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fredd [130]
3 years ago
11

I have two coins. One is fair, and has a probability of coming up heads of.5. \text {The second is bent, and has a probability o

f coming up heads of} .75The second is bent, and has a probability of coming up heads of.75. If I toss each coin once, what is the probability that at least one of the coins will come up tails?
Mathematics
1 answer:
aniked [119]3 years ago
6 0

Answer:0.625

Step-by-step explanation:

Given

There are two coins one is fair and another is bent

Probability of getting a head on fair coin is 0.5

Probability of getting a head on bent coin is 0.75

If both coins are tossed once

probability of getting at least one of the coin will come up tails is

P=1-P(both\ heads)

P=1-0.5\times 0.75

P=1-0.375

P=0.625

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\bf \qquad \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 &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
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\bf \qquad \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 &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396

add both amounts, and that's her investment for the 11 years.
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