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hodyreva [135]
3 years ago
9

A box contains 3 coins, one fair, one weighted with a 2 3 chance of coming up heads, and one weighted with a 2 3 chance of comin

g up tails. You take a coin at random from the box, flip it 20 times, and get 11 heads. What is the probability you have the fair coin?
Mathematics
1 answer:
ddd [48]3 years ago
7 0

Answer:

The probability that we have a fair coin is 0.2717

Step-by-step explanation:

Let Ci be the event that the coin i is used for while we let i = 1,2,3 (representing the three coin)

We also let H be the event that the coin which is flipped lands (which is head)

Therefore, the problem arise:

That:  P(H/C1) = 1/2.............fair coin

          p(H/C2) = 2/3..........Chance of head

          p(H/C3) = 2/3............Chance of tail

Now, noting that the coin was picked at random,

We have, : p(Ci) = P(C1) = P(C2) = P(C3) = 1/3

We can then say that or calculate thus:

P( C2 | H ) = P (H ∩ C2) ÷ P(H)

Where P (H ∩ C2) means the probability of event intersection

= P(H ∩ C2) ÷ P(H ∩ C1) + P(H ∩ C2) + P(H ∩ C3)

= P(H | C2) P(C2) ÷ P(H | C1) P(C1) + P(H | C2) P(C2) + P(H | C3) P(C3)

= (1 / 2) (1 / 3) ÷  (1/2)(1/3) + (2/3)(1/3) + (2/3) (1/3)

0.5 × 0.3 ÷ (0.5 × 0.3) + (0.67 × 0.3) + (0.67 × 0.3)

0.15 ÷ 0.15 + 0.201 + 0.201

=0.15 / 0.552

= 0.2717

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Answer:

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Step-by-step explanation:

The given geometric sequence is

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The general form of a geometric sequence is given by

a_{n} = a_{1}\cdot r^{n-1}

Where n is the nth term that we want to find out.

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r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

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You can try other consecutive numbers too, you will get the same common ratio

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So the common ratio is 4 in this case.

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This is the general form of the given geometric sequence.

Therefore, the correct option is B

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Verification:

a_{n} = 2\cdot 4^{n-1}

Lets find out the 2nd term

Substitute n = 2

a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8

Lets find out the 3rd term

Substitute n = 3

a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32

Lets find out the 4th term

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a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128

Lets find out the 5th term

Substitute n = 5

a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512

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Answer:

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The expect average profit is $–16,000.

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Step-by-step explanation:

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3 years ago
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zzz [600]
\bf \qquad  \qquad  \textit{Annual Yield Formula}
\\\\
~~~~~~~~~~~~\textit{4.0784\% compounded monthly}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0784\%\to \frac{4.0784}{100}\to &0.040784\\
n=
\begin{array}{llll}
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\end{cases}
\\\\\\
\left(1+\frac{0.040784}{12}\right)^{12}-1\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}\\\\


\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
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\end{cases}
\\\\\\
\left(1+\frac{0.040798}{2}\right)^{2}-1\\\\
-------------------------------\\\\


\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0730\%\to \frac{4.0730}{100}\to &0.040730\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
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\\\\\\
\left(1+\frac{0.040730}{365}\right)^{365}-1
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3 years ago
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