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**0123** · November 3rd 2007, 08:04 AM

This should be the last.

A college basketball player who sinks 75% of his free throws comes to the line to shoot a "one and one" (if the first shot id successful he is allowed a second shot, but no second shot is taken if the first is missed; one point is scored for each successfull shot). Assume that the outcome of the second shot, if any, is independent of that of the first. Find the expected number of points resulting from the "one and one". Compare this with the expected number of points from a "two-shot foul", where a second shot is allowed, irrespective of the outcome of the first.

I do not know. I do not know.

**Soroban** · November 3rd 2007, 09:29 AM

Hello, 0123!

A college basketball player who sinks 75% of his free throws
comes to the line to shoot a "one and one".
(If the first shot is successful he is allowed a second shot,
but no second shot is taken if the first is missed.
One point is scored for each successfull shot).

Assume that the outcome of the second shot, if any, is independent of that of the first.

Find the expected number of points resulting from the "one and one".

There are three possible outcomes:

[1] He misses the first shot: 0 points
[2] He makes the first shot and misses the second: 1 point
[3] He makes both shots: 2 points

$\begin{array}{cccccc} \text{Points} & & & \text{Probability} \\ \hline 0 & P(\sim\!1^{st}) & = & & & 0.25\\ 1 & P(1^{st} \wedge \sim\!2^{nd}) & = & (0.75)(0.25) & = & 0.1875\\ 2 & P(1^{st} \wedge 2^{nd}) & = & (0.75)(0.75) & = & 0.5625 \end{array}$

$E \;=\;(0)(0.25) + (1)(0.1875) + (2)(0.5625) \;=\;{\color{blue}\boxed{1.3125}}$

Compare this with the expected number of points from a "two-shot foul",
where a second shot is allowed, irrespective of the outcome of the first.

There are four possible outcomes:

[1] He misses both shots: 0 points
[2] He make the first, misses the second: 1 point
[3] He misses the first, makes the second: 1 point
[4] He makes both shots: 2 points

$\begin{array}{cccccc} \text{Points} & & & \text{Probability} \\ \hline 0 & P(\sim\!1^{st} \wedge \sim\!2^{nd}) & = & (0.25)(0.25) & = & 0.0625\\ 1 & P(1^{st} \wedge \sim\!2^{nd}) & = & (0.75)(0.25) & = & 0.1875\\ 1 & P(\sim\!1^{st} \wedge 2^{nd}) & = & (0.25)(0.75) & = & 0.1875 \\ 2 & P(1^{st} \wedge 2^{nd}) & = & (0.75)(0.75) & = & 0.5625 \end{array}$

$E \;=\; (0)(0.0625) + (1)(0.1875) + (1)(0.1875) + (2)(0.5625) \;=\;{\color{blue}\boxed{1.5}}$

**0123** · November 3rd 2007, 10:43 AM

Thank you so much Soroban it appears to clear and clean now you've made it.

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