[ASK] Probability with Factors

[Monox D. I-Fly]

In a bag there are m white balls and n red balls with mn = 200 and there are more white balls than red balls. If two balls are taken randomly at once and the probability of taking two different colored balls is [tex]\frac{40}{87}[/tex] then the value of 2m + 3n is ....

A. 30

B. 45

C. 50

D. 70

E. 80


Okay, so the possibility of m and n are like this:

m = 200 and n =1

m = 100 and n = 2

m = 50 and n = 4

m = 40 and n = 5

m = 25 and n = 8

m = 20 and n = 10


Do I need to count their probability one by one then adding them up to make [tex]\frac{40}{87}[/tex]? Or am I not supposed to do that?

[romsek]

[tex]\text{Let the number of white balls be $W$ and the number of red balls be $R$}[/tex]

[tex]P[\text{choosing 1 white and 1 red when choosing two balls}] = \dfrac{\dbinom{W}{1}\dbinom{R}{1}}{\dbinom{W+R}{2}} = \dfrac{40}{87}[/tex]

[tex]P[\text{choosing 1 white and 1 red when choosing two balls}] = \dfrac{WR}{\dbinom{W+R}{2}} = \dfrac{40}{87}[/tex]

[tex]P[\text{choosing 1 white and 1 red when choosing two balls}] = \dfrac{200}{\dbinom{W+\frac{200}{W}}{2}} = \dfrac{200}{435}[/tex]

[tex]\dbinom{W+\frac{200}{W}}{2}= 435[/tex]

At this point it's probably easiest to just start testing some possiblities of [tex]W[/tex]

[tex]200 = 2^3 5^2[/tex]

If we dump the likely suspects into some software we find that [tex](W,R) = (10,20) \text{ or } (W,R) = (20,10)[/tex]

We're told that [tex]m>n[/tex] so it must be that [tex](W,R)=(20,10)[/tex]

[tex]2m+3n=40+30=70[/tex]

[Monox D. I-Fly]

Thanks.

[Guest]

[tex]P(W _{1 } \cap W _{2 })=P(W_{1})\cdot P(W_{2}|W_{1})= (m/(m+n)) \cdot ((m-1)/(m+n-1))[/tex]
[tex]P(W_{1 } \cap R _{2 })=P(W_{1})\cdot P(R_{2}|W_{1})= (m/(m+n)) \cdot (n/(m+n-1))[/tex]
[tex]P(R _{1 } \cap W _{2 })=P(R_{1})\cdot P(W_{2}|R_{1})= (n/(m+n)) \cdot (m/(m+n-1))[/tex]
[tex]P(R _{1 } \cap R _{2 })=P(R_{1})\cdot P(R_{2}|R_{1})= (n/(m+n)) \cdot ((n-1)/(m+n-1))[/tex]
[tex]S=(W _{1 } \cap W _{2 }) \cup (W _{1 } \cap R _{2 })\cup (R _{1 } \cap W _{2 })\cup (R _{1 } \cap R _{2 })[/tex]
[tex]P(S)=P((W _{1 } \cap W _{2 }) \cup (W _{1 } \cap R _{2 })\cup (R _{1 } \cap W _{2 })\cup (R _{1 } \cap R _{2 }))=P(W _{1 } \cap W _{2 })+P(W _{1 } \cap R _{2 })+P(R _{1 } \cap W _{2 })+P(R _{1 } \cap R _{2 })=(m/(m+n)) \cdot ((m-1)/(m+n-1))+ (m/(m+n)) \cdot (n/(m+n-1))+ (n/(m+n)) \cdot (m/(m+n-1))+ (n/(m+n)) \cdot ((n-1)/(m+n-1)) = (m ^{2 }-m)/(m^{2 }+ n^{2 }+2mn-m-n)+mn/(m^{2 }+ n^{2 }+2mn-m-n)+mn/(m^{2 }+ n^{2 }+2mn-m-n)+(n ^{2 }-n)/(m^{2 }+ n^{2 }+2mn-m-n)=( m^{2 }-m+mn+mn+n^{2 } -n)/(m^{2 }+ n^{2 }+2mn-m-n)=(m^{2 }+ n^{2 }+2mn-m-n)/(m^{2 }+ n^{2 }+2mn-m-n)=1[/tex]
[tex]E= (W _{1 } \cap W _{2 })\cup (R _{1 } \cap R _{2 })[/tex]
[tex]P(E)= P((W _{1 } \cap W _{2 })\cup (R _{1 } \cap R _{2 }))=(m/(m+n)) \cdot ((m-1)/(m+n-1))+(n/(m+n)) \cdot ((n-1)/(m+n-1))= (m^{2}-m)/(m^{2 }+n^{2 }+2mn-m-n)+(n^{2}-n)/(m^{2 }+n^{2 }+2mn-m-n)=(m^{2 }+n^{2}-m-n)/(m^{2 }+ n^{2}+2mn-m-n)[/tex]
[tex]F=E'[/tex]
[tex]P(F)=P(E')=P(S)-P(E)=1-(m^{2 }+n^{2}-m-n)/(m^{2 }+ n^{2}+2mn-m-n)=(m^{2 }+ n^{2}+2mn-m-n)/(m^{2 }+ n^{2}+2mn-m-n)-(m^{2 }+n^{2}-m-n)/(m^{2 }+ n^{2}+2mn-m-n)=2mn/(m^{2 }+ n^{2}+2mn-m-n)[/tex]
[tex]P(F)=40/87[/tex]
[tex]2mn/(m^{2 }+ n^{2}+2mn-m-n)=40/87[/tex]
[tex]2 \cdot 200/(m^{2 }+ n^{2}+2mn-m-n)=40/87[/tex]
[tex]400/(m^{2 }+ n^{2}+2mn-m-n)=40/87=(40/87)\cdot1=(40/87)\cdot(10/10)=400/870[/tex]
[tex]m^{2 }+ n^{2}+2mn-m-n=870[/tex]
[tex]m=200[/tex]
[tex]n=1[/tex]
[tex]m^{2 }+ n^{2}+2mn-m-n=40000+1+400-200-1=40200 \ne870[/tex]
[tex]m=100[/tex]
[tex]n=2[/tex]
[tex]m^{2 }+ n^{2}+2mn-m-n=10000+4+400-100-2=10302\ne870[/tex]
[tex]m=50[/tex]
[tex]n=4[/tex]
[tex]m^{2 }+ n^{2}+2mn-m-n=2500+16+400-50-4= 2862\ne870[/tex]
[tex]m=40[/tex]
[tex]n=5[/tex]
[tex]m^{2 }+ n^{2}+2mn-m-n=1600+25+400-40-5= 1980\ne870[/tex]
[tex]m=25[/tex]
[tex]n=8[/tex]
[tex]m^{2 }+ n^{2}+2mn-m-n=625+64+400-25-8= 1056\ne870[/tex]
[tex]m=20[/tex]
[tex]n=10[/tex]
[tex]m^{2 }+ n^{2}+2mn-m-n=400+100+400-20-10= 870[/tex]
[tex]2 m+ 3 n=2 \cdot20+3 \cdot10=40+30=70[/tex]