Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Probability of Independent Events - Problem 2

Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Finding the probability of independent events, and events that don't have any bearing on each other. So what we're looking at now is rolling a dice three times and if you think about it you roll a dice once, you pick it up you roll it again, the outcome from the first roll has no bearing on the outcome from the second, just like that similarly the second and first have no bearing on the third.

So the question we're looking at is the probability we roll just one odd number. So what we can look at that is we have the probability we roll an even and an even and an odd, we actually have three events that have to occur. They don't have any bearing on each other so really what we're looking for is probability of an even times probability of an even times probability of an odd.

So thinking about our die, there are six sides, 2, 4 and 6 occur once each so what we end up with is a three six a one-half chance of rolling it even. That is the same thing for our second even and it's actually the same thing for our odd as well.

So what we have is one-half times one-half times one-half, but what I did in this case is I've put my odd being my last roll sort of to assign it to a spot what we