anonymous
  • anonymous
Lara tossed a fair coin 3 times. What is the probability of getting heads in the first two trials? 1 over 8 2 over 8 3 over 8 4 over 8 will fan and medal
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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terenzreignz
  • terenzreignz
Okay, this might not make sense NOW, but it will later. Independent events are events that don't influence one another, and the probability of both happening is simply the product of their respective probabilities.
terenzreignz
  • terenzreignz
There, I said it
terenzreignz
  • terenzreignz
Now let's begin. with just one coin toss, what's the probability of getting heads? :D

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More answers

anonymous
  • anonymous
um idk
terenzreignz
  • terenzreignz
Really? C'mon, at least give it a guess :)
terenzreignz
  • terenzreignz
If it's easier to understand, what are the chances that you get heads with one toss?
anonymous
  • anonymous
i honestly do not know
terenzreignz
  • terenzreignz
Half. Fifty percent. 1/2 ^^
terenzreignz
  • terenzreignz
Now, the second coin toss would be the same, half or 50% chance of getting heads, yes?
anonymous
  • anonymous
yes
terenzreignz
  • terenzreignz
Are the first and second coin tosses independent of each other?
anonymous
  • anonymous
no..?
rational
  • rational
coins dont have memory
terenzreignz
  • terenzreignz
in other words... does the result of the second coin toss depend on the result of the first one?
anonymous
  • anonymous
yes
anonymous
  • anonymous
idk
rational
  • rational
as if the coin remembers what happened in the previous toss and change its mind to flip the other side in next toss that can never happen
anonymous
  • anonymous
can i just get the answer
rational
  • rational
if you just blindly guess the answer, you will get it correct with a probability of 1/4 because there are 4 options
anonymous
  • anonymous
is the answer 2/8 ?
terenzreignz
  • terenzreignz
Which is a coincidence XD
anonymous
  • anonymous
so 2/8 is the answer?
rational
  • rational
Haha your luck is at its best today!
anonymous
  • anonymous
yay lol
anonymous
  • anonymous
can you help me with some more problems
anonymous
  • anonymous
?
anonymous
  • anonymous
\[\frac{2}{8}?\]
anonymous
  • anonymous
\[\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\] which is in fact \(\frac{2}{8}\) but not what you want as a final answer
anonymous
  • anonymous
oh
anonymous
  • anonymous
so does anybody know the answer
rational
  • rational
2/8 is correct
anonymous
  • anonymous
At a game booth, a student gets a box of candy as the prize for winning a game. The boxes come in four colors: white, red, green, and blue. There are 10 boxes of each color. All the boxes are equally likely to be given away as prizes. Which expression shows the probability of the first winner receiving a white box and the second winner also receiving a box of the same color? 10 over 40 multiplied by 9 over 39 10 over 40 multiplied by 10 over 39 10 over 40 plus 9 over 39 10 over 40 plus 10 over 39
rational
  • rational
Alright, read the question and see if you can tell me what we need to find out exactly
anonymous
  • anonymous
2/8 is silly
rational
  • rational
Haha not more sillier than 200/800 I think the teacher wants the kids to solve it by listing out all the possibilities and pick the favorable ones
rational
  • rational
rational numbers trip all kids in the start because they have infinitely many representations : (2, 8) = (1, 4) = (4, 16) = ...
ikram002p
  • ikram002p
thats why i hate them :)
terenzreignz
  • terenzreignz
hehe forgetting all sentimental attachment to certain box colours, the colour of the candy boxes given away to separate winners don't affect each other, right? IE... they're indpendent? ;)
anonymous
  • anonymous
ok
terenzreignz
  • terenzreignz
Okay, so this is a bit of a stretch, so I'm going to lay it out for you. In general, probability is: number of desirable outcomes divided by total number of possible outcomes. With forty boxes and ten white boxes, what's the probability of getting a white box?
anonymous
  • anonymous
idk
terenzreignz
  • terenzreignz
\[\Large \frac{\text{number of desired outcomes}}{\text{total number of outcomes}}\] how about now?
anonymous
  • anonymous
still dont know
terenzreignz
  • terenzreignz
Give it your best guess? I bet the choices give you some kind of clue ^^
anonymous
  • anonymous
Why don't we try process elimination? C and D seem silly.
terenzreignz
  • terenzreignz
lol yes, let's :D My professors were evil with these multiple choice questions, having the correct answer and the deceptively similar common mistakes among the choices o.O
anonymous
  • anonymous
Lmao. ^-^
anonymous
  • anonymous
@Skielerlucas04 are you still mad at me

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