the other two dies is given by: =P1(3)*nP(3)*nP(3) =1/6*5/6*5/6 =25/216 As there are 3 dies,so similarly for the rest of the two dies probability will be 25/216 Summing up all the 3 cases we get: =3*25/216 =25/72 "A" die is rolled 3 times.. There aren't 3 dies... which is why I can't seem to understand why not 25/216



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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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13 Apr 2018, 01:44
saumya2805 wrote: "A" die is rolled 3 times.. There aren't 3 dies... which is why I can't seem to understand why not 25/216 Hey saumya2805 , There isn't any difference in rolling the same die thrice or rolling three different does. Hence, for each die we have the total outcomes = 6. Therefore, total possibilities= 6*6*6 = 216. Does that make sense?
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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13 Apr 2018, 01:52
EMPOWERgmatRichC wrote: Hi All, You can approach the math in this question in a couple of different ways, so you have to think about what would be the easiest way for YOU to organize your work. The questions tells us to roll a 6sided dice 3 times. We're asked for the probability of rolling EXACTLY one '3' on those three rolls. Here's a way to break the calculation down into 3 smaller calculations: (first roll is 3)(second roll is NOT 3)(third roll is NOT 3) = (1/6)(5/6)(5/6) = 25/216 (first roll is NOT 3)(second roll is 3)(third roll is NOT 3) = (5/6)(1/6)(5/6) = 25/216 (first roll is NOT 3)(second roll is NOT 3)(third roll is 3) = (5/6)(5/6)(1/6) = 25/216 Total = 3(25/216) = 75/216 = 25/72 Final Answer: GMAT assassins aren't born, they're made, Rich Hi, Can you please help me? I completely understand the logic you've followed... What I don't understand is why the need.. I mean, the question simply asks the probability of 3 appearing exactly once. It doesn't specify to find for "on which of these throws..." There's a probability of 3 appearing exactly once, 25 out of 216 times. If the question had been that 3 dies are rolled simultaneously 3 times... it would've made more sense to me.. Since you're an expert, surely I'm missing something. Could you please help me understand what I'm missing?



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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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13 Apr 2018, 02:55
AaronPond wrote: Here is the full answer, from the ground up. To understand the mathematics behind this question we need to understand three basic probability rules:
(1) A probability is calculated by dividing the number of options when the event occurs by the total number of possibilities.
(2) When calculating the probability of multiple events that must ALL occur to meet a certain condition, we must multiply the separate probabilities of each event together. Thus, the probability of all the events \(A\), \(B\), and \(C\) occurring would be: \(P_{combined} = (P_A)*(P_B)*(P_C)\)
(3) When calculating the probability of a series of mutuallyexclusive events, where ANY event occurring would be sufficient, we must add the separate probabilities of each event together. Thus, the probability of any of the events \(A\), \(B\), and \(C\) occurring would be: \(P_{combined} = (P_A)+(P_B)+(P_C)\)
Let's leverage these three basic rules to solve this problem.
First, the probability of rolling a single value (in this case 3) on a fair 6sided die would be one out of six (or \(\frac{1}{6}\)).
The probability of NOT rolling that number would be five out of six (or \(\frac{5}{6}\)).
So, if you need to roll a single value on one die, while simultaneously not rolling that value on the other two dice, then we invoke rule #2 from above (i.e., we multiply the probabilities.) Thus,
\(P=(\frac{1}{6})*(\frac{5}{6})*(\frac{5}{6})=\frac{25}{216}\)
Notice that this is answer choice A. However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work. There are three possibilities:
\(3, N, N\) \(N, 3, N\) \(N, N, 3\)
The probability of each of these orders is the same (\(\frac{25}{216}\)), but each is separate and independent. We now need to invoke rule #3 from above and add the separate probabilities together:
\(P_{Total} = \frac{25}{216} + \frac{25}{216} + \frac{25}{216} = \frac{3*25}{216}\)
Recognizing common factors in the top and bottom of this fraction allow us to get to the answer without any messy math. \(216 = 6*6*6 = 2*3*6*6\). Thus,
\(\frac{3*25}{216}=\frac{3*25}{2*3*6*6}=\frac{25}{2*6*6}=\frac{25}{72}\)
The answer is C. You mention the below: "Notice that this is answer choice A. However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work." Plz refer the bold colored portion.. Is it understood by default, that the order always matters? Even if it isn't mentioned explicitly in the question? This is exactly what I'm always confused about in probability related questions.



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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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13 Apr 2018, 15:43
saumya2805 wrote: You mention the below: "Notice that this is answer choice A. However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work."
Plz refer the bold colored portion.. Is it understood by default, that the order always matters? Even if it isn't mentioned explicitly in the question? This is exactly what I'm always confused about in probability related questions. Thank you, Saumya2805, for the question. Yes, when calculating the probability of multiple situations, the order always matters. This isn't just the "default", this is how the mathematics of probability work. Even if the events are simultaneous, think about the solution one event at a time. Each different arrangement has a separate probability.
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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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14 Apr 2018, 00:24
AaronPond wrote: saumya2805 wrote: You mention the below: "Notice that this is answer choice A. However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work."
Plz refer the bold colored portion.. Is it understood by default, that the order always matters? Even if it isn't mentioned explicitly in the question? This is exactly what I'm always confused about in probability related questions. Thank you, Saumya2805, for the question. Yes, when calculating the probability of multiple situations, the order always matters. This isn't just the "default", this is how the mathematics of probability work. Even if the events are simultaneous, think about the solution one event at a time. Each different arrangement has a separate probability. Thanks Aaron, that helps!



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Re: If a fair 6sided die is rolled three times, what is the probability t [#permalink]
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15 Apr 2018, 18:15
saumya2805 wrote: niel1989 wrote: shyind wrote: If a fair 6sided die is rolled three times, what is the probability that exactly one 3 is rolled?
A. 25/216
B. 50/216
C. 25/72
D. 25/36
E. 5/6 Probability if getting 3 on a die is: P(3)=1/6 Probability of not getting 3 on a die is nP(3)=5/6 Let the three dies be denoted by P1,P2 and P3 Probability of getting 3 on the first die and not getting 3 on the other two dies is given by: =P1(3)*nP(3)*nP(3) =1/6*5/6*5/6 =25/216 As there are 3 dies,so similarly for the rest of the two dies probability will be 25/216 Summing up all the 3 cases we get: =3*25/216 =25/72 "A" die is rolled 3 times.. There aren't 3 dies... which is why I can't seem to understand why not 25/216 How do you get 216? You say the FIRST roll could be 1/2/3/4/5/6 i.e. 6 ways The SECOND roll could be 1/2/3/4/5/6 i.e. 6 ways The THIRD roll could be 1/2/3/4/5/6 i.e. 6 ways Hence rolling a 3 on FIRST roll and any of the other 5 numbers of other two rolls is different from rolling a 3 on SECOND roll and any of the other 5 numbers on other two rolls because we have counted these cases apart {3, 1, 6}, {1, 3, 6}. That is how we get 75 and not just 25. This is same as saying  I have 3 dice of different colours  Red, Yellow and Blue  and I roll them together The Red could have outcome in 6 ways, Yellow in 6 and Blue in 6 so 216 ways Same as case above. How about saying I have 3 identical dice and I roll them together. How many outcomes do I have in that case? Is (1, 2, 3) different from (3, 1, 2)? No. Then I do not have 216 outcomes in this case. In this case, there is no "order".
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