Trillian's Mice

Let $b$ be the number of black mice. Since there are two white mice, the probability that two of the four mice are white is $\frac{{b \choose 2}}{{b + 2 \choose 4}}$ And the probability that none of the four mice are white is $\frac{{b \choose 4}}{{b + 2 \choose 4}}$ Hence, we must have $\begin{aligned} {b \choose 2} &= 2 {b \choose 4} \\ \frac{b(b-1)}{2} &= \frac{b(b-1)(b-2)(b-3)}{12} \\ b(b-1)(b-2)(b-3) - 6b(b-1) &= 0 \\ b(b-1)( b^2 - 5b + 6 - 6) &= 0 \\ b^2(b-1)(b-5) &= 0 \\ \end{aligned}$ Therefore, $b = 0$ , $b = 1$ , or $b = 5$ . In order for the probabilities used in the question to be defined (i.e. not $\frac00$ ), we must have $b \ge 2$ ; hence $b = 5$ . The number of mice is $5 + 2 = \boxed{7}$ .

Let the number of mice be $n$ . Only two of them are white. This implies that $n-2$ are not white. See in 4 selections for white mice we have 2 selections fixed and we have to choose $2$ out of $n-2$ which is given by $(n-2)C2$ . For none are white we have to select $4$ out of number of mice that are not white which is $n-2$ and it is given by $(n-2)C4$ . It is given that $2*(n-2)C4$ = $(n-2)C2$ . On simplification we get $6=(n-4)(n-5)$ . By solving this we get $n=7$
Note: In this problem probablity was given but the total number of selections i.e. $nC4$ will get cancelled out from our equation.

Probability of 2 white rats = 2 * Probability of 0 white rats

Possiblities of 2 white rats = 2 * Possiblities of 0 white rats

(n-2)!/ (2! (n-2-2)!) = 2 * (n-2)!/ (4! (n-2-4)!)

(n-4)!/(n-6)! = 6

(n-4)*(n-5) = 6

This means n-4 and n-5 are two consecutive integers whose product is 6, obviously they are 2 and 3.

so n-5 = 2

n = 7

assuming that total number of mice is n, the probability of getting two white mice out of four is [(n-2)C2]/[nC4] and getting no white mice at all is presumably [(n-2)C4]/[nC4]. After making a equation by using the relation of this two arguments, the value of n implies 7.

Let x be the total no. of mice .

The no. of ways of choosing 4 mice with 2 white mice included are ( x - 2 ) C ( 2 )

The no. of ways of choosing 4 mice which are not white are ( x - 2 ) C ( 4 )

Therefore ( from the given of the problem ) : ( x - 2 ) C ( 2 ) = 2 ( ( x - 2 ) C ( 4 ) )

Then : ( ( x - 2 ) ! ) / ( ( 2 ! ) ( ( x - 4 ) ! ) ) = ( 2 ( ( x - 2 ) ! ) ) / ( ( 4 ! ) ( ( x - 4 ) ! ) )

After simplifying this , you will end up with a quadratic equation giving you two answers : x = 2 or 7 The right answer is 7 because there are obviously more than 2 mice in the problem ;) :)

Let us take number of mice that are not white as 'x' . Given , Number of white mice Trillian has are 2. Total number of mice then becomes : x+2. Number of ways of selecting 4 mice randomly without replacement can be done in (x+2)C4 ways . Probable number of ways of getting Exactly two white mice in the four chosen in 2C2 * xC2. So the probabaility of getting exactly two white mice out of 4 randomly picked is given as xC2 / (x+2)C4. [ Since 2C2 is 1.] Now ,number of probable ways of getting no white at all ,out of 4 chosen is xC4. So probabilty of getting no white mice out of 4 randomly picked is given as xC4 / (x+2)C4. Given that ,xC2/(x+2)C4 = 2 * xC4 / (x+2)C4. solving above we getas x= 5, whcih is number of mice that are not white. So,total number of mice will be 7

