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12 Days of Christmas GMAT Competition 2025 - Day 4: A pair of fair six

1 2 3 4

jkkamau

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19 Dec 2025, 04:04

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Total outcomes is 6x6= 36
Both prime = 3x3 =9
At least one primer = Total minus no prime = 36-(3x3)= 27 (1,4,6)

So all prime given at least one shows prime is 9/27 = 1/3 (Conditional probability)
Ans B

Bunuel

A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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rutikaoqw

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Favourable outcomes: (22,23,25,32,33,35,52,53,55)
Total outcomes: (12,13,15,21,22,23,24,25,26,31,32,33,34,35,36,42,43,45,51,52,53,54,55,56,62,63,65)
=9/27=1/3

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19 Dec 2025, 04:53

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This question is based on conditional Probability!

P(B) = Prob that both dice shows prime #
P(C) = Prob that at least once die shows a prime#

Q: P(B/C)

P(B) = Prime # (2,3,5) on both die/Total = 3*3/36 = 9/36

P(C) = 1 - P(B) = 27/36

P(B/C) = 9/36 * 36/27 = 9/27 = 1/3 (Option B)

Bunuel

A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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Zeus_

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19 Dec 2025, 05:12

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Total outcome = 6*6 = 36

Favourable outcome = 2,3,5 = 3*3 = 9

9/36 = 1/4

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19 Dec 2025, 05:16

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Bunuel

A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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To find P(both prime | atleast one prime) = P(both prime and atleast one prime) / P(atleast one prime) = P(both prime) / P(atleast one prime)

Total primes = 3 (2,3,5)
Non primes = 3 (1, 4, 6)

Total occurrences of both prime = 9 (2,2) (2,3) (2,5) (3,2) ...

Total occurences of atleast one prime = 36 - None prime = 36 - 9 = 27

P(both prime) / P(atleast one prime) = 9/36 / 27/36 = 1/3

Correct Answer: B

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19 Dec 2025, 06:04

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Total cases with at least one die showing a prime number:-
When one die shows 2- 6 cases
When one die shows 3- 6 cases
When one die shows 5- 6 cases
Total cases= 6+6+6=18

Both dice showing prime numbers cases= (2,2) ; (2,3) ; (2,5) ; (3,2) ; (3,3) ; (3,5) ; (5,2) ; (5,3) ; (5,5)= 9 cases in total

So, Probability of both dice showing prime numbers given at least one die shows a prime number= 9/18=1/2

Option C

Bunuel

A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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kapoora10

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19 Dec 2025, 06:09

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We want => P (both prime/ at least one prime)

Total Outcomes = 6*6 = 36
Both dice show prime = 3*3 (primes are 2,3 and 5) = 9
At least one prime = total - both primes non prime (1,4,6) = 36 - 3*3= 27

Conditional probability = both prime/at least one prime = 9/27 = 1/3

Answer => B

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19 Dec 2025, 06:19

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There three prime numbers, i.e. 2,3 & 5
Two dice are rolled. At least one shows prime number, Probability for that dice showing prime is 1
Probability of the second dice showing prime is 3/6 = 1/2

Probability that both dice show prime= 1 * 1/2 = 1/2

Bunuel

A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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sanjitscorps18

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19 Dec 2025, 06:28

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At least one dice will show a prime number = 1 - Both dice will show a non-prime number

Primes are [2, 3, 5]
Non-prime numbers are [1, 4, 6]

Cases for two dice rolling = 6*6 = 36
Cases where at least one dice is prime = 36 - 3*3 = 27
Cases where both dice are prime = 3*3 = 9

Probability = 9/27 = 1/3

Option B

Bunuel

A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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redandme21

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19 Dec 2025, 06:44

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Prime numbers: 2 3 5
Non prime numbers: 1 4 6

