Probability
- Thread starter Nelson01
- Start date
Assuming the balls are indistinguishable to the person drawing them,
also assuming we are calculating the probability on the first draw,
then in order for the probability of drawing a blue ball being
twice the probability of drawing a red ball initially,
there must be twice as many blue balls in the bag.
also assuming we are calculating the probability on the first draw,
then in order for the probability of drawing a blue ball being
twice the probability of drawing a red ball initially,
there must be twice as many blue balls in the bag.
Hello, "Nelson01!
chrisr is absolutely correct . . . and has the best solution!
I'll do it the Long Way . . .
\(\displaystyle \text{The bag contains: }\:\begin{Bmatrix} \text{5 red balls} \\ b\text{ blue balls} \end{Bmatrix}\)
There is a total of \(\displaystyle b+5\) balls.
. . \(\displaystyle \begin{array}{ccc}P(\text{blue}) &=& \dfrac{b}{b+5} \\ \\[-3mm] P(\text{red}) &=& \dfrac{5}{b+5} \end{array}\)
\(\displaystyle P(\text{blue}) \;=\; 2\times P(\text{red}) \quad\Rightarrow\quad \frac{b}{b+5} \;=\;2\cdot\frac{5}{b+5}\)
\(\displaystyle \text{Therefore: }\;\frac{b}{b+5} \:=\:\frac{10}{b+5} \quad\Rightarrow\quad\boxed{ b\,=\,10}\)
chrisr is absolutely correct . . . and has the best solution!
I'll do it the Long Way . . .
\(\displaystyle \text{The bag contains: }\:\begin{Bmatrix} \text{5 red balls} \\ b\text{ blue balls} \end{Bmatrix}\)
There is a total of \(\displaystyle b+5\) balls.
. . \(\displaystyle \begin{array}{ccc}P(\text{blue}) &=& \dfrac{b}{b+5} \\ \\[-3mm] P(\text{red}) &=& \dfrac{5}{b+5} \end{array}\)
\(\displaystyle P(\text{blue}) \;=\; 2\times P(\text{red}) \quad\Rightarrow\quad \frac{b}{b+5} \;=\;2\cdot\frac{5}{b+5}\)
\(\displaystyle \text{Therefore: }\;\frac{b}{b+5} \:=\:\frac{10}{b+5} \quad\Rightarrow\quad\boxed{ b\,=\,10}\)
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learnerjack is just winsome wasting everybody's time, again.
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And so is Jhonson9009.
If I had the power to commit virtual murder at freemathhelp.com, then winsome et al would have all died a long time ago.