A basketball player attempts a free throw...

**ceasar_19134** · January 29th 2007, 12:30 PM

A basketball player attempts a free throw. If he is successful, he can attempt a second free throw. If P is his probability of a successful free throw, and if his probability of making 0 points is equal to that of making 2 points, what is P? (Round the answer to the nearest thousandth.)

**Quick** · January 29th 2007, 01:00 PM

Originally Posted by ceasar_19134

A basketball player attempts a free throw. If he is successful, he can attempt a second free throw. If P is his probability of a successful free throw, and if his probability of making 0 points is equal to that of making 2 points, what is P? (Round the answer to the nearest thousandth.)

Let's say that: $P=\frac{1}{x}$

So the odds of making a shot are: $\frac{1}{x}$

The odds of making two shots are: $\frac{1}{x}\times \frac{1}{x}=\frac{1}{x^2}$

The odds of missing the first shot are: $1-\frac{1}{x}=\frac{x-1}{x}$

Which means that: $\frac{x-1}{x}=\frac{1}{x^2}$

Multiply both sides by $x$ to get: $x-1=\frac{1}{x}$

Mutiply again: $x^2-x=1$

Subtract 1 from both sides: $x^2-x-1=0$

Use the quadratic equation: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute: $x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

Solve: $x=\frac{1\pm\sqrt{5}}{2}$

That gives that: $x=1.61803399, -0.618033989$

Since it can't be negative, we know that: $x=1.61803399$

Remember: $P=\frac{1}{x}$

Substitute: $P=\frac{1}{1.61803399}$

Solve: $P=0.618033988$

Round to get: $\boxed{P\approx 0.618}$

BUT CHECK MY ARITHMETIC

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