There are $20$ balls in a bag. $a$ of them are red, $b$ of them are white, and $c$ of them are black. It is known that $ \bullet$ if we double the white balls, the probability of drawing one red ball is $\dfrac 1{25}$ less than the probability of drawing one red ball at the beginning, and $ \bullet$ if we remove all red balls, the probability of drawing one white ball is $\dfrac 1{16}$ more than the probability of drawing one white ball at the beginning. Find $a,b,c$.
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Tags: probability
Farenhajt
18.01.2013 21:40
We have the system \begin{align}a+b+c &= 20\\ {a\over 20+b} &={a\over 20}-{1\over 25}\\ {b\over 20-a} &={b\over 20}+{1\over 16}\end{align} Equation $(2)$ yields ${a\over 20}-{a\over 20+b}={1\over 25}\iff {ab\over 20+b}={4\over 5}\quad(4)$ Equation $(3)$ yields ${b\over 20-a}-{b\over 20}={1\over 16}\iff {ab\over 20-a}={5\over 4}\quad(5)$ ${(4)\over (5)}\implies {20-a\over 20+b}={16\over 25}\iff b={180-25a\over 16}$ Hence $(5)\implies {180a-25a^2\over 16}={100-5a\over 4}\iff a^2-8a+16=0$ Therefore $a=4\iff b={180-25a\over 16}=5\iff c=20-a-b=11$
adatta0517
31.10.2021 02:17
Sorry for the 8 year bump but I am confused on what they did after getting b=(180-25a)/16.. can someone explain to me what happened after that? Thank you
xeroxia
09.10.2024 22:04
adatta0517 wrote: Sorry for the 8 year bump but I am confused on what they did after getting b=(180-25a)/16.. can someone explain to me what happened after that? Thank you We know $ab=\dfrac{5(20-a)}{4}$ then rewrite $b$ in terms of $a$.
xeroxia
09.10.2024 22:14
After $\dfrac{20-a}{20+b}=\dfrac{16}{25}$, we can realize that $a=4$ and $b=5$ is a solution. Since it is a proof based exam, it is not enough to guess. But with simple operations, we can also get the proof. $320+16b=500-25a\Longrightarrow 25a+16b=180$, $a=0$ or $b=0$ is not a solution. It should be $a=4k$ and $b=5m$ where $k$ and $m$ positive integers. $100k+80m=180$. So $k=1$ and $m=1$, then $a=4$, $b=5$, $c=11$.