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Actually, I was able to prove to myself that without brackets my answer would always be negative, no matter what. (which was the reason why I started this topic)
But yeah, a last one, just to be sure ! thank you
2^6 /-1^5 * 2^5
2^1 /-1
-2
WOuld this be correct ? (Im trying it without the brackets)
And could you show me an example where the answer would'nt be negative ? (WITHOUt brackets...thanks )
I don't think I made it clear what I wanted, the way you showed it to me bob, that I know. But I'm talking about the rules of exponenets. ex :
2^6/-2^5 =
I know it will give me a 2^1 in my answer, but I don't know if it will stay positive or not. I could do the calculations like you did bob, but how would I know directly without doing them ?
And why would it be that way ?
Nah, no brackets. But I'm not sure to have understood. Will there be a negative sign or not ??? And could you explain me why ? THanks
Hi,
If I had : 2^6/-2^5
How would I know if the base will keep it's negative sign or not ? Thanks
Ok, bye and thank you again. If it happens again, I'll remember to do a drawing like yours !
YEs, I've included the m's in the equation above. ALso, I got 0,392 for two persons hitting !
AH......... ok, Sh... I should have done a drawing to help me see it. Ok, now, I know how to finish do it.
Here's, in the ordrer of the diagramm, the probabilities :
0.096+0.024+0.224+0.056+0,144+0,036+0,336+0,084=1
GOOD!
Thank you bobbym, I've learned from it !
Ok, must I see it that that way
1-No one gets-2 persons don't get it-1 person don't get it- every one gets it- 2 persons gets it- 1 person gets it = 0
???
could you show me in details my error pls ?
For any 2 hitting the target .392 and for 1 hitting the target .428
How did you get .392 ? Here's how I got for any two : 0,4 * 0,3= 0,12 or 0,4*0,8= 0,32 or 0,3*0,8=0,24
Could you show me ?
Well, here's how I see it :
If 1 is our omega, the we SHOULD get this :
14% (Only one person) + 68 % (Only two persons, with the different variations of the two persons, because this is omega) + 9,6 % (The three of them hitting) + 8,4 % (none hitting) = 100 %
... What are those answers ?
Ah, no I don't, I only had to find the probability of atleast 1 person hitting the target. But would the 0,14 that only one person hits the target be good, would it be the answer you would have chosen ?
Well, I don't see what else it could be.... So...
Which one ? The 0,14 ?
Here's what I did :
0,82-0,68=0,14
and 0,14-0,14= 0
0,14, the probability of only one person hitting the target
Is there any corrections to be made ? Thanks
Well, I got three of them
Two person hitting : 0,4 * 0,3= 0,12 or 0,4*0,8= 0,32 or 0,3*0,8=0,24
0,12+0,32+0,24= 0,68
Yes, so we can get to 0.
I'm trying to get to 0 by substracting every probability. To be able to see in depth ^^
Also, would this be corrrect :
1-0,084 = 0,916
0,916- 0,096 = 0,82 0,096(all the 3 hitting) = 0,4*0,3*0,8
Two person hitting = 0,4 * 0,3= 0,12 or 0,4*0,8= 0,32 or 0,3*0,8=0,24
0,12+0,32+0,24= 0,68
0,82-0,68=0,14
Is this any good ?
AHHHHHHH.. Thank you. THe problem was, I wasn't sure that atleast one person would also mean all the three like you showed me(It didn't make sense that it would mean ONLY one person, this is why I asked the question about it). A vocabulary problem ?
Anyway, thank you very much ! Im happy that I can count on this forum my problems ![]()
Hi, I need someone to help me make sense of something :
We have three person shooting one arrow.
P(Thomas hits the target) = 0,4 >> 0,6 (He doesn't hit it)
P(Alicia hits the target) = 0,3 >> 0,7
P(Julie hits the target)= 0,8 >> 0,2
Now, the question is : What is the probability of atleast one person out of the three hits the target if they shoot one arrow each.
A' = No one hitting the target
0,6 * 0,7 * 0,2 = 0,084 or 8,4 %
Now, we know that :
A = Atleast one person hitting the target
1- 0,084 = 0,916 or 91,6%
There is one LITTLE thing which I do not understand ! It concerns the A, what does the 91,6 % stand for ? Yes, it's said to be " Atleast one person hitting the target" But I don't understand why. How can you substract the probability of no one getting the target and get the probability of atleast one person hitting it ? I feel it doesn't have any link together ? COuld someone enlighten me to see how it's possible ? Thank you ( If you do not understand what I mean, I will reformulate
)
anyway, ill go through what you posted tomorrow, i dont have the energy tonight to think about it. thanks for your efforts to help.*curiously, when you get stuck on problems, how do you operate to solve them? any principles,etc. ?