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yawa3891 [41]
2 years ago
8

A computer is used to pick four letters, one after the other, from {P,Q}. Each letter can be picked more than once and has the s

ame chance of being picked.
The possible outcomes for this experiment are shown in the table.

PPPP PPPQ PPQP PQPP
QPPP PPQQ QQPP PQQP
QPPQ QPQP PQPQ QQQP
QQPQ QPQQ PQQQ QQQQ

P( all P or all Q) =

A.0.0250
B.0.1250
C.0.5000
D.0.5625
E.0.6250
Mathematics
2 answers:
kumpel [21]2 years ago
3 0
Ok so your answer is gonna be B. 0.1250
rodikova [14]2 years ago
3 0

Answer: B. 0.1250

Step-by-step explanation:

Given: A computer is used to pick four letters, one after the other, from {P,Q}.

From the given possible outcomes for this experiment, the number of outcomes having all P or all Q (favorable outcomes)= 2

Total number of outcomes = 16

Now, the probability of picking four letters such that they are all P or all Q is given by :-

\text{P( all P or all Q)}=\frac{\text{favorable outcomes}}{\text{total outcomes}}\\\\=\frac{2}{16}=0.1250

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Step-by-step explanation:

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also solving

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\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill

\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill

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4 0
3 years ago
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