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?
+118609 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.
+118609 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.
