excans:Ee168c35b6: Difference between revisions
From Stochiki
(Created page with "'''Solution: A''' For i = 1, 2, let <math>R_i</math> = event that a red ball is drawn form urn i, and <math>B_i</math> = event that a blue ball is drawn from urn i . Then if...") |
m (Added notice) |
||
| Line 21: | Line 21: | ||
\end{align*} | \end{align*} | ||
</math> | </math> | ||
{{soacopyright | 2023}} | |||
Latest revision as of 10:10, 28 April 2023
Solution: A
For i = 1, 2, let [math]R_i[/math] = event that a red ball is drawn form urn i, and [math]B_i[/math] = event that a blue ball is drawn from urn i .
Then if x is the number of blue balls in urn 2,
[[math]]
\begin{align*}
0.44 &= \operatorname{P}[( R1 ∩ R2 ) ∪ ( B1 ∩ B2 )] \\ &= \operatorname{P}[ R1 ∩ R2 ] + \operatorname{P}[ B1 ∩ B2 ] \\ &= \operatorname{P}[ R1 ] \operatorname{P}[ R2 ] + \operatorname{P}[ B1 ] \operatorname{P}[ B2 ] \\
&= \frac{4}{10} \frac{16}{x+16} + \frac{6}{16} \frac{x}{x+16}
\end{align*}
[[/math]]
Therefore,
[[math]]
\begin{align*}
2.2 &= \frac{32}{x+16} + \frac{3x}{x+16} \\
2.2 x + 35.2 &= 3 x + 32 \\
0.8 x &= 3.2 \\
x &= 4.
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.