More Probability

**hansel13** · June 29th 2009, 12:29 AM

I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.

A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.

(c) What is the probability that one bulb of each type is selected?

$ P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264 $

(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?

$ P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017 $

Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.

**mr fantastic** · June 29th 2009, 01:01 AM

Originally Posted by hansel13

I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.

A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.

(c) What is the probability that one bulb of each type is selected?

$ P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264 $

(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?

$ P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017 $

Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.

(c) looks OK.

To do (d), I'd suggest calculating 1 - Pr(select 5 or less bulbs to get a 75 watt).

Note that Pr(select 5 or less bulbs to get a 75 watt) = Pr(1 bulb) + Pr(2 bulbs) + Pr(3 bulbs) + Pr94 bulbs) + Pr(5 bulbs).

Pr(1 bulb) = 6/15.

Pr(2 bulbs) = (9/15)(6/14).

Pr(3 bulbs) = (9/15)(8/14)(6/13)

Pr(4 bulbs) = (9/15)(8/14)(7/13)(6/12)

etc.

**Soroban** · June 29th 2009, 05:52 AM

Hello, hansel13!

For part (d), I had to crank out an exhaustive list.

A box contains four 40w lightbulbs, five 60w bulbs, and six 75w bulbs.

(d) Suppose now that bulbs are to be selected one by one until a 75w bulb is found.
What is the probability that it is necessary to examine at least six bulbs?

There are six 75w bulbs and nine Others.

We want the probability of finding a 75w bulb in 6, 7, 8, 9 or 10 draws.

Six draws
The first five draws are Others: . ${9\choose5}\text{ ways }\hdots \:P(\text{5 Others}) \:=\:\frac{{9\choose5}}{{15\choose5}}$
And the 6th is a 75w: . $\frac{6}{10}$
. . Hence: . $P(\text{6 draws}) \;=\;\frac{{9\choose5}}{{15\choose5}}\left(\frac{6 }{10}\right)$

Seven draws
The first six draws are Others: . ${9\choose6}\text{ ways } \hdots\:P(\text{6 Others}) \:=\:\frac{{9\choose6}}{{15\choose6}}$
And the 7th is a 75w: . $\frac{6}{9}$
. . Hence: . $P(\text{7 draws}) \;=\;\frac{{9\choose6}}{{15\choose6}}\left(\frac{6 }{9}\right)$

Eight draws
The first seven draws are Others: . ${9\choose7}\text{ ways }\hdots\:P(\text{7 Others}) \:=\:\frac{{9\choose7}}{{15\choose7}}$
And the 8th is a 75w: . $\frac{6}{8}$
. . Hence: . $P(\text{8 draws}) \;=\;\frac{{9\choose7}}{{15\choose7}}\left(\frac{6 }{8}\right)$

Nine draws
The first eight draws are Others: . ${9\choose8}\text{ ways }\hdots\:P(\text{8 Others}) \:=\:\frac{{9\choose8}}{{15\choose8}}$
And the 9th is a 75w: . $\frac{6}{7}$
. . Hence: . $P(\text{9 draws}) \;=\;\frac{{9\choose8}}{{15\choose8}}\left(\frac{6 }{7}\right)$

Ten draws
The first nine draws are Others: . ${9\choose9} \:=\:1\text{ way }\hdots \:P(\text{9 Others}) \:=\:\frac{1}{{15\choose9}}$
And the 10th is a 75w: . $\frac{6}{6} \:=\:1$
. . Hence: . $P(\text{10 draws} \:=\:\frac{1}{{15\choose9}}$

And add these probabilities.

Note: I hope my reasoning is correct.
. . . . .I'm quite able to overcount and/or miss cases.
.

**hansel13** · June 29th 2009, 11:47 AM

Thanks for the input.

I got .0503496 from mr fantastic.

and .04195804 from Soroban.

Still a little confused as well.

Wouldn't the following just work?:
$ P = {{\binom{5}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042 $

**Plato** · June 29th 2009, 12:47 PM

Originally Posted by hansel13

Still a little confused as well.
Wouldn't the following just work?:
$ P = {{\binom{5}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042 $

Actually that does work!

It may have been clearer to say $ P = {{\binom{6}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042 $ .

Or even better $ P = {\binom{9}{5}}/{\binom{15}{5}} = .042 $

Than is the probability that none of the 75's is found among the first 5.

**hansel13** · June 29th 2009, 12:58 PM

Originally Posted by Plato

Actually that does work!

It may have been clearer to say $ P = {{\binom{6}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042 $ .

Or even better $ P = {\binom{9}{5}}/{\binom{15}{5}} = .042 $

Than is the probability that none of the 75's is found among the first 5.

Right, I meant to put ${\binom{6}{0}}$ instead of ${\binom{5}{0}}$ .

Thanks for the help, I think I'm getting this stuff now.

**mr fantastic** · June 29th 2009, 02:24 PM

Originally Posted by hansel13

Thanks for the input.

I got .0503496 from mr fantastic. Mr F says: Then you made an arithmetic error because it does give 0.042 (correct to three decimal places).

and .04195804 from Soroban.

Still a little confused as well.

Wouldn't the following just work?:
$ P = {{\binom{5}{0}}{\binom{9}{5}}}/{\binom{15}{5}} = .042 $

..

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