simple probability question

**lllll** · April 12th 2008, 03:44 PM

twenty students are asked to select an integer between 1 and 10. Eight choose either 4, 5, or 6.

if the students make their choices independently and each is as likely to pick one integer as any other, what is the probability that 8 or more will select 4, 5, or 6?

the solution is 0.2277, but I don't know how you get to it.

**mr fantastic** · April 12th 2008, 03:52 PM

Originally Posted by lllll

twenty students are asked to select an integer between 1 and 10. Eight choose either 4, 5, or 6.

if the students make their choices independently and each is as likely to pick one integer as any other, what is the probability that 8 or more will select 4, 5, or 6?

the solution is 0.2277, but I don't know how you get to it.

Let X be the random variable number of students who 4, 5 or 6.

X ~ Binomial(n = 20, p = 0.3).

You require $\Pr(X \geq 8 | X \geq 8)$ . This is just the same as the unconditional probability $\Pr(X \geq 8)$ , which is equal to 0.2277 (correct to four decimal places).

**mr fantastic** · April 12th 2008, 05:23 PM

Originally Posted by mr fantastic

Let X be the random variable number of students who 4, 5 or 6.

X ~ Binomial(n = 20, p = 0.3).

You require $\Pr(X \geq 8 | X \geq 8)$ . This is just the same as the unconditional probability $\Pr(X \geq 8)$ , which is equal to 0.2277 (correct to four decimal places).

I hate to say - I made a mistake. But so has the book!

It is true that $\Pr(X \geq 8) = 0.2277$ but this is not the same as $\Pr(X \geq 8 | X \geq 8)$ ......

$\Pr(X \geq 8 | X \geq 8) = \frac{\Pr(X \geq 8 \, \text{and} \, \Pr(X \geq 8)}{\Pr(X \geq 8)} = \frac{\Pr(X \geq 8)}{\Pr(X \geq 8)} = 1$ !!!!

Furthermore, $\Pr(X \geq 8 | X = 8) = 1$ too.

I'm not sure what the book intended the question to be if the answer is 0.2277, but it cannot be as stated ..... Can you re-check the wording of the question?

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