Favorable cases:
Prime number is first dice, prime number in second dice: 3x3=9

Total cases:
Prime number is first dice, non prime number in second dice: 3x3=9
Non prime number is first dice, prime number in second dice: 3x3=9
Prime number is first dice, prime number in second dice: 3x3=9
9+9+9=27

Favorable cases/Total cases = 9/27 = 1/3

IMO B

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19 Dec 2025, 06:52

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We know that in a fair six sided dice the outcomes possible are {1,2,3,4,5,6}
In this, atleast one die shows a prime number.
=> Possible outcomes are {2,3,5}
P(Shows prime) = 3/6 = 1/2

Let say that,
A = Both dice show prime
B = Atleast one of the dice shows prime number

P(A) = (1/2)(1/2) = 1/4

P(B) = 1 - P(No primes) = 1 - (1/2)(1/2) = 3/4

We need to know the probability that both dice show prime numbers which is a conditional probability here
=> P(A|B) = P(A) / P(B) = (1/4) / (3/4) = 1/3

B. 1/3

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19 Dec 2025, 06:54

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A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

Prime number on a dice = 2,3,5
Non prime numbers on a dice = 1,4,6
Probability of one dice showing a prime = 3/6= 1/2
Probability of both dice showing a prime = (1/2)*(1/2) = 1/4
Probability of no dice showing a prime = (1/2)*(1/2)= 1/4
Probability of atleast 1 dice showing a prime number = 1-(1/4) = 3/4

Probability (both prime| atleast 1 prime) = (1/4)/(3/4)= 1/3

B. 1/3

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19 Dec 2025, 07:16

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There are 3 prime numbers (2,3,5) and 3 non prime numbers (1,4,6) in a dice

The two dices show a prime number: 3*3=9 outcomes
At least one dice shows a prime number: 6*6 - the two dices show a non prime number = 36 - 3*3 = 36 - 9 = 27 outcomes

9/27 = 1/3

Answer B

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19 Dec 2025, 07:17

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Answer = C

If atleast 1 shows prime number = (2, 3, 5)
outcomes if Dice A showing prime number = 3
outcomes if Dice B showing prime number = 3
outcomes if Dice A&B both showing prime number = 6
Total outcomes = 3+3+6 = 12

Favourable outcome if both show prime number = 6

Therefore, probability = 6/12 = 1/2

Bunuel

A pair of fair six-sided dice is rolled once. If at least one die shows a prime number, what is the probability that both dice show prime numbers?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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topgmat25

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19 Dec 2025, 07:38

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The possible combinations are:
prime - prime
prime - non prime
non prime - prime
non prime - non prime

As in a dice there are equal number of primes (2, 3 and 5) and non primes (1, 4 and 6), all the combinations have the same probability.

favorable -> 1 combination (prime - prime)
total -> 3 combinations (prime - prime, prime - non prime, non prime - prime)

favorable/total = 1/3

The answer is B

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19 Dec 2025, 07:59

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prime = 2,3,5
non prime = 1,4,6

3*3 = 9 cases with prime numbers in both dices
3*3 = 9 cases with non prime numbers in both dices
3*3 = 9 cases with prime number in 1st dice and non prime number in 2nd dice
3*3 = 9 cases with non prime number in 1st dice and prime number in 2nd dice

cases with prime numbers in both dices / (cases with prime numbers in both dices + cases with prime number in 1st dice and non prime number in 2nd dice + cases with non prime number in 1st dice and prime number in 2nd dice) = 9/(9+9+9) = 9/27 = 1/3

The correct answer is B

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19 Dec 2025, 09:07

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This is a conditional probability question.

We are asked to find the probability for both dice being prime when at least one of them is prime.

I see it like this: P(both prime) / P(at least one is prime)

P (both prime) = (3/6) x (3/6) = 1/4

P(at least one prime) = 1 - P(no prime) = 1 - (1/2 x 1/2) = 3/4

Therefore, answer is (1/4) / (3/4) = 1/3