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Mathematics
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each time 2 outcomes are possible, namely head and tail

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so 8 outcomes right?
yes!
so for part one, it would be 2*2*2*2=8?
now for part 2
please wait: it is 2*2*2*2=16
yeah but isn't it a total of 8 outcomes
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
if we add: 2+2+2+2 we get 8 nevertheless that number is not interesting for your exercise
okay so 2*2*2*2=16 :]
correct! I'm sorry for my previous error
It is alright!
Let's move on to part II
ok!
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: \[\Large \begin{gathered} HH \hfill \\ HT \hfill \\ \end{gathered} \]
Yes
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: \[\Large \begin{gathered} TH \hfill \\ TT \hfill \\ \end{gathered} \]
okay
in total we got these four outcomes: \[\Large \begin{gathered} 1)HH \hfill \\ 2)HT \hfill \\ 3)TH \hfill \\ 4)TT \hfill \\ \end{gathered} \]
So that is part II? HH,HT, TH, and TT
no, we have to finish :)
okay
next, let's start from the outcome 1) then next flip can be a head or a tail, so we get these outcomes: \[\Large \begin{gathered} HHH \hfill \\ HHT \hfill \\ \end{gathered} \]
okay
now, let's start from outcome 2), then next flip can be a head ot a tail, so we get these new outcomes: \[\Large \begin{gathered} HTH \hfill \\ HTT \hfill \\ \end{gathered} \]
Yes, :} Would the next be, THH and THT
correct!
and I'm guessing the next is TTH and TTT
and the next two new outcomes are: \[\Large \begin{gathered} TTT \hfill \\ TTH \hfill \\ \end{gathered} \] since I have started from outcome 4)
Does it matter what order?
no, the order of the outcomes, is not important so, in total we got these outcomes: \[\Large \begin{gathered} a)HHH \hfill \\ b)HHT \hfill \\ c)HTH \hfill \\ d)HTT \hfill \\ e)THT \hfill \\ f)THH \hfill \\ g)TTT \hfill \\ h)TTH \hfill \\ \end{gathered} \]
Okay :}
now, we have to start from outcome a) and, with the same procedure we get these two new outcomes: \[\Large \begin{gathered} HHHH \hfill \\ HHHT \hfill \\ \end{gathered} \]
Yup :}
and then it would be HHTH and HHTT
starting from outcome b) I get these ones: \[\Large \begin{gathered} HHTH \hfill \\ HHTT \hfill \\ \end{gathered} \]
and we have to repeat the same procedure for the remaining outcomes c), d), e), f), g) and H)
so we get these outcomes: \[\Large \begin{gathered} 1)HHHH \hfill \\ 2)HHHT \hfill \\ 3)HHTH \hfill \\ 4)HHTT \hfill \\ 5)HTHH \hfill \\ 6)HTHT \hfill \\ 7)HTTT \hfill \\ 8)HTTH \hfill \\ 9)THTT \hfill \\ 10)THTH \hfill \\ 11)THHT \hfill \\ 12)THHH \hfill \\ 13)TTTT \hfill \\ 14)TTTH \hfill \\ 15)TTHT \hfill \\ 16)TTHH \hfill \\ \end{gathered} \]
that is the answer for part II
as you can see there are 16 possible outcomes, namely the number which we have computed in part I
Yes :}
now, for part III, we have highlight those outcomes which contains two T, namely TT
how many outcomes contain two (T)
2 or more or just two
I got six outcomes, precisely the subsequent outcomes: #4, #6, #8 ,#10, #11, #16
Okay :}
so the probability to have two flips which comes up with (T) is: p=favorable outcomes/possible outcomes=6/16 am I right?
Yes
But would't 6/16 be reduced
it is: \[\Large p = \frac{6}{{16}} = \frac{3}{8}\]
for part II right?
for part III is right!
oh okay part III?
next is part IV
ok!
here we have to highlight those outcomes which contain three (H). How many outcomes do contain three (H) ?
Okay so #2, #3, #5, and #12
correct! we have 5 outcomes, so the requested probability, is: p=favorable outcomes/possible outcomes=\[\Large p = \frac{5}{{16}}\]
oops.. sorry we have 4 outcomes, so the requested probability is: \[\Large p = \frac{4}{{16}} = \frac{1}{4}\]
okay :}
Thank you sooo much !!!!!!!
:) :)

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