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Ivenika [448]
3 years ago
10

12 yellow balls and 9 red balls are placed in an urn. Two balls are then drawn in succession without replacement. What is the pr

obability that the first ball drawn is a red ball if the second ball drawn is yellow?
Mathematics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

\frac{9}{35}

Step-by-step explanation:

When the first draw is done there are 9 red balls in a sample size of 21. So there probability of drawing a red ball will be \frac{9}{21}

When the second draw is done, there will be 12 yellow balls in a sample size of 20 since the first ball will not have been replaced into the bag. So the chance of someone drawing the second ball in the second draw is \frac{12}{20}

The probability of them happening in this order is the product of both probabilities:

\frac{9}{21}  * \frac{12}{20}  = \frac{9}{35}

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Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.
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\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6

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