How plz explain

Pls explain the answer. I think most of the people chose the answer B.

Thank you very much

What do you mean by "\(=" in the last line? I am confused.

Is the Latex code that sometimes does not show properly

Actually is this

\(1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}= 1 - \frac{5}{18} = \frac{13}{18}\) = 0.72

Regards

Thank you very much for your detailed answer?

How are you so good at both the GRE Quant and Verbal?

Thank you so much for your heart words

Start reading this https://gre.myprepclub.com/forum/gre-all-y ... -8898.html

I hope soo complete the rest.

Regards

Answer: B
Four people each roll a fair die once.

A. The probability that at least two people will roll the same number
B. 70%

When it says at least two people it means two, three or four people out of four.
We know that P(x>=2) = P(2) + P(3) + P(4)

P(a) = a people roll the same number

We know P(x) = 1 - P(x’)
P(x>=2) = 1 - P(1)

P(1) = a person roll the same number or/ nobody will roll the same number, it means if the first one roll 1, second can roll between 2 and 6 and will have 5 choices not 6, the third person will have 4 choices and so on.

P(1) = 6/6 * 5/6* 4/6 * 3/6 = 5/6* 1/3 = 5/18
P(x>=2) = 1 - P(1) = 1 - 5/18 = 13/18 = 72% which is more than B

Hello @Carcass @chetan2u

Just for a different perspective i tried to solve this question case by case but not getting the right answer.
Can u point out the mistakes in the solution? Thanks

case 1 two people throw the same number and two throw different e.g 6654

6/6*1/6*5/6*4/6 * 4!/3!2! =40/216

I am multiplying by 4! to get all the arrangements and then divide by 3! and 2! to remove the double counting.

Case 2

Three throw the same number and one throws different e.g 6664

6/6 *1/6*1/6*5/6 * 4!/3!2! =10/216

Case 3 two throw the same number and the other two throw the same but different number e.g 6655


6/6 * 1/6* 5/6*1/6 *4!/2! 2! 2! = 15/216
2! each for removing the duplicates e.g 66 and 55 and 2! more for removing the double counting of other arrangements

Case 4 All throw the same number

6/6* 1/6*1/6*1/6 =1/216

adding all cases gives 66/216 which is way off 0.72. Clearly I am missing something.

Hi,

honestly I get lost in your approach. Sorry.

However, from what I see is a process prone to errors.

The easy approach is what I showed above.

Kudos to you because is always worth to think the same question through the lens of a different approach. Nonetheless, The approach must be the simplest and fastest. Not cumbersome.

regards

In case-1 solution ,you are trying to arrange the number which isn't necessary, that's where you are going wrong.

In this case it doesn't matter if its 6645 or 6456 or 6465... all are considered same

PS: Arrangement is necessary only if the order is important

the ways to produce are numbers are =6x6x6x6=1296
the way to produce deferent number each =6x5x4x3 =360
the probability to different number each = 360/1296=.2777
the probability to produce at least 2 similar = 1-.2777=0.722222

I did "1- P(none)- p(Exactly 1)". Since P(1) is same as P(none), do we need to consider the subtrahend only once ?

Please note that there 4 possible different cases:
Case 1: Anyone will theow the same number (e.g., 6543)
Case 2: Two people will throw the same number (e.g., 6654)
Case 3: Three people will throw the same number (e.g., 6665)
Case 4: Everyone will throw the same number (e.g, 6666)

Since P(at least two people will roll the same number) is equal to the sum of the probabilities of case 2, case 3 and case 4, the easiest way to solve this problem is to calculate the probability of case 1 and get the difference to 1. In this way, we calculate only one probability instead of three.

Thus, P(at least two people will roll the same number) = 1 - P(anyone will throw the same number)

P(anyone will throw the same number) = 5/6*4/6*3/6 = 0,27777...

You can do it this way because it doesn't matter which will be the value of the first throw. Another way to do it would be:
Case 1st throw = 1: 1/6*5/6*4/6*3/6
Case 1st throw = 2: 1/6*5/6*4/6*3/6
Case 1st throw = 3: 1/6*5/6*4/6*3/6
Case 1st throw = 4: 1/6*5/6*4/6*3/6
Case 1st throw = 5: 1/6*5/6*4/6*3/6
Case 1st throw = 6: 1/6*5/6*4/6*3/6

If you sum these 6 cases you will get 6/6*5/6*4/6*3/6, which is equal to 5/6*4/6*3/6 = 0,27777...

Finally, if we do the subtraction 1 - 0,2777 we get 0,72

Answer A since 0,72 > 0,70

subtract the probability of no two people getting the same number on die from 1 to get the required output.

(6C1/6)*(5C1/6)*(4C1/6)*(3C1/6) = 360/1296 = 0.27

Hence required result = 1-0.27 = 0.73. Hence 73%.

That's why A>B

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.