[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 51 KB, 396x385, sad_frog.jpg [View same] [iqdb] [saucenao] [google]
5370670 No.5370670 [Reply] [Original]

/sci/ this isn't a troll post I am genuinely confused what is probability of throwing at least one 6 in a pair of die.

Sample Space : 36
Number of events: 11

But if you sum the chance of a 6 of 1 die and addd another it is 2/6 = 1/3

so what is the correct answer: 11/36 or 1/3 ? google wont give me the fucking answer.

>> No.5370675

>>5370670
You can only add probabilities is they are disjoint right ? So would you then subtract the extra 6,6 roll ?

>> No.5370679

Your post is a bit confusing

>> No.5370684

>>5370670

(1/6) + (1/6) = (2/6) = (1/3)

>> No.5370688

>>5370679
You throw a pair of dice, what is the probability that at least one die is a 6 ?

>> No.5370691

11 out of 36 situations have a 6 in them.

>> No.5370692

>>5370684
Name one dice D1 and another dice D2

Aren't you double counting because wouldn't this consider D1 = 6 and D2 = 6 as one event. And then count it again just because D2 = 6 and then D1 = 6, which is really the same thing ?

>> No.5370703

>>5370670
>Sample Space : 36
>Number of events: 11
If your sample space has 36 elements then you have 2^36 events.

>> No.5370711

1 2 3 4 5 6
X X X X X X 6
X 5
X 4
X 3
X 2
X 1

11/36

>> No.5370712

The chance of not getting a 6 at all is 1-(5/6)^2 = 11/36
The chance of getting two sixes is (1/6)^2 = 1/36
Therefore the chacne of getting at least one six is 1/3

>> No.5370719

>>5370670
OP I think I have an answer that should help you.
The three events that you are summing together are as follows:

You roll the first die and get a 6 and you roll the other and do not get a 6. ((1/6) * (5/6)) = 5/36

You roll the first die and do not get a 6and roll the other and get a 6. Once again ((1/6) * (5/6)) = 5/36

Finally you have the probability of rolling 6 on both die. (1/6 * 1/6) = 1/36

Sum the probabilities to get the probability of "getting a six with at least one of the die"
(1/36 + 5/36 + 5/36) = 11/36

>> No.5370737

>>5370719
thanks man!

>> No.5370913

>>5370670
You don't add them OP

By that logic once you have six die there's no chance of NOT rolling a six. Which of course could happen.