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

1 2 3 4

canopyinthecity

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18 Dec 2025, 13:51

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Let X be the event that prime number comes on the dice

P(X) = 1/2
P(not X) = 1/2

P(at least one X) = 1- P(neither X)
= 1 - P(not X) * P(not X) = 1 - (1/4) = 3/4

P(both X) = P(X) * P(X) = 1/4

P(both X given at least one X) = (1/4)/(3/4) = 1/3

Hence (B) is the answer

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18 Dec 2025, 14:14

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there are 3 cases here
c-1 D1 prime * D2 non prime
=(3/6) * (3/6) =1/4

c-2 D1 non prime * D2 prime
=(3/6) * (3/6) = 1/4

c-3 both prime
1/4

now we need both prime given at least one prime
so p (at least one prime) / p (both prime)
1/3

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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Fisayofalana

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18 Dec 2025, 15:48

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Because in a six-sided die there are 3 prime numbers, the probability of getting a prime on one die is 3/6 or 1/2. If the first AND second die show prime numbers, multiply. I got 1/4 as the answer.

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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Noise1st

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18 Dec 2025, 17:44

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If one dice has already a prime number, the probability that the other one has a prime is 1/3. Three, because the prime numbers from 1 to 6 are 2, 3, and 5.

asperioresfacere

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18 Dec 2025, 18:59

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On a Fair dice (1-6) , Prime numbers are 2,3,5 .
They asked atleast one dice shows a prime number = Excluding options (Both Dices are non prime).
All possible cases
P,P
P,N
N,P
N,N
We remove N, N
we get total outcomes = 3 and favourable outcome = 1 , which is P,P
Correct ans is 1/3

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18 Dec 2025, 19:30

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Let's take cases one by one. First we will find total outcomes. If the outcome of first die is 1 , then 3 cases are possible (2,3,5) such that atleast one die shows a prime number. Similarly , if first die shows 2 , then 6 cases are possible , if first die shows 3 , again 6 cases , for 4 , 3 cases , for 5 , 6 cases , for 6 , 3 cases. Total outcomes will be 3+6+6+3+6+3 = 27.

Now let's find favourable outcomes. To have both die as prime numbers , first die can show on (2 ,3 ,5) and second die can show (2,3,5) so total cases will be 3x3 =9. So Probability that both dice show prime numbers is 9/27 = 1/3

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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officiisestyad

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18 Dec 2025, 20:26

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Total outcomes of 2 dice: 36
Case 1: Atleast one dice will give us prime number P(A):
Number of non prime numbers: {1,4,6}
So total non prime numbers : 3*3= 9
So, atleast one non prime number is P(A) = 36-9 = 27
Case 2: Both the dice prime number {2,3,5} --> P(B) = 3*3 = 9 (Option B)
So, P(B)/P(A) = 9/27 = 1/3

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18 Dec 2025, 20:37

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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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both dice are prime = 3 * 3 = 9 (prime no on a dice are 2, 3, 5)

if at least one die shows a prime => die 1 can have prime or die 2 can have prime not both at the same time
total combination for 2 dice = 6 * 6 = 36
total combination where at least one shows prime = 36 - 9 = 27

=> prob of both dice showing prime = 9/27 = 1/3

truedelulu

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18 Dec 2025, 21:18

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Prime numbers in a dice: 2,3,5 => Probability that a dice shows a prime number = 3/6 = 1/2
=> Prob of two dices show prime numbers = P(A) = 1/2*1/2 = 1/4

Prob that a dice doesn't show a prime number = 3/6 = 1/2
=> Prob of two dices dont show prime numbers = 1/4
=> Prob of at least one dice shows a prime number = P(B) = 1 - 1/4 = 3/4

Conditional probability: P(A|B) = P(A)/P(B) = 1/3

Answer: 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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jai169

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18 Dec 2025, 21:34

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so in question each dice can have prime no as {2,3,5}
its a problem of conditional probability in which atleast one die shows a prime number is already taken place
so our sample space is reduce from 6*6=36 to 36-(none showed prime no)
none showed prime no= there are three nonprime no (1,4,6) in each die so there comibination 3*3=9
sample space is 36-9=27
now our faviourable outcome is if both dice take no from set {2,3,5}=so total 3*3=9
ans is = option B 9/27=1/3

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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Nitish03

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18 Dec 2025, 22:58

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Probability of getting prime when a dice is rolled(2,3,5) = 3/6 =1/2
So For two dies probability of getting both primes will be 1/2*1/2 =1/4 ...(1)
So , probability of getting atleast one prime will be 1-1/4= 3/4......(2)
Hence conditional probability will be 1/4 ÷ 3/4 = 1/3
Hence, option B

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19 Dec 2025, 00:01

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This is conditional probability.
P(both dice show prime given at least one is prime) is what is asked.

