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Category results for 'probability'.

Vusi from South Africa wants to know - if you roll a blue die and a yellow die, why are these considered to be independent events?

Well, Vusi, let's start off by making sure we understand the terms "dependent events" and "independent events."

Dependent Events
Two or more events are dependent if the outcome of one of the events will affect the outcome of the other.

For example, if you have a jar that contains 100 blue jelly beans and 100 yellow jelly beans, and you draw one jelly bean and eat it. The probability that the jelly bean will be blue is 0.5 (100/200, because there are 100 blue jelly beans and a total of 200 jelly beans all together). However, if you then draw a second jelly bean from the jar, the probability of it being blue is not 0.5. In fact, we really don't know what the probability is. Why? Because if the first draw was a blue, the probability will be 99/199 (because we have one less blue, and one less total), but if the first draw was yellow, the probability of a blue on the second draw will be 100/199 (because we still have 100 blues, but one fewer yellow). Thus, the probability of the second event can't be calculated without knowing the result of the first event. The first event affects the outcome of the second event. These are therefore dependent events.

Independent Events
Two or more events are independent if the outcome of one event has no effect on the outcome of another.

To keep this simple, let's talk about the same jar. Only this time, instead of eating the jelly bean, you put it back in the jar after drawing it out. The probability that the first jelly bean drawn is blue is 0.5. But what about the second drawing? In the second drawing, you still have 100 blue jelly beans and a total of 200, because the first jelly bean drawn was put back in. Thus, it makes no difference what you draw the first time; the probability for the second draw being blue is still 0.5. The first draw does not affect the second draw, so these are independent events.

Hopefully, with this in mind, you can answer your own question. Are the rolls of a blue die and a yellow die independent? Yes they are! Because the roll of the blue die is not going to affect the roll of the yellow die. The yellow die isn't going to think, "Oh, the blue die is showing a 3, so I better not land on 3," or "The blue die is showing an even number, so I should show an odd," or "Hey, the blue die is showing a 6, so if I land on 6 we'll have doubles!" Whatever happened with the blue die has absolutely no bearing on what happens with the yellow die. Therefore, these are independent events.

By the way, we have a reference unit on probability on this site, which you can find here: Probability concepts and problems. This unit contains a section on independent events and dependent events, so you can read more on that here: Independent Events, Dependent Events.

Mathi, from Vellore, wants to know how to figure out the following game show probability:

"On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains \$1, the second \$100, and the third \$1,000. The boxes are randomly lined up and the contestant gets to assign each key to one of the boxes. The contestant wins the amount of money contained in each box that is opened by the key he assigns to it. What is the probability that a contestant will win more than \$1,000?"

I'm not going to answer that exact question, because I think we can make it more challenging and interesting by changing the numbers a bit. Let's do this one instead:

"On a game show, a contestant is given four keys, each of which opens exactly one of four identical boxes. The first box contains \$250, the second \$500, the third \$750, and the fourth \$1,000. The boxes are randomly lined up and the contestant gets to assign each key to one of the boxes. The contestant wins the amount of money contained in each box that is opened by the key he assigns to it. What is the probability that a contestant will win more than \$1,000?"

First, we need to figure out how many ways the keys can be arranged. The first key can be assigned in 4 ways, the second one in 3 ways (since one key has already been placed), the third key in 2 ways, and then there's only one way to place the last key. That gives us a total of 4 x 3 x 2 x 1 = 24 ways. So if we can figure out how many possibilities are wins, all we need to do is divide that by 24 to get the answer.

It's a win if you place all the keys in the right position. There is one way to do that.

It's a win if you place \$1000 and any one of the others correctly. (Note that you can't place \$1000 and two others correctly, because if you place three of them correctly, the fourth one must be correct as well, and we've already counted that possibility!). So there are three ways to do that.

It's a win if you place the \$500 and \$750 correctly (but not the other two, since we've already counted that possibility!). You can do that in one way.

And there are no other combinations that work. Thus, we have a probability of (1 + 3 + 1)/24 = 5/24.

Now that we've gone through that, you'll be surprised at how easy the other problem is - there are far fewer combinations to consider!

Good luck!
Professor Puzzler

Today's question is a probability question from Sue.  Her question is about how to calculate the probability of winning a game at least a certain number of times.

Let's say the probability of Sue winning Fraction Concentration against the computer is 3/4.  If she plays 8 times, how do you calculate the probability that she'll win at least 6 times?

If she wins at least  six times, that means she could win 6, 7, or 8 times.

So we're going to have to work out the probability that she wins exactly 6 times, the probability that she'll win exactly 7 times, and the probability that she'll win exactly 8 times.  Once we've found those three probabilities, we add them together to get the total probability.

Note that in the equations below, 3/4 is the probability of a win, and 1/4 is the probability of a loss (since 1 - 3/4 = 1/4).

Six wins: (3/4)^6 x (1/4)^2

Seven wins: (3/4)^7 x (1/4)

Eight wins: (3/4)^8

Now, depending on who your teacher is, and whether or not they like to torture you with fractions, you'll either have an ugly fraction, or a decimal answer.  I came up with approximately 0.1446.

Now that you know the process, you can fill in any numbers you like. Make the probiblity of winning 0.05.  Make the number of games 20. Whatever you like.

Helpful tip: Suppose you played 20 games, and wanted the probability of winning at least 5.  Using this method, you'd have to find the probability of winning 5, 6, 7, 8,...18, 19, or 20 games. Not pretty.

So instead, Find the probability of LOSING no more than 4.  That's a much easier probability, and it comes out to the same thing!

Happy computing!
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