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# For one roll of a certain die, the probability of rolling a two is 1/6

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Math Expert
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For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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03 Jul 2015, 02:51
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Question Stats:

63% (01:52) correct 37% (02:04) wrong based on 146 sessions

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For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) $$(\frac{1}{6})^4$$

(B) $$2(\frac{1}{6})^3 + (\frac{1}{6})^4$$

(C) $$3(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(D) $$4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(E) $$6(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

Kudos for a correct solution.

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For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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03 Jul 2015, 03:56
1
Bunuel wrote:
For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) $$(\frac{1}{6})^4$$

(B) $$2(\frac{1}{6})^3 + (\frac{1}{6})^4$$

(C) $$3(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(D) $$4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(E) $$6(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

Kudos for a correct solution.

Outcome 2 must come atleast 3 times

i.e.Outcome 2 must come either 3 times or 4 times

Case-1:Outcome 2 comes 3 times

Probability = 4C3*(1/6)^3*(5/6) = 4*(1/6)^3*(5/6)

(1/6) represents the probability of each throw when outcome 2 is obtained
(5/6) represents the probability of one throw out of four when outcome 2 is not obtained
4C3 identifies 3 places out of 4 which represent outcome 2

Case-2:Outcome 2 comes 4 times

Probability = 4C4*(1/6)^4) = (1/6)^4

(1/6) represents the probability of each throw when outcome 2 is obtained
4C4 identifies 4 places out of 4 which represent outcome 2

Total favourable probability = [4*(1/6)^3*(5/6)] + [(1/6)^4]

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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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03 Jul 2015, 04:04
1
Bunuel wrote:
For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) $$(\frac{1}{6})^4$$

(B) $$2(\frac{1}{6})^3 + (\frac{1}{6})^4$$

(C) $$3(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(D) $$4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(E) $$6(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

Kudos for a correct solution.

Solution -
Below are the chances of getting a two at least 3 times after die rolled 4 times:

One chance is not favorable, three chances are favorable -
NSSS - (5/6)*(1/6)^3
SNSS - (1/6)*(5/6)*(1/6)^2
SSNS - (1/6)^2*(5/6)*(1/6)
SSSN - (1/6)^3*(5/6)

All the chances are favorable -

YYYY - (1/6)^4

Sum up all the above possibilities = $$4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

ANS D
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For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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03 Jul 2015, 09:55
This follows Bernoulli's formula. The difference between the answer choices C, D, and E is how many $$2$$s were rolled with another number and how many $$2$$s without another number (thus, four times)

IMO C

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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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03 Jul 2015, 10:00
Ah, dang. I see my mistake. The answer is indeed D.

Thanks,
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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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03 Jul 2015, 20:31
2
1
Case 1: Dice rolls out 2 four times
P1=(1/2)*(1/2)*(1/2)*(1/2)

Case 2:Dice rolls out 2 at least three times
Observe that there are 4 favorable cases when this can appear i.e
A222
2A22
22A2
222A

So Probability P2 =(4!/3!) *(1/6)*(1/6)*(1/6)*(5/6)

Sum of Both the cases =Probability of 2 occurring at-least three times

Option(D) is correct.

Note: For some reason,I am unable to use the maths function to type out my answers.

Regards,
Manish Khare
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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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04 Jul 2015, 09:31
Option D

3 two's
Probability=4!/3!) *(1/6)*(1/6)*(1/6)*(5/6)
4 two'2
Probability=4!/4! * (1/6)*(1/6)*(1/6)*(1/6)

Sum of these two probabilities gives answer D
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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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04 Jul 2015, 10:04
1
Bunuel wrote:
For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) $$(\frac{1}{6})^4$$

(B) $$2(\frac{1}{6})^3 + (\frac{1}{6})^4$$

(C) $$3(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(D) $$4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(E) $$6(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

Kudos for a correct solution.

The first line specifies that the die is unbiased. P(T) = 1/6 and thus probability of 'NOT 2' = P(N)=1-1/6 = 5/6

Per question, probability of atleast 3 tosses = Prob of 3 2s + Prob of 4 2s = Probability of TTTN + Probability of TTTT = $$\frac{4!}{3!} * (\frac{1}{6})^3*\frac{5}{6}+\frac{4!}{4!} (\frac{1}{6})^4$$ . Thus the answer is D.
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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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06 Jul 2015, 06:15
Bunuel wrote:
For one roll of a certain die, the probability of rolling a two is 1/6. If this die is rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times?

(A) $$(\frac{1}{6})^4$$

(B) $$2(\frac{1}{6})^3 + (\frac{1}{6})^4$$

(C) $$3(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(D) $$4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

(E) $$6(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4$$

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Unfortunately, you cannot easily use the 1 - x trick here, so you must express the probability directly. You must regard the desired outcome in two separate parts: first, rolling a two exactly 4 times, and second, rolling a two exactly 3 times out of four attempts. First, the probability of rolling a two exactly 4 times is (1/6)(1/6)(1/6)(1/6) = (1/6)^4.

Next, if you roll a two exactly 3 times out of 4 attempts, then on exactly one of those attempts, you do not roll a two. Hence, the probability of rolling a two exactly 3 times out of 4 attempts is the sum of the following four probabilities:

Notice that there are 4 rearrangements of 3 “Twos” and 1 “Not a two”. In other words, you have to count as separate outcomes the 4 different positions in which the “Not a two” roll occurs: first, second, third, or fourth.

There is no way to roll a two exactly 4 times AND exactly 3 times, so you can now just add up these probabilities. Thus, the desired probability is 4(1/6)^3(5/6) + (1/6)^4.

Attachment:

2015-07-06_1712.png [ 28.05 KiB | Viewed 4505 times ]

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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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18 Jun 2017, 13:46
D

This actually an easy question. It is the sum of probabilities of rolling exactly 3 and exactly 4.
For 4 it is simply (1/6)^4, for 3 it is 4*(1/6)^3*(5/6). The logic is the following: You roll a dice and the probability of success is 1/6, then you roll a dice again and the probability again is 1/6, then again 1/6 and finally there must be some other outcome, and for another outcome, the probability is 5/6. However, we must not forget that there are actually several ways to roll 3 out of 4.
TTTF
FTTT
TFTT
TTFT
In this problem it is easy to count with the hand, however in some problems the combinatorics may provide the faster solution and the general formula is C= n!/((n-x)!*x!), where X denotes the number of success. If you are want to better understand the concept just google "Binomial Distribution"
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Re: For one roll of a certain die, the probability of rolling a two is 1/6  [#permalink]

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22 Nov 2019, 09:36
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Re: For one roll of a certain die, the probability of rolling a two is 1/6   [#permalink] 22 Nov 2019, 09:36
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