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A die is rolled 4 times. Find the probability of two 1s, one 2, and one 3.
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If a fair die is rolled three times, what is the probability that a 3 [#permalink]
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If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?A) 25/36B) 125/216C) 91/216D) 11/36
E) 36/216
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
zxcvbnmas wrote:
If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?A) 25/36B) 125/216C) 91/216D) 11/36
E) 36/216
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. _________________
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If a fair die is rolled three times, what is the probability that a 3 [#permalink]
If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?a 25/36b 125/216c 91/216d 11/36e 36/216
thanks !!!
Originally posted by geetgmat on 14 Oct 2016, 03:33.
Last edited by abhimahna on 14 Oct 2016, 04:28, edited 1 time in total.
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Re: a fair die [#permalink]
geetgmat wrote:
If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?a 25/36b 125/216c 91/216d 11/36e 36/216can someone please explain how the answer is 91/216?
thanks !!!
probability that a 3 occurs on one roll= 1/6--> probability that a 3 does not occurs on one roll= 1- (1/6) = 5/6Now, 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) = 91/216Hence C is the answer.~Consider Kudos if it helped
Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
zxcvbnmas wrote:
If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?A) 25/36B) 125/216C) 91/216D) 11/36
E) 36/216
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
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
I'm brainfarting right now - how would you go about calculating this the other way, i.e. that the 3 appears at least once?
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
zxcvbnmas wrote:
If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?A) 25/36B) 125/216C) 91/216D) 11/36
E) 36/216
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 = 125/216Thus, P(at least one three) is 1 - 125/216 = 91/216.Answer: C _________________
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
Probability of obtaining 3 = 1/6Probability of not getting 3 = 1-1/6 = 5/6Probability 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 _________________
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
Probability of atleast 1 3's = 1 - probability of No 3'sProbability of getting No 3's when rolled out a dice = 5/6Probability of not getting when dice is rolled 3 times is = 1 - (5/6) (5/6) (5/6) = 91/216
Ans: C
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
zxcvbnmas wrote:
If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?A) 25/36B) 125/216C) 91/216D) 11/36
E) 36/216
Hi I did it like this:P(E)=1/6P(E)'=5/6So probablity of getting in first roll at any position =1/6*5/6*5/6*3also this can happen in 2 rolls at any position= 1/6*1/6*5/6*3and finally at all the 3 positions 1/6*1/6*1/6So adding these 3 will give us75/216+15/216+1/216=91/216
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
zxcvbnmas wrote:
If a fair die is rolled three times, what is the probability that a 3 occurs on at least one roll?A) 25/36B) 125/216C) 91/216D) 11/36
E) 36/216
P of not getting 3 ; 5/6so5/6 *5/6 * 5/6 = 125/216P of 31-125/21691/216 ; IMO C
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
bfistein wrote:
I'm brainfarting right now - how would you go about calculating this the other way, i.e. that the 3 appears at least once?
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
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
\(1 - \frac{5}{6} * \frac{5}{6} * \frac{5}{6} = \frac{91}{216}\)
P.S. If you have to guess in these types of question, go for one of the "complementary fractions". It is very likely that the test-makers put a final trap for those distracted/tired who can maybe forget to subtract from 1.
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
Probability of getting a 3: 1/6Probability of not getting a 3: 5/6Probability of not getting a 3 * 3 die = 5/6 * 5/6 * 5/6 = 125/216Probability of getting 3 in dice rolls = 1 - Probability of not getting 3 in 3 dice rolls = 1 - 125/216 = 91/216 _________________
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
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
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
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 = \(6^3\) = 216P(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
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Re: If a fair die is rolled three times, what is the probability that a 3 [#permalink]
13 Oct 2022, 09:24