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When a random experiment is conducted, the probability that [#permalink]
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26 Sep 2010, 08:38
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When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice? A. 5/243 B. 25/243 C. 64/243 D. 80/243 E. 16/17 took more time than I hoped for in the tests ( Kaplan). Don't forget to KUDOS if you like the question. Tx.
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Re: Coin Toss [#permalink]
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26 Sep 2010, 08:48
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hemanthp wrote: When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice? 5/243 25/243 64/243 80/243 16/17 took more time than I hoped for in the tests ( Kaplan). Don't forget to KUDOS if you like the question. Tx. The probability that event A occurs is \(\frac{1}{3}\); The probability that event A will not occur is \(1\frac{1}{3}=\frac{2}{3}\). We want to calculate the probability of event YYNNN (Yoccurs, Ndoes not occur). \(P(YYNNN)=\frac{5!}{2!3!}*(\frac{1}{3})^2*(\frac{2}{3})^3=\frac{80}{243}\), we are multiplying by \(\frac{5!}{2!3!}\) as event YYNNN can happen in # of times: YYNNN, YNYNN, NNNYY, ... basically the # of permutations of 5 letters YYNNN out of which 2 Y's and 3 N's are identical (\(\frac{5!}{2!3!}\)). Answer: D. Hope it's clear.
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Re: Coin Toss [#permalink]
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26 Sep 2010, 08:56
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This can be broken down into two parts:
1. The ways in which two of these events can occur  Combination 2. The probability of the event occurring.
For part 1: We have 2 events out of 5 events and hence the ways you can choose two out of five events is: \(\frac{5!}{2! * 3!} = 10\)
For part 2: Probability of correct coin toss \(= \frac{1}{3}\) and probability of not happening = \(1  \frac{1}{3} = \frac{2}{3}\)
Hence required probability \(= \frac{1}{3} * \frac{1}{3} * \frac{2}{3}* \frac{2}{3}* \frac{2}{3} * 10 = \frac{80}{243}\)
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Re: Coin Toss [#permalink]
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26 Sep 2010, 19:31
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1. number of ways to choose 2 occasion in which A would occur is 5c2=10 2. out of 5 times, the probability of occurring A is (1/3)(1/3)(12/3)(12/3)(12/3)=8/243
So the final probability is 10*(8/243)=80/243



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Re: Coin Toss [#permalink]
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27 Sep 2010, 00:28
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Thank you guys! I felt this question was interesting.



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Re: Coin Toss [#permalink]
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27 Sep 2010, 12:00
Awesome question, it tests important fundamentals



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Re: Coin Toss [#permalink]
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28 Sep 2010, 02:18
Bunuel wrote: hemanthp wrote: When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice? 5/243 25/243 64/243 80/243 16/17 took more time than I hoped for in the tests ( Kaplan). Don't forget to KUDOS if you like the question. Tx. The probability that event A occurs is \(\frac{1}{3}\); The probability that event A will not occur is \(1\frac{1}{3}=\frac{2}{3}\). We want to calculate the probability of event YYNNN (Yoccurs, Ndoes not occur). \(P(YYNNN)=\frac{5!}{2!3!}*(\frac{1}{3})^2*(\frac{2}{3})^3=\frac{80}{243}\), we are multiplying by \(\frac{5!}{2!3!}\) as event YYNNN can happen in # of times: YYNNN, YNYNN, NNNYY, ... basically the # of permutations of 5 letters YYNNN out of which 2 Y's and 3 N's are identical (\(\frac{5!}{2!3!}\)). Answer: D. Hope it's clear. Great explanation.Thanks!



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Re: When a random experiment is conducted, the probability that [#permalink]
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18 Jul 2015, 05:41
hemanthp wrote: When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice? A. 5/243 B. 25/243 C. 64/243 D. 80/243 E. 16/17 took more time than I hoped for in the tests ( Kaplan). Don't forget to KUDOS if you like the question. Tx. Probability of The event to occur = 1/3 i.e. Probability of The event to NOT occur = 1(1/3) = (2/3) Let Occurrence is represented by 'O' and NonOccurrence is represented by 'N' Then the total ways of two occurrences out of 5 trials is same as arrangement of 5 letters 'OONNN'Arrangement of 'OONNN' = 5C2 or 5!/(2!)(3!) = 10 i.e. The probability of two occurrences and 3 nonoccurrences for one arrangement = (1/3)(1/3)(2/3)(2/3)(2/3) = (1/3)^2*(2/3)^3 i.e. The probability of two occurrences and 3 nonoccurrences for all the arrangement = 5C2*(1/3)^2*(2/3)^3 = 10*(1/3)^2*(2/3)^3 = 80/243 Answer: option D
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Re: When a random experiment is conducted, the probability that [#permalink]
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20 Jul 2017, 15:48
hemanthp wrote: When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?
A. 5/243 B. 25/243 C. 64/243 D. 80/243 E. 16/17 We can let y = event A occurs and n = event A does not occur. Thus, the probability that event A occurs two times followed by three nonoccurrences of event A is: P(yynnn) = 1/3 x 1/3 x 2/3 x 2/3 x 2/3 = 8/243 However, we also have to consider that the outcome of 2 Ys and 3 Ns can occur in other orderings. For example, we can have (nynnyn) or (nnnyy), and so forth. The number of ways for these rearrangements to occur can be calculated by using the formula for permutations of indistinguishable objects: 5!/(3! x 2!) = 10. Thus, the probability is 8/243 x 10 = 80/243. Answer: D
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Re: When a random experiment is conducted, the probability that [#permalink]
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16 Sep 2017, 01:38
Bunuel wrote: hemanthp wrote: When a random experiment is conducted, the probability that event A occurs is \(\frac{1}{3}\). If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice? 5/243 25/243 64/243 80/243 16/17 took more time than I hoped for in the tests ( Kaplan). Don't forget to KUDOS if you like the question. Tx. The probability that event A occurs is \(\frac{1}{3}\); The probability that event A will not occur is \(1\frac{1}{3}=\frac{2}{3}\). We want to calculate the probability of event YYNNN (Yoccurs, Ndoes not occur). \(P(YYNNN)=\frac{5!}{2!3!}*(\frac{1}{3})^2*(\frac{2}{3})^3=\frac{80}{243}\), we are multiplying by \(\frac{5!}{2!3!}\) as event YYNNN can happen in # of times: YYNNN, YNYNN, NNNYY, ... basically the # of permutations of 5 letters YYNNN out of which 2 Y's and 3 N's are identical (\(\frac{5!}{2!3!}\)). Answer: D. Hope it's clear. Is applying the binomial formula correct in this problem? i.e, (5c2)(1/3)^2(2/3)^3.. the results are same, but is it the right approach?



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Re: When a random experiment is conducted, the probability that [#permalink]
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16 Sep 2017, 05:54
P(event A occurs) = 1/3 P (event A does not occur) = 2/3 P(event A will occur exactly 2 times)= 1/3*1/3*(2/3)^2 = 8/243 Now two experiments can occur in 5!/2!*3! ways as 2 events are similar and the rest are similar too. = 8/243*10 = 80/243 (D)




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