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If a fair 6-sided die is rolled three times, what is the probability t

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If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

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If a fair 6-sided 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
[Reveal] Spoiler: OA

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Re: If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

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shyind wrote:
If a fair 6-sided 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 6-sided 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 6-sided die is rolled three times, what is the probability t [#permalink]

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New post 04 Nov 2015, 20:22
shyind wrote:
If a fair 6-sided 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 6-sided die is rolled three times, what is the probability t [#permalink]

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New post 05 Nov 2015, 00:22
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


If a fair 6-sided 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 6-sided die is rolled three times, what is the probability t [#permalink]

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shyind wrote:
If a fair 6-sided 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 6-sided die is rolled three times, what is the probability t [#permalink]

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New post 06 Jan 2018, 16:32
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 mutually-exclusive 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 6-sided 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 6-sided die is rolled three times, what is the probability t   [#permalink] 06 Jan 2018, 16:32
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