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

- anonymous

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- chestercat

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- 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

There, I said it

- 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

um idk

- terenzreignz

Really? C'mon, at least give it a guess :)

- terenzreignz

If it's easier to understand, what are the chances that you get heads with one toss?

- anonymous

i honestly do not know

- terenzreignz

Half. Fifty percent.
1/2
^^

- terenzreignz

Now, the second coin toss would be the same, half or 50% chance of getting heads, yes?

- anonymous

yes

- terenzreignz

Are the first and second coin tosses independent of each other?

- anonymous

no..?

- rational

coins dont have memory

- terenzreignz

in other words... does the result of the second coin toss depend on the result of the first one?

- anonymous

yes

- anonymous

idk

- 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

can i just get the answer

- 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

is the answer 2/8 ?

- terenzreignz

Which is a coincidence XD

- anonymous

so 2/8 is the answer?

- rational

Haha your luck is at its best today!

- anonymous

yay lol

- anonymous

can you help me with some more problems

- anonymous

?

- anonymous

\[\frac{2}{8}?\]

- 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

oh

- anonymous

so does anybody know the answer

- rational

2/8 is correct

- 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

Alright, read the question and see if you can tell me what we need to find out exactly

- anonymous

2/8 is silly

- 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 numbers trip all kids in the start because they have infinitely many representations :
(2, 8) = (1, 4) = (4, 16) = ...

- ikram002p

thats why i hate them :)

- 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

ok

- 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

idk

- terenzreignz

\[\Large \frac{\text{number of desired outcomes}}{\text{total number of outcomes}}\]
how about now?

- anonymous

still dont know

- terenzreignz

Give it your best guess? I bet the choices give you some kind of clue ^^

- anonymous

Why don't we try process elimination? C and D seem silly.

- 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

Lmao. ^-^

- anonymous

@Skielerlucas04 are you still mad at me

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