P(A given B) = P(A | B) = (P(A) intersection P(B)) / P(B)

P(B = at least one is prime) = 3/4, why? Because we have three cases, Prime-NonPrime, NonPrime-Prime, and Prime-Prime, and the probability of selecting a prime is 1/2 and a non-prime is 1/2, each of the three cases gives 1/4, so we multiply it by 3.

Now, the numerator is the probability of selecting Prime-Prime = 1/4

(1/4) / (3/4)
Gives 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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AbhishekP220108

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19 Dec 2025, 00:22

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Given- A fair die means its sides have 1,2,3,4,5,6. In which 3 are prime, and 3 are non-prime.
Total outcome 6*6=36 because it has been mentioned that the dice are in pairs.

To get total events of outcome as prime, we have to find outcome without prime which will be 3*3=9
Total outcome as prime=36-9=27

Favourable events as both prime=3*3=9

the required probability =9/27=1/3

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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MANASH94

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19 Dec 2025, 00:24

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On a dice there are 1,2,3,4,5 and 6.
Prime numbers are 2,3 and 5.
Non Primes= 1,4 and 6.
P(Both prime | atleast 1 prime) to be found.
P(Both Prime) = (3/6)*(3/6) = 1/4
P(Atleast 1 die shows a prime number) = 1 - (P(No Primes)) = 1 - (1/4) = 3/4
P(Both Prime | Atleast 1 Prime) = (1/4) / (3/4) = 1/3
And B = 1/3

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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chasing725

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19 Dec 2025, 01:31

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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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P(prime) = 1/2

Probability that a prime number appears on both dice = 1/2 * 1/2 = 1/4

Probablity that a prime number appears on atleast one dice = 1 - 1/4 = 3/4

Using formula of conditional probablity, required probablity = 1/4 / 3/4 = 1/3

Option B

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19 Dec 2025, 02:52

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For conditional probability, we divide the probability of intersection of events A&B by the probability of the event A.

Event A: at least one shows a prime is the opposite of none showing a prime. Primes are 2,3,5; non-rimes are 1;4;6. Therefore, the probability of none showing a prime is $1/2*1/2=1/4.$ And then, at least one showing a prime is $P(A) = 1-1/4 = 3/4$

Event B: Since probability of a prime is 1/2, then both showing a prime is $P(B) = 1/2*1/2 = 1/4$

Intersection of A and B: in case both dice show a prime, at least one of them must show a prime. Therefore, the intersection simply equals to event B, and $P(BnA)=P(B)$

Finally, our conditional probability is $P = P(BnA) / P(A) = 1/4 : 3/4 = 1/3$. The right answer is B.

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19 Dec 2025, 03:07

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We have a locked condition here: Out of the two six-sided dice, we need at least one of the dice show a prime number. It isn't what's going to happen each time, as far as I can tell, but we'd like to only find and consider conditions in which it will inevitably happen. Which means, we can eliminate cases in which both dice will not show a prime number.

Non-primes: 1, 4, and 6. Hence, 1/2 (die 1) * 1/2 (die 2) or 1/4 probability of both showing non-prime. Which we can then deduct from 1 to get 1 - 1/4 = 3/4. This 3/4 will include cases in which Die 1 is prime, Die 2 is prime, and both Die are prime.

Now, we need to see among these conditions, how likely are both dice to show prime numbers, and see how that compares to all three conditions (3/4). For that, the probability is, again - for 2, 3, 5 or 3 of the 6 - 1/2, which multiplied by 1/2 for both dice to show primes, becomes 1/4.

But now, this 1/4 or 0.25 will need to occur out of 3/4 or 0.75 or 0.25 / 0.75 = 1/3 times the answer.

This was tricky!!!

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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raffaeleprio

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19 Dec 2025, 03:19

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So let "2P" the event of BOTH NUMBERS ARE PRIME and
event "P" at least one number is prime

so we have P(2P | P ) = [P ( 2P AND P)][/P( P ) ] = [P (2P) ][/( 1 - P(NOONE PRIME)] = [0.25][/1 - 0.25] = [1][/3]

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19 Dec 2025, 03:21

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Primes between 1 and 6=2,3,5 3out of 6 are prime.
at least one prime=p(both prime/p(at least one prime
6x6=36
Both primes: per die 3 prime faces
3x3=9

p(both prime)=9/36=1/4
p(At least one prime)=1-1/4=3/4

P(both prime| at least one prime)=1/4/3/4=1/3

B.1/3

p(no rpime)=9

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The Prime numbers on six dice are:2,3 and 5
The total sample space for two rolling dice =36
Let,
Event A=atleast one dice shows prime number
Event B=both dice shows Prime number.
Number of outcomes were both are prime numbers=3X3=9 so P(b)=9/36
the no of dice where neither is prime=1-9/36=27/36
So by using conditional formula=p(b)|p(a)=(9/36)/(27/36)=9/27=3/9=1/3
so answer is 1/3