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What is the probability of rolling two fair dice and having both dice

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What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 29 Mar 2017, 04:53
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What is the probability of rolling two fair dice and having both dice show distinct odd numbers?

A. 1/36
B. 1/12
C. 1/6
D. 1/2
E. 35/36
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 29 Mar 2017, 08:45
2
odd numbers = 1,3,5

two dice will give 3 x 3 = 9 combination of odd nos... but
combination of 2 same odd nos = (1,1) (3,3) (5,5)

hence 9-3 =6

prob = 6/36 = 1/6

ans 1/6
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 30 Mar 2017, 11:16
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(1,3), (1,5), (3,5), (3,1), (5,1), (5,3)

6 x 6 = 36

6/36 = 1/6 which is answer C
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 30 Mar 2017, 13:00
total possible 1 3 5 (3) and 1 3 5 (3) = 9

cases excluded - 11 33 55

9-3 = 6

6/36

C
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 30 Mar 2017, 16:49
With the regard to the popular 1/6 answer, do you guys think it is problematic that you are counting both "1,3" and "3,1" as part of your 6 positive outcomes? Does sequence matter in a dice roll?
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 30 Mar 2017, 22:58
ar500 wrote:
With the regard to the popular 1/6 answer, do you guys think it is problematic that you are counting both "1,3" and "3,1" as part of your 6 positive outcomes? Does sequence matter in a dice roll?



IMO , yes sequence matters

1st throw having a 1 on the dice can lead to six options on the second dice.

Assume this situation

In a Football (soccer) match , the final score is 3-1
now first player X of a team scores a hattrick , then in the dying minutes of the game , a player from the opposition scores a goal and makes it 3-1
now this score can be achieved in multiple ways and we notice that each way is different

I'm not the best person around here to explain it in a lucid way but i hope you got my point


Regards,
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 30 Mar 2017, 23:38
The probability of one dice having odd side is 3/6=1/2. The same thing with the second dice. The probability of the second dice having distinct number is 2/3 (we have 3 possible numbers but only 2 that satisfy "distinct" condition).
Therefore general probability equals 1/2*1/2*2/3=1/6.
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 31 Mar 2017, 03:27
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ar500 wrote:
With the regard to the popular 1/6 answer, do you guys think it is problematic that you are counting both "1,3" and "3,1" as part of your 6 positive outcomes? Does sequence matter in a dice roll?


Since the problem doesn't say that the dice are identical, think about them as two different coloured dice - one red and one yellow.
A 1 on the red die and 3 on the yellow die is different from a 3 on the red die and 1 on the yellow die.

So, there are 3 ways of getting an odd outcome on the red die and 2 ways of getting an odd outcome on the yellow die. In all there are 3*2 = 6 ways of satisfying the condition.

Of course, total cases are 6*6 = 36

Probability = 6/36 = 1/6
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 31 Mar 2017, 12:42
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Bunuel wrote:
What is the probability of rolling two fair dice and having both dice show distinct odd numbers?

A. 1/36
B. 1/12
C. 1/6
D. 1/2
E. 35/36


We need to determine the probability that the two dice show distinct odd numbers.

Let’s start with die #1: There are 3 odd numbers, and thus the probability of rolling an odd number is 3/6 = 1/2. When rolling die #2, since we must roll a different odd number, the probability of doing so is 2/6 = 1/3.

Thus, the probability is 1/2 x 1/3 = 1/6.

Answer: C
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What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 31 Mar 2017, 13:55
Bunuel wrote:
What is the probability of rolling two fair dice and having both dice show distinct odd numbers?

A. 1/36
B. 1/12
C. 1/6
D. 1/2
E. 35/36


Solution



    • Odd numbers on a die are (\(1,3\) or \(5\))

    • We need distinct odd numbers on the two dice.

      o Thus, the first die can show any one of the 3 odd numbers, which is \(= {}^3C_1\)

      o The second die can then show the other 2 odd numbers, which is \(= {}^2C_1\)

      o Thus, favorable cases of two dice showing distinct odd numbers \(= 3 * 2 = 6\)


    • Total number of cases are \(6*6 = 36\)

    • Thus, Probability \(= \frac{6}{36} = \frac{1}{6}\)

    • Correct Answer is Option C.

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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 03 Aug 2017, 01:35
Distinct odd values from dice 1 can have 3 outcomes.
Similarly from dice 2.
Total 6 favorable outcomes from 2 dices.
Total outcomes =36
Probability of favorable outcomes=6/36=1/6
Option C.
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Re: What is the probability of rolling two fair dice and having both dice  [#permalink]

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New post 13 Aug 2019, 19:48
EgmatQuantExpert wrote:
Bunuel wrote:
What is the probability of rolling two fair dice and having both dice show distinct odd numbers?

A. 1/36
B. 1/12
C. 1/6
D. 1/2
E. 35/36


Solution



    • Odd numbers on a die are (\(1,3\) or \(5\))

    • We need distinct odd numbers on the two dice.

      o Thus, the first die can show any one of the 3 odd numbers, which is \(= {}^3C_1\)


      o The second die can then show the other 2 odd numbers, which is \(= {}^2C_1\)

      o Thus, favorable cases of two dice showing distinct odd numbers \(= 3 * 2 = 6\)


    • Total number of cases are \(6*6 = 36\)

    • Thus, Probability \(= \frac{6}{36} = \frac{1}{6}\)

    • Correct Answer is Option C.

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Hi E-GMAT Team,
I request reference to your "Must Read Articles & Practice Questions to score 51". In the probability section one of the key takeaways you have emphasized upon is: "If there are more than one arrangements possible then we will find the probability of only one case and multiply it by the total number of possible arrangements".

By that logic, no. of possible arrangements in the question under discussion are 6:
1,3;1,5;3,5;3,1;5,1;5,3

So shudn't the probability be be then:

6*3/6*2/6=1

I know i am not understanding it right. but cud u plzz guide me where??
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Re: What is the probability of rolling two fair dice and having both dice   [#permalink] 13 Aug 2019, 19:48
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