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16 Jul 2026, 02:52
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Bunuel
we know the first 11 primes are :
{2,3,5,7,11,13,17,19,23,29,31}
We need to draw 4 numbers, and the range of these 4 numbers is 12
that means, maximum(the selected 4 numbers) - minimum(the selected 4 numbers) = 12
lets find all the valid pairs that satisfy this condition :
lets add 12 to every prime of the set and check if its valid :
2+12 = 14 ( not prime)
3+12 = 15(not prime)
5+12=17(prime- valid)
7+12 = 19(prime, valid)
11+12 = 12(prime, valid)
13+12=25(not prime)
17+12=29(prime, valid)
19+12=31(prime, valid)
since the next primes will exceed 31 which is the highest prime in our set , we can stop here.
now lets count the combinations possible for each valid case:
1. 5,17- primes between them : 7,11,13 - we only need to choose 2 out of the 3 then we can get the required 4 numbers - 3C2 ways = 3 ways
similarly:
2. 7,19 - primes in between : 11,13,17 - 3C2 ways = 3 ways
3. 11,23 - primes in between : 13,17,19 - 3C2 ways = 3 ways
4. 17,29 - primes in between : 19,23 - 2C2 ways = 1 way
5. 19,31 - primes in between : 23,29 - 2C2 ways = 1 way
so no of possible combinations : 3+3+3+1+1 = 11 ways
to calculate probability, the total no of possible outcomes : 11C4 = 330
probability = 11/330 = 1/30
B
16 Jul 2026, 03:31
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Probability Questions are an interesting puzzle. Here is the way I solved them:
Basics: Probability = Favourable Outcomes / Total Outcomes.
The first 11 primes are - 2,3,5,7,11,13,17,19,23,29,31
Total outcomes = We have 11 slips of paper of which 4 are picked without replacement. Visualize that 4 are taken out at the same time - Hence you just need to pick 4 out of 11 which can be done in 11C4 ways.
Total = 11C4 or 330 outcomes
Now for the numerator, i.e., favourable outcomes
We need the pairs of primes which have the range at 12: - These are
(17,5)
(19,7)
(23,11)
(29,17)
(31,19)
Now between the first three pairs, they have three numbers each. For Example: 17 and 5 has 7,11 and 13 respectively.
For the next pairs, there are only 2 numbers in between and so they have to be picked compulsorily. For example: 29,17 have 19 and 23 in between respectively.
So the number of favourable pairs are 3C2 + 3C2 + 2C2 + 2C2 = 11
Hence probability = Favourable / Total = 11/330 or 1/30 ->B
Basics: Probability = Favourable Outcomes / Total Outcomes.
The first 11 primes are - 2,3,5,7,11,13,17,19,23,29,31
Total outcomes = We have 11 slips of paper of which 4 are picked without replacement. Visualize that 4 are taken out at the same time - Hence you just need to pick 4 out of 11 which can be done in 11C4 ways.
Total = 11C4 or 330 outcomes
Now for the numerator, i.e., favourable outcomes
We need the pairs of primes which have the range at 12: - These are
(17,5)
(19,7)
(23,11)
(29,17)
(31,19)
Now between the first three pairs, they have three numbers each. For Example: 17 and 5 has 7,11 and 13 respectively.
For the next pairs, there are only 2 numbers in between and so they have to be picked compulsorily. For example: 29,17 have 19 and 23 in between respectively.
So the number of favourable pairs are 3C2 + 3C2 + 2C2 + 2C2 = 11
Hence probability = Favourable / Total = 11/330 or 1/30 ->B
16 Jul 2026, 05:12
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Ans Choice D:
The first 11 prime numbers : 2,3,5,7,11,13,17,19,23,29,31
Total no. of cases for selecting 4 slips from 11 = 11C4 = 330
Range of 4 slips drawn is 12 means H - L = 12 and rest 2 numbers are between H & L.
H= 17,19,23,29,31 and their respective L = 5,7,11,17,19
Number of Cases for H=17,19,23 and L=5,7,11 : 6 + 6 + 6 = 18 (As 3 prime no. to be arranged for 2 places for each case)
Number of Cases for H=29,31 and L=17,19 = 2 + 2 = 4 (As 2 prime no. to be arranged for 2 places for each case)
Therefore, Req Prob = (18+4) / 330 = 1/15
The first 11 prime numbers : 2,3,5,7,11,13,17,19,23,29,31
Total no. of cases for selecting 4 slips from 11 = 11C4 = 330
Range of 4 slips drawn is 12 means H - L = 12 and rest 2 numbers are between H & L.
H= 17,19,23,29,31 and their respective L = 5,7,11,17,19
Number of Cases for H=17,19,23 and L=5,7,11 : 6 + 6 + 6 = 18 (As 3 prime no. to be arranged for 2 places for each case)
Number of Cases for H=29,31 and L=17,19 = 2 + 2 = 4 (As 2 prime no. to be arranged for 2 places for each case)
Therefore, Req Prob = (18+4) / 330 = 1/15
Bunuel
