Simple (not to me!) probability problem

It’s been a LONG time since probability class, and I’m struggling with this this morning.

You have a bag with 1000 balls in it. Of those, 250 are red, 750 are black.
If you pick one ball, the probability is .25 that it will be red.

If you pick two balls, the probability is roughly .25 *.25 = .0625 that they BOTH will be red.

But, if you choose two balls, what is the probability that at least ONE of them will be red?

Part of me says it’s .25 +.25 = .5, but that doesn’t seem right because it implies that if I choose 4 balls, the odds of one being red is approaching 1.

To find the probability that at least one of them is red just calculate the posibility that both are black and subtract from 1.

1 - (0.75*0.75) = 0.4375

To get the probablithy that exactly 1 is red subtract the chances of both being red from the above.

0.4375 - (0.25*0.25) = 0.375

The above is rough as the odds will change slightly after you’ve removed a ball.

Thanks, that sounds better!

And why is this very important to understand?

Say we are talking about two sports teams, team RED and team BLACK.

And BLACK wins 3 out of four games between them.

But still in a two game series RED will win at least one game nearly half the time.

Assuming the two teams are evenly matched, yes. But that’s not a valid assumption when it comes to sports.

While I understand the point your making, a better example is if you flip 3 heads in a row, the the probability of flipping a head on the 4th toss is still .5.

Not really.

The point I am making is that, even if the teams are far from evenly matched there is no guarantee that the better team will finish on top over a short series of competitions.

Most people fail to see this and invoke nonsense theories.

Turek, ColdPhoenix’ analysis is of course correct.

What’s wrong with your .25 + .25 = .5 analysis is that you double-counted some outcomes : .25 that ball *one *is red already counts outcomes where ball *two *is red as well as black, and counting .25 that ball two is red counts *all *outcomes where ball two is red.

In other words, you double-counted the outcomes where both balls are red.

So your analysis requires subtracting that probability - which is 1/4 *1/4 = .0625.

.25 + .25 - .0625 = .4375. :cool:

It seems like there are 4 possibilities if you pick 4 balls. Both red, both black, first red second black, first black second red. In three of those cases you have at least one red ball. So 75%? But why do I have the feeling I am missing something?

Your calculation assumes the probabilities of picking red or black ball on any pull are the same, which they aren’t in the OP (three times as many black as red balls).

Thank you! I guess I was confusing it with the four kids, what’s the probability one is a girl question.

That is basically right though.

There are four possibilties:

RR
RB
BR
BB

and the chances are

.25 * .25 = .0625
.25 * .75 = .1875
.75 * .25 = .1875
.75 * .75 = .5625

Though of course, it isn’t actually that difficult to arrive at the exact probabilty, as opposed to a good approximattion:

(250/1000 * 249/999) + (250/1000 * 750/999) + (750/1000 * 250/999) =

1- (750/1000 * 749/999) =

437250/999000 =

4.37̅6̅8̅

Are you putting the first ball back before you pick the second one? If you are, you’re right; otherwise, you have to account for the fact you have on fewer red ball and one fewer ball total.

Anyway, you’re looking at a hypergeometric distribution there and we know lots and lots about them.

No, I’m not putting the first ball back, which is why I said “roughly” .25 * .25.

This is my favorite probability error. Similarly, if I buy a lottery ticket, either it will win or it won’t win: 50/50 chance.

If only!

I often would represent it to my students this way: say I’m in a boxing match with Mike Tyson. One of us will win, right? Therefore, I have a 50% chance of winning, right?

Most of them already know what when I say something, and then say, “Right?” I purposefully was wrong and I’m hoping they’ll figure out the mistake.

Which is why I always buy two, to cover my bases.

One thing not specified in the OP is whether the picks are with replacement or without replacement. That is, after you pick the first ball, do you put it back in and randomize the bag before picking out the second ball? Or do you leave the first ball out when you reach in to pick a second ball?

Also, one of the ways you can analyze the problem is to calculate the probability that both balls will be black, and then subtract that from 1 to calculate the probability that at least won’t be black; i.e., that one ball will be red.

E.g., in the first scenario you’ve got a 750/1000 chance each time to pick a black ball, and 1 - 0.75 * 0.75 = 0.4375, just as ColdPhoenix said. In the second scenario, the probability is 1 - 750/1000 * 749/999 = 0.437̅6̅8̅, like These are my own pants said.

I didn’t specify the replacement scheme in the OP, but I did specify a few posts later. :slight_smile:

The actual, real life scenario is this: My wife and I are going to a reading by a famous author. For buying a ticket, you get to attend the reading and you get a copy of the book. 1000 tickets were sold. Randomly mixed into the 1000 books being distributed are 250 signed copies. I was trying to figure out the probability of us getting at least one signed copy.