If a fair die is rolled three times, what is the probability that a 3

Question

If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?

A) 25/36

B) 125/216

C) 91/216

D) 11/36

E) 36/216

ENGRTOMBA2018 · Accepted Answer

Questions such as these that talk about "at least" or "maximum" or "minimum" in probability questions should make sure realize that probability of any event (N) to occur = 1- P(Not N)

Thus, the probability of at least 1 roll = 1- Probability of NO 3s = 1- (5/6)(5/6)(5/6) = 1-125/216 = 91/216.

5/6 is the probability of NOT getting a 3 in any 1 roll with 5 allowed numbers (=1,2,4,5,6) out of a total of 6 possibilities.

C is thus the correct answer.

Hope this helps.

@p00rv@ · Answer

probability that a 3 occurs on one roll= 1/6 --> probability that a 3 does not occurs on one roll= 1- (1/6) = 5/6 Now, probability that a 3 occurs on at least one roll= 1- (probability that a 3 occurs on none of the 3 rolls) =1- (5/6 * 5/6 * 5/6) = 1- (125/216) = 91/216 Hence C is the answer. ~Consider Kudos if it helped

geetgmat · Answer

If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?

a 25/36
b 125/216
c 91/216
d 11/36
e 36/216

can someone please explain how the answer is 91/216?

thanks !!!

ydmuley · Answer

Probability of 3 occurs  = \frac{1}{6}

Probability of 3 does not occurs  = \frac{5}{6}

Probability of not getting 3 in three dice rolls  = \frac{5}{6} * \frac{5}{6} * \frac{5}{6}

Probability of not getting 3 in three dice rolls  = \frac{125}{216}

Probability of getting 3 in three dice rolls  = 1 - \frac{125}{216}

= \frac{(216 - 125)}{216}

= \frac{91}{216}

Hence, Answer is C

bfistein · Answer

I'm brainfarting right now - how would you go about calculating this the other way, i.e. that the 3 appears at least once?

JeffTargetTestPrep · Answer

We can use the following formula:

1 = P(at least one three) + P(no threes)

Let’s determine P(no threes):

5/6 x 5/6 x 5/6 = 125/216

Thus, P(at least one three) is 1 - 125/216 = 91/216.

Answer: C

GmatBawse · Answer

Probability of obtaining 3 = 1/6

Probability of not getting 3 = 1-1/6 = 5/6

Probability of getting 3 at least once = 1- Probability of never getting 3 (Getting 3 once, twice or thrice and Never getting 3 are mutually exclusive events)

Thus, Probability of getting 3 at least once in 3 rolls of die= 1- (Probability of never getting 3 in 3 rolls of die

= 1- (5/)*(5/6)*(5/6)
=1-125/216
=91/216
= Ans

akadiyan · Answer

Probability of atleast 1 3's = 1 - probability of No 3's
Probability of getting No 3's when rolled out a dice = 5/6

Probability of not getting when dice is rolled 3 times is = 1 - (5/6) (5/6) (5/6) = 91/216

Ans:   C

suramya26 · Answer

Hi 
I did it like this:
P(E)=1/6
P(E)'=5/6
So probablity of getting in first roll at any position =1/6*5/6*5/6*3
also this can happen in 2 rolls at any position= 1/6*1/6*5/6*3
and finally at all the 3 positions 1/6*1/6*1/6

So adding these 3 will give us
75/216+15/216+1/216=91/216

Archit3110 · Answer

P of not getting 3 ; 5/6
so
5/6 *5/6 * 5/6 = 125/216
P of 3
1-125/216
91/216 ; IMO C

MalvikaNS · Answer

So to calculate it the other way round,

probability of getting a 3 atleast once - (1/6 * 5/6 * 5/6) * 3 (because it can appear in the 1st 2nd or 3rd roll)
probability of getting two 3s - (1/6 * 1/6 * 5/6) * 3 (can appear in 1st & 2nd, 2nd & 3rd or 1st & 3rd)
probability of getting three 3s - (1/6 * 1/6 * 1/6)

hence ((25*3) + (5*3) + 1)/216 = 91/216

Basshead · Answer

Probability of getting a 3: 1/6
Probability of not getting a 3: 5/6

Probability of not getting a 3 * 3 die = 5/6 * 5/6 * 5/6 = 125/216

Probability of getting 3 in dice rolls = 1 - Probability of not getting 3 in 3 dice rolls = 1 - 125/216 = 91/216

Mona2019 · Answer

Probability of getting 3 at least in 1 roll = 1 - probability of not getting 3 in any of the rolls.

I.e 1- 5/6*5/6*5/6
= 91/216

Posted from my mobile device

BrushMyQuant · Answer

Given that a fair die is rolled three times and We need to find what is the probability that a 3 occurs on at least one roll?

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

P(Getting at least one 3)  = 1 - P( zero 3) = 1 - P(Each outcome should result in a number other than 3).

We have 6 numbers in total so getting a number other than 3 can happen in 5 ways.
Doing this three times can happen in 5*5*5 = 125 ways

=>  P(Getting at least one 3)  = 1 - P(Each outcome should result in a number other than 3) = 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

bumpbot · Answer

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If a fair die is rolled three times, what is the probability that a 3

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If a fair die is rolled three times, what is the probability that a 3

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