zarkam21
  • zarkam21
Help please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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linn99123
  • linn99123
with?
zarkam21
  • zarkam21
1 Attachment
Michele_Laino
  • Michele_Laino
each time 2 outcomes are possible, namely head and tail

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zarkam21
  • zarkam21
so 8 outcomes right?
Michele_Laino
  • Michele_Laino
yes!
zarkam21
  • zarkam21
so for part one, it would be 2*2*2*2=8?
zarkam21
  • zarkam21
now for part 2
Michele_Laino
  • Michele_Laino
please wait: it is 2*2*2*2=16
zarkam21
  • zarkam21
yeah but isn't it a total of 8 outcomes
Michele_Laino
  • 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
  • Michele_Laino
if we add: 2+2+2+2 we get 8 nevertheless that number is not interesting for your exercise
zarkam21
  • zarkam21
okay so 2*2*2*2=16 :]
Michele_Laino
  • Michele_Laino
correct! I'm sorry for my previous error
zarkam21
  • zarkam21
It is alright!
zarkam21
  • zarkam21
Let's move on to part II
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • 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
  • zarkam21
Yes
Michele_Laino
  • 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
  • zarkam21
okay
Michele_Laino
  • 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
  • zarkam21
So that is part II? HH,HT, TH, and TT
Michele_Laino
  • Michele_Laino
no, we have to finish :)
zarkam21
  • zarkam21
okay
Michele_Laino
  • 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
  • zarkam21
okay
Michele_Laino
  • 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
  • zarkam21
Yes, :} Would the next be, THH and THT
Michele_Laino
  • Michele_Laino
correct!
zarkam21
  • zarkam21
and I'm guessing the next is TTH and TTT
Michele_Laino
  • 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
  • zarkam21
Does it matter what order?
Michele_Laino
  • 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
  • zarkam21
Okay :}
Michele_Laino
  • 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
  • zarkam21
Yup :}
zarkam21
  • zarkam21
and then it would be HHTH and HHTT
Michele_Laino
  • Michele_Laino
starting from outcome b) I get these ones: \[\Large \begin{gathered} HHTH \hfill \\ HHTT \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
and we have to repeat the same procedure for the remaining outcomes c), d), e), f), g) and H)
Michele_Laino
  • 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
  • Michele_Laino
that is the answer for part II
Michele_Laino
  • Michele_Laino
as you can see there are 16 possible outcomes, namely the number which we have computed in part I
zarkam21
  • zarkam21
Yes :}
Michele_Laino
  • Michele_Laino
now, for part III, we have highlight those outcomes which contains two T, namely TT
Michele_Laino
  • Michele_Laino
how many outcomes contain two (T)
zarkam21
  • zarkam21
2 or more or just two
Michele_Laino
  • Michele_Laino
I got six outcomes, precisely the subsequent outcomes: #4, #6, #8 ,#10, #11, #16
zarkam21
  • zarkam21
Okay :}
Michele_Laino
  • 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
  • zarkam21
Yes
zarkam21
  • zarkam21
But would't 6/16 be reduced
zarkam21
  • zarkam21
@Michele_Laino
Michele_Laino
  • Michele_Laino
it is: \[\Large p = \frac{6}{{16}} = \frac{3}{8}\]
zarkam21
  • zarkam21
for part II right?
Michele_Laino
  • Michele_Laino
for part III is right!
zarkam21
  • zarkam21
oh okay part III?
zarkam21
  • zarkam21
next is part IV
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
here we have to highlight those outcomes which contain three (H). How many outcomes do contain three (H) ?
zarkam21
  • zarkam21
Okay so #2, #3, #5, and #12
Michele_Laino
  • Michele_Laino
correct! we have 5 outcomes, so the requested probability, is: p=favorable outcomes/possible outcomes=\[\Large p = \frac{5}{{16}}\]
Michele_Laino
  • Michele_Laino
oops.. sorry we have 4 outcomes, so the requested probability is: \[\Large p = \frac{4}{{16}} = \frac{1}{4}\]
zarkam21
  • zarkam21
okay :}
zarkam21
  • zarkam21
Thank you sooo much !!!!!!!
Michele_Laino
  • Michele_Laino
:) :)

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