What is the probability that, on three rolls of a single fair die, AT

Question

What is the probability that, on three rolls of a single fair die, AT LEAST ONE of the rolls will be a 6?

A. 1/216
B. 1/18
C. 91/216
D. 1/2
E. 125/216

ShahNYC · Answer

For each of the die the probability is 1/6. So 1/6 + 1/6 + 1/6 = 1/2.

D is the answer.

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Kinshook · Answer

Probability of 6 in 1 dice = 3C1 1/6 * 5/6 * 5/6 = 75/216.
Probability of 6 in 2 dices = 3C2 * 1/6 * 1/6 * 5/6 = 15/216
Probability of 6 in all 3 dices = 1/6 * 1/6 * 1/6 = 1/216

Probability of 6 in AT LEAST ONE dice = (75+15+1)/216 = 91/216

Alternatively, probability of 6 in NONE of the dice= 5/6 * 5/6 * 5/6 = 125/216
Probability of 6 in AT LEAST ONE dice = 1 - probability of 6 in NONE of the dice = 1-125/216 =91/216

IMO C

RJ7X0DefiningMyX · Answer

IMO, the answer to this should be as follows:

Probability (AT LEAST ONE of the rolls to be a 6) = 1 - Probability (No 6 on either of the rolls)

Probability (No 6 on either of the rolls) = 5/6*5/6*5/6 = 125/216

1 - (125/216) = 91/216 (C)

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Lampard42 · Answer

Given "AT LEAST ONE " so reverse work = 1- Prob of not getting a 6 in all three rolls

1-5/6*5/6*5/6 = 91/216 IMO C

Direct method will be as follows (Long approach):
6 In one of the die = 1/6*5/6*5/6*3(as that 1/6 can be on the 1,2, or 3 die) = 75/216
6 in two dies = 1/6*1/6*5/6*3 (as 5/6 can be on 1,2 or 3 die) = 15/216
6 in all three dies = 1/6*1/6*1/6=1/216
So prob = 91/216

Archit3110 · Answer

P not 6 in 3 rolls; 5/6*5/6*5/6 ; 125/216
and atleast 1 6 ; 1-125/216 ; 91/216
IMO C

CareerGeek · Answer

Prob (atleast 1 six) = 1 - prob (0 six)
= 1 - 5/6*5/6*5/6
= 1 - 125/216
= (216 - 125)/216
= 91/216

IMO Option C

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BrushMyQuant · Answer

Given that a fair die is rolled three times and We need to find on three rolls of a single fair die, AT LEAST ONE of the rolls will be a 6?

As we are rolling three dice => Number of cases =  6^3  = 216

P(Getting at least one 6) = 1 - P( zero 6) = 1 - P(Each outcome should result in a number other than 6).

We have 6 numbers in total so getting a number other than 6 can happen in 5 ways.
Doing this three times can happen in 5*5*5 = 125 ways

=> P(Getting at least one 6) = 1 - P(Each outcome should result in a number other than 6) = 1 -  \frac{125}{216}  =  \frac{216 - 125}{216}  =  \frac{91}{216}

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

dave13 · Answer

Bunuel  i wonder is there a list of probability problems when you need to find probability of certain multiple / or factors of a number ?

What is the probability that, on three rolls of a single fair die, AT

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What is the probability that, on three rolls of a single fair die, AT

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