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

Question

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

mohshu · Answer

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

matthewsmith_89 · Answer

(1,3), (1,5), (3,5), (3,1), (5,1), (5,3)

6 x 6 = 36

6/36 = 1/6 which is answer C

Hatakekakashi · Answer

total possible 1 3 5 (3) and 1 3 5 (3) = 9

cases excluded - 11 33 55

9-3 = 6

6/36

C

ar500 · Answer

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?

Hatakekakashi · Answer

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,
HK

malevg · Answer

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.

KarishmaB · Answer

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

ScottTargetTestPrep · Answer

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

EgmatQuantExpert · Answer

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  .

Thanks,
Saquib
Quant Expert
e-GMAT

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FB2017 · Answer

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.

pankajpatwari · Answer

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??

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=1xu3QqY2LYA

We need to find What is the probability of rolling two fair dice and having both dice show distinct odd numbers?

As we are rolling two dice => Number of cases =  6^2  = 36

Now we have three odd numbers in 1 to 6 -> 1, 3, 5. Lets start writing the possible cases where two dice will show distinct odd numbers

- If we get 1 in the first roll then the second roll can give us other two odd numbers which are 3 and 5
=> (1,3), (1,5) => 2 cases

Similarly, we will get following cases
(3,1), (3,5)
(5,1), (5,3)

=> 6 cases

=> Probability that rolling two fair dice and having both dice show distinct odd numbers =  \frac{6}{36}  =  \frac{1}{6}

So,  Answer will be C 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

Nikkz99 · Answer

Simpler way to solve this question is, 
No : of odd faces in a dice -> {1,3,5} therefore, 3 numbers

Question states that we there must be "2 distinct odd faces".

Therefore we mustn't get -> {(1,1), (3,3), (5,5)}

Probability of getting odd number in die 1 -> 1/2 
Probability of getting  Distinct Odd  number in die 2 -> 2/6 => 1/3

P(Getting two distinct odd number on both the dice) = 1/2 * 1/3 = 1/6

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

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