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# Wu rolls two fair, six-sided dice. What is the probability that Wu

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Math Expert
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Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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01 Mar 2017, 01:00
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Question Stats:

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Wu rolls two fair, six-sided dice. What is the probability that Wu rolls at least one five but no sixes?

A. 5/36

B. 275/1296

C. 2/9

D. 1/4

E. 5/18
[Reveal] Spoiler: OA

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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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01 Mar 2017, 02:55
1
KUDOS

Approach :-

Cases to get at least one 5 and No sixes

Case 1 :- 5_(this blank can be filled by 1,2,3,4 except 5 and 6) hence 4 ways

Case 2 :- _5(this blank can be filled by 1,2,3,4 except 5 and 6) hence 4 ways

Case 3:- 55 = 1 ways

Favorable Outcomes :- Case 1 + Case 2 + Case 3 = 9 ways

Total Outcomes
= 36

Probability(Wu rolls at least one five but no sixes) = 9/36 = 1/4
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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu rol [#permalink]

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01 Mar 2017, 10:02
Total sample size = 36, if on the first dice only 5 comes after rolling then in the second dice we will have 5 possibilities 1,2,3,4,5(no sixes as per the question) same will be repeated with second dice. Total possibilities 10, hence probability is 10/36 or 5/18 correct answer E

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Posts: 551
Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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01 Mar 2017, 22:42
sravankumar0911 wrote:
Total sample size = 36, if on the first dice only 5 comes after rolling then in the second dice we will have 5 possibilities 1,2,3,4,5(no sixes as per the question) same will be repeated with second dice. Total possibilities 10, hence probability is 10/36 or 5/18 correct answer E

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u r counting the possibility of 5,5 two times....

favourable outcomes will 9
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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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03 Mar 2017, 00:44
sravankumar0911 wrote:
Total sample size = 36, if on the first dice only 5 comes after rolling then in the second dice we will have 5 possibilities 1,2,3,4,5(no sixes as per the question) same will be repeated with second dice. Total possibilities 10, hence probability is 10/36 or 5/18 correct answer E

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You have to subtract 1 from 10 possibilities cos you are counting 5,5 twice. Actual no of possibilities is 9 out of 36 which is 1/4 D
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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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03 Mar 2017, 05:43
Got it thanks for pointing out the mistake

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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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06 Mar 2017, 17:06
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Expert's post
Bunuel wrote:
Wu rolls two fair, six-sided dice. What is the probability that Wu rolls at least one five but no sixes?

A. 5/36

B. 275/1296

C. 2/9

D. 1/4

E. 5/18

We can can break this up into scenarios:

Scenario 1:

P(5 on the first die and 1,2,3, or 4 on the second die)

1/6 x 4/6 = 4/36 = 1/9

Scenario 2:

P(1,2,3, or 4 on the first die and 5 on the second die)

4/6 x 1/6 = 4/36 = 1/9

Scenario 3:

P(5 on both dies)

1/6 x 1/6 = 1/36

Thus, the final probability is:

1/9 + 1/9 + 1/36 = 4/36 + 4/36 + 1/36 = 9/36 = 1/4.

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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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07 Mar 2017, 17:09
Bunuel wrote:
Wu rolls two fair, six-sided dice. What is the probability that Wu rolls at least one five but no sixes?

A. $$\frac{5}{36}$$

B. $$\frac{275}{1296}$$

C. $$\frac{2}{9}$$

D. $$\frac{1}{4}$$

E. $$\frac{5}{18}$$

Official solution from Veritas Prep.

As is often the case with probability problems, there are multiple approaches to finding the right answer.

One such method is to directly count both the good outcomes and the total outcomes for rolling two fair, six-sided dice.

The total outcomes are straightforward: each die has six possible outcomes, so there are $$6∗6=36$$ total outcomes.

The good outcomes are those involving either:

a 5 on the first roll and a non-5, non-6 on the second roll – 4 such outcomes, from (5,1) through (5,4); or
a non-5, non-6 on the first roll and a 5 on the second roll – 4 such outcomes, from (1,5) through (4,5); or
a 5 on the first roll and a 5 on the second roll – one such outcome (5,5).

This is an “or” case, so the total good outcomes are $$4+4+1=9$$.Overall, the probability is $$\frac{9}{36}=\frac{1}{4}$$, and the correct answer is D.
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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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08 Mar 2017, 02:22
Way to get atleast one 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6).
Total possible cases = 11. Since we dont want any 6, we remove (5,6), (6,5). Hence the total cases will be 9.
number of possibilities = 6*6 = 36

Therefore probability = 9/36 = 1/4.
Option D
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Re: Wu rolls two fair, six-sided dice. What is the probability that Wu [#permalink]

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13 Mar 2017, 21:59
probability of getting at least one 5 and no six will be such as :

1/6*5/6 + 1/6*5/6 - 1/6*1/6= 9/36 =1/4

can someone clarify my method
Re: Wu rolls two fair, six-sided dice. What is the probability that Wu   [#permalink] 13 Mar 2017, 21:59
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