Let n be the no of black mice No of ways she can choose 2 white mice = 2C2 X nC2 No of ways she can choose 0 white mice = nC4 => nC2 = 2 X nC4 => nC2 > nC4 => n-4 > 2-0 => n = 5 (SO we don't need to actually solve the equation)

Let number of mouse all together = n

The probability that two are white , p2 = $\frac{{2 \choose 2} \times {n-2 \choose 2}} {{n \choose 4}}$

The probability that none is white , p0 = $\frac{{2 \choose 0} \times {n-2 \choose 4}} {{n \choose 4}}$

Now it's given that, p2 = $2 \times p0$

After solving this we get n = 2 or n = 7

But here 4 mice are chosen, so n must be greater than or equal to 4. So answer is 7.

{1x C( n-2 , 2)} / C(n,4) = 2 x {C( n-2 , 4 )} / C(n,4)

Solving, we get,

(n-4)(n-5) = 3 x 2

By comparing, we get n = 7

I did this by drawing a tree diagram of all the possible combinations. Picking out the combinations of four containing 2 white mice would be: WWNN, WNWN, WNNW, NWWN, NWNW, NNWW. Given that there are two white mice, the first branch of W is $2/x$ and the first branch of N $x-2/x$ .

Continuing this, obtain the probability of getting two white mice would be \( \frac{2}{x*(x-1)} \times 6

Probability of getting no white mice = \frac{(x-4)(x-5)}{x(x-1)}

Put that into the equation with \frac{(x-4)*(x-5)}{x*(x-1)} \times 2 = \frac{2}{x*(x-1)} \times 6

Solve by expanding and factorizing. There should be two brackets of (x-2) and (x-7), but as four mice are chosen there cannot be less than four in total. Therefore the answer is 7 mice. \)

Let, No: of white mice be w, No: of other mice be r and Total no: of mice be n

Given, C(w,2)*C(r,2) = 2 * C(r,4)

Also given, w=2

C(r,2)=2*C(r,4)

Solving it we get,

r=5

Now n=w+r n=2+5=7

Let there be N mice out of which, there are 2 white mice. There are 6 ways of drawing 2 white mice from 4 mice. In each case the probability of drawing 2 white mice is 2/(N (N-1). Therefore the probability of drawing 2 white mice out of four is 12/N(N-1). The probability of not drawing any white mice from 4 mice is ((N-2)/N) (N-3)/(N-1) (N-4)/(N-2) (N-5)/(N-3). Since the first probability is twice the second, we have (N-4)*(N-5) = 6 which implies N= 7 or N=2.since there are 2 white mice, N=2 is discarded. Hence N= 7

Let Trillian have $n$ mice.

Now, the probability that the $2$ out of the $4$ drawn mice are white is The mice can be drawn in $6$ ways. $W$ indicates a white mouse and $N$ indicates a mouse that isn't white.

• WNWN
• WWNN
• NNWW
• NWNW
• WNNW
• NWWN

Each of these outcomes has a probability of $\dfrac{2}{n \left(n-1\right)}$ .

Thus, the total probability of drawing exactly $2$ white mice is $6 \times \dfrac{2}{n \left(n-1\right)}$

or $P(\text{exactly 2 mice are white}) = 6 \times \dfrac{2}{n \left(n-1\right)}$

Now, the probability that all the four mice are not white is $\dfrac{{n-2 \choose 4}}{{n \choose 4}}$

or $P(\text{none are white}) = \dfrac{{n-2 \choose 4}}{{n \choose 4}}$

According to the given condition,

$P(\text{exactly 2 mice are white}) = 2 \times P(\text{none are white})$

Thus,

$6 \times \dfrac{2}{n \left(n-1\right)} = 2 \times \dfrac{{n-2 \choose 4}}{{n \choose 4}}$

Solving this equation, we get $\boxed{n = 7}$