pic posted
there's 9 that are less than 6 (not including the 2)
@Psymon help?
The chart is more confusing than it needs to be. I'm drawing my own on some scratch is all :P
but DO U know how to do it...
sorry my limit is 4 minutes for this question
Yeah, I dont think I know itwhen it comes to the term and being thrown in. Too rusty with it all.
Without the and, its just 4/36, but I know thats not the answer.
so u don't know?? ugh! :(
Im not confident if I know and Id rather not give an answer and say its correct.
Otherwise Id say 4/36 reduces to 1/9
Amount of occurence divided by possibilities. 4 occurences of a 2 showing once and being less than 6 divided by 36 overall outcomes means 4/36. Thats what I got, but Im not confident in it.
its not 4/36 in the 1st place..
Well, itd reduce to 1/9. But like I said, Im not confident in my answer *shrugs*
not including the 2 there are 9 numbers less than 6
Right. Which is why I said the and throws me off. Im no good with probability. Sorry no-one else came.
@vinnv226
even without the "and" its 9/36... :(
There are 5 outcomes, out of 36, so 5/36
what???
It only wants one 2, though.
It doesn't say that
there is only one 2!
It just says one die shows a 2. I take that to mean just whatit says.
If one die shows a 2, the other die could be anything. It does not say "exactly one" die shows a 2. Based on the current wording of the problem, it would be 5/36
wouldn't it be 9/36 * 1/6 ???
No, because those are simply only the sums that are less than 6. It needs to satisfy the condition of having a 2 as well.
oh cuz it says AND?
Yeah.
so then... I don't get this
You need to find the outcomes where one die has a 2, and the sum is less than 6. So eliminate all the sums that are 6 or over. Then, which of the remaining outcomes has a 2 in it? See my screenshot i uploaded, I marked the 5 outcomes.
they don't have a 2 in it... :( ?
oh wait I see it cuz to the left/right it can go above 6 but theyre all in the "2" section...
oh ok.. keep going actually
but it says "and ONE die shows a 2" ??
Right, so at least one die needs to show a 2, so you eliminate all outcomes that don't have a 2 in them
so why is it in both of the dice there is a 2? it says "one" right?
ohh. ok.. so its 5/36?
Right, so they're talking about one of the 2 dice. The other die is irrelevant
but the answer is still 5/26..
5/36 I mean?
So what if there was a problem like this: You toss 3 coins. What is the probability that you get exactly 3 heads given that you get at least one head???
@vinnv226
I'm pretty sure I answered that exact problem earlier today
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