a ranch has 7 geese and x ducks.2 animals are selected one after another without replacement. given that the probability that a geese and a duck are seleced is 7/15, and there are more geese than ducks, what is the value of x?

Guest Sep 2, 2014

#1**+10 **

Number of geese = 7

Number of ducks = x

More geese than ducks so x<7

Probability of a goose and a duck chosen = $$2(\frac{7}{x+7}\times \frac{x}{x+7-1})$$

I times by 2 because the probability of choosing a duck and then a goose is the same.

$$\\2(\frac{7}{x+7}\times \frac{x}{x+7-1})=\frac{7}{15}\\\\

2(\frac{7}{x+7}\times \frac{x}{x+6})=\frac{7}{15}\\\\

\frac{14x}{(x+7)(x+6)}=\frac{7}{15}\\\\

\mbox{multiply both sides by 15(x+7)(x+6) to get rid of all the fractions}\\\\

15*14x=7(x+7)(x+6)\\\\

30x=x^2+13x+42\\\\

0=x^2-17x+42\\\\

0=(x-3)(x-14)\\\\

x=3\; or\; x=14\\

but\; x<7 \;therefore\\

x=3$$

So there are 3 ducks and 7 geese.

Melody
Sep 2, 2014

#1**+10 **

Best Answer

Number of geese = 7

Number of ducks = x

More geese than ducks so x<7

Probability of a goose and a duck chosen = $$2(\frac{7}{x+7}\times \frac{x}{x+7-1})$$

I times by 2 because the probability of choosing a duck and then a goose is the same.

$$\\2(\frac{7}{x+7}\times \frac{x}{x+7-1})=\frac{7}{15}\\\\

2(\frac{7}{x+7}\times \frac{x}{x+6})=\frac{7}{15}\\\\

\frac{14x}{(x+7)(x+6)}=\frac{7}{15}\\\\

\mbox{multiply both sides by 15(x+7)(x+6) to get rid of all the fractions}\\\\

15*14x=7(x+7)(x+6)\\\\

30x=x^2+13x+42\\\\

0=x^2-17x+42\\\\

0=(x-3)(x-14)\\\\

x=3\; or\; x=14\\

but\; x<7 \;therefore\\

x=3$$

So there are 3 ducks and 7 geese.

Melody
Sep 2, 2014