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andreyandreev [35.5K]
3 years ago
13

An unfair coin with​ Pr[H] = 0.6 is flipped. If the flip results in a​ head, a marble is selected at random from a urn containin

g 4 red and 6 blue marbles.​ Otherwise, a marble is selected from a different urn containing three red and five blue marbles. If the selected marble selected is​ red, what is the probability that the flip resulted in a​ head?
Mathematics
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

0.6154

Step-by-step explanation:

Using Baye's theorem, the probability that the flip resulted in a​ head If the selected marble selected is​ red is given by;

P( H | R ) = (P( R | H ) × P( H ))/[(P( R | H ) × P( H)) + (P( R | -H ) × P( -H ))]

We can deduce that;

P( R | H ) = 4/10 = 0.4

P( R | -H ) = 3/8 = 0.375

P( -H ) = 1 - P( H )

We are given P(H) = 0.6

Thus;

P( -H ) = 1 - P( H ) = 1 - 0.6 = 0.4

Plugging in the derived values into the derived Baye's theorem, we have;

P( H | R ) = [0.4 × 0.6] /[(0.4 × 0.6) + (0.375 × 0.4)]

P( H | R ) = 0.6154

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Let's move like a crab, backwards some.

after 2 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
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\begin{array}{llll}
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\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
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after 3 years?

\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3

is that enough to pay the $1100?


now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000   <---- standard form.
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The correct option is;

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The given information are;

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