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]

Answer: Option D
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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

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

Thanks

Ah, dang. I see my mistake. The answer is indeed D.

Thanks,

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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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

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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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.

Answer: D.

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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"

A doubt here: Why are we considering four different cases as to where does the dice not show "2"? We shouldn't be considered about the order, right? As long as there are 3 2s that show up, shouldn't we be fine irrespective of which throw of dice do they show up in?

ArvindCrackVerbal

CrackVerbalGMAT

ArvindCrackVerbal

Even I have the same doubt. Can someone please answer the question

ArvindCrackVerbal CrackVerbalGMAT

Posted from my mobile device

embyforyou

RULE 1: EITHER BOTH NUMERATOR AND DENOMINATOR SHOULD BE CALCULATED BY COMBINATION OR BOTH BY ARRANGEMENT
RULE 2: WHEN THE NUMBERS ARE MULTIPLIED TOGETHER TO FIND TEH TOTAL OUTCOME THEN IT CONSIDERS TEH ARRANGEMENT

OBSERVATION: DENOMINATOR IS \(6^4\) I.E. 6 MULTIPLIED 4 TIME I.E ARRANGEMENT HENCE NUMERATOR ALSO MUST TAKE CARE OF ARRANGEMENTS

I HOPE THIS HELP!
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Given probability of getting a two is 1/6
so not two ; 5/6
case 1 ; getting two thrice; (1/6)^3 and 1 not two ; (5/6)
this can be arranged in 4c3 ways ; 4*(1/6)^4*(5/6)
case 2 ; all are two ; (1/6)^4
total possible cases
\(4(\frac{1}{6})^3(\frac{5}{6}) + (\frac{1}{6})^4\)
OPTION D

Embyforyou,

It’s very important to understand that in a topic like Probability where you are not certain of a lot of things, you need to take into consideration everything that can happen.
How does that relate to your question, you ask?

A dice is being thrown 4 times one after the other (as opposed to what you are thinking – you are thinking that there are 4 dice being thrown simultaneously). We are trying to find the probability that the number 2 shows up at least thrice.

Now, do we have control over when the 2s will turn up? No, we don’t.
If you throw the dice 4 times, 2 may turn up on the first three throws, whereas when your friend throws the dice 4 times, the 2s may turn up in the last three throws. So, can we disregard the order in which the 3s appear?

The answer is, we cannot, because we would want to take into consideration all possible ways in which at least 3 TWOs can be obtained. That's the reason why the order is important.

Hope that helps!
Arvind
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