Help please

- zarkam21

Help please

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- linn99123

with?

- zarkam21

##### 1 Attachment

- Michele_Laino

each time 2 outcomes are possible, namely head and tail

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## More answers

- zarkam21

so 8 outcomes right?

- Michele_Laino

yes!

- zarkam21

so for part one, it would be 2*2*2*2=8?

- zarkam21

now for part 2

- Michele_Laino

please wait:
it is 2*2*2*2=16

- zarkam21

yeah but isn't it a total of 8 outcomes

- Michele_Laino

more precisely:
we have 2 possible outcomes for each flip, so when we multiply, we get:
\[\Large 2 \times 2 \times 2 \times 2 = 16\]
that number expresses the possible outcomes

- Michele_Laino

if we add:
2+2+2+2 we get 8
nevertheless that number is not interesting for your exercise

- zarkam21

okay so 2*2*2*2=16 :]

- Michele_Laino

correct!
I'm sorry for my previous error

- zarkam21

It is alright!

- zarkam21

Let's move on to part II

- Michele_Laino

ok!

- Michele_Laino

here we have to do the subsequent reasoning:
lets suppose to start from a head (H)
then the next flip can be another head or a tail, so we have these outcomes:
\[\Large \begin{gathered}
HH \hfill \\
HT \hfill \\
\end{gathered} \]

- zarkam21

Yes

- Michele_Laino

whereas if we start from a tail (T), then next flip can be another tail or a head, so we get these other possible outcomes:
\[\Large \begin{gathered}
TH \hfill \\
TT \hfill \\
\end{gathered} \]

- zarkam21

okay

- Michele_Laino

in total we got these four outcomes:
\[\Large \begin{gathered}
1)HH \hfill \\
2)HT \hfill \\
3)TH \hfill \\
4)TT \hfill \\
\end{gathered} \]

- zarkam21

So that is part II? HH,HT, TH, and TT

- Michele_Laino

no, we have to finish :)

- zarkam21

okay

- Michele_Laino

next, let's start from the outcome 1)
then next flip can be a head or a tail, so we get these outcomes:
\[\Large \begin{gathered}
HHH \hfill \\
HHT \hfill \\
\end{gathered} \]

- zarkam21

okay

- Michele_Laino

now, let's start from outcome 2), then next flip can be a head ot a tail, so we get these new outcomes:
\[\Large \begin{gathered}
HTH \hfill \\
HTT \hfill \\
\end{gathered} \]

- zarkam21

Yes, :} Would the next be, THH and THT

- Michele_Laino

correct!

- zarkam21

and I'm guessing the next is TTH and TTT

- Michele_Laino

and the next two new outcomes are:
\[\Large \begin{gathered}
TTT \hfill \\
TTH \hfill \\
\end{gathered} \]
since I have started from outcome 4)

- zarkam21

Does it matter what order?

- Michele_Laino

no, the order of the outcomes, is not important
so, in total we got these outcomes:
\[\Large \begin{gathered}
a)HHH \hfill \\
b)HHT \hfill \\
c)HTH \hfill \\
d)HTT \hfill \\
e)THT \hfill \\
f)THH \hfill \\
g)TTT \hfill \\
h)TTH \hfill \\
\end{gathered} \]

- zarkam21

Okay :}

- Michele_Laino

now, we have to start from outcome a) and, with the same procedure we get these two new outcomes:
\[\Large \begin{gathered}
HHHH \hfill \\
HHHT \hfill \\
\end{gathered} \]

- zarkam21

Yup :}

- zarkam21

and then it would be HHTH and HHTT

- Michele_Laino

starting from outcome b) I get these ones:
\[\Large \begin{gathered}
HHTH \hfill \\
HHTT \hfill \\
\end{gathered} \]

- Michele_Laino

and we have to repeat the same procedure for the remaining outcomes c), d), e), f), g) and H)

- Michele_Laino

so we get these outcomes:
\[\Large \begin{gathered}
1)HHHH \hfill \\
2)HHHT \hfill \\
3)HHTH \hfill \\
4)HHTT \hfill \\
5)HTHH \hfill \\
6)HTHT \hfill \\
7)HTTT \hfill \\
8)HTTH \hfill \\
9)THTT \hfill \\
10)THTH \hfill \\
11)THHT \hfill \\
12)THHH \hfill \\
13)TTTT \hfill \\
14)TTTH \hfill \\
15)TTHT \hfill \\
16)TTHH \hfill \\
\end{gathered} \]

- Michele_Laino

that is the answer for part II

- Michele_Laino

as you can see there are 16 possible outcomes, namely the number which we have computed in part I

- zarkam21

Yes :}

- Michele_Laino

now, for part III, we have highlight those outcomes which contains two T, namely TT

- Michele_Laino

how many outcomes contain two (T)

- zarkam21

2 or more or just two

- Michele_Laino

I got six outcomes, precisely the subsequent outcomes:
#4, #6, #8 ,#10, #11, #16

- zarkam21

Okay :}

- Michele_Laino

so the probability to have two flips which comes up with (T) is:
p=favorable outcomes/possible outcomes=6/16
am I right?

- zarkam21

Yes

- zarkam21

But would't 6/16 be reduced

- zarkam21

@Michele_Laino

- Michele_Laino

it is:
\[\Large p = \frac{6}{{16}} = \frac{3}{8}\]

- zarkam21

for part II right?

- Michele_Laino

for part III is right!

- zarkam21

oh okay part III?

- zarkam21

next is part IV

- Michele_Laino

ok!

- Michele_Laino

here we have to highlight those outcomes which contain three (H).
How many outcomes do contain three (H) ?

- zarkam21

Okay so #2, #3, #5, and #12

- Michele_Laino

correct! we have 5 outcomes,
so the requested probability, is:
p=favorable outcomes/possible outcomes=\[\Large p = \frac{5}{{16}}\]

- Michele_Laino

oops.. sorry we have 4 outcomes, so the requested probability is:
\[\Large p = \frac{4}{{16}} = \frac{1}{4}\]

- zarkam21

okay :}

- zarkam21

Thank you sooo much !!!!!!!

- Michele_Laino

:) :)

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