Answer :
Answer:
The value of n is 5.
Step-by-step explanation:
It is provided that in there are n black and n white balls in an urn.
Three balls are selected at random without replacement.
And if all the three balls are white the probability is [tex]\frac{1}{12}[/tex]
The probability of selecting 3 white balls without replacement is:
[tex]P(3\ white\ balls)=\frac{n}{2n}\times \frac{n-1}{2n-1}\times \frac{n-2}{2n-2} \\\frac{1}{12}=\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} \\\frac{1}{12}=\frac{(n-1)(n-2)}{2(2n-1)(2n-2)} \\2(4n^{2}-6n+2)=12(n^{2}-3n+2)\\8n^{2}-12n+4=12n^{2}-36n+24\\4n^{2}-24n+20=0\\n^{2}-6n+5=0\\[/tex]
Solve the resultant equation using factorization as follows:
[tex]n^{2}-6n+5=0\\n^{2}-5n-n+5=0\\n(n-5)-1(n+5)=0\\(n-1)(n-5)=0[/tex]
So the value of n is either 1 or 5.
Since 3 white balls are selected the value of n cannot be 1.
So the value of n is 5.
Check:
[tex]P(3\ white\ balls)=\frac{n}{2n}\times \frac{n-1}{2n-1}\times \frac{n-2}{2n-2} \\=\frac{5}{2\times5}\times \frac{5-1}{(2\times5)-1}\times \frac{5-2}{(2\times5)-2}\\=\frac{1}{2}\times\frac{4}{9}\times\frac{3}{8} \\=\frac{1}{12}[/tex]
Thus, the value of n is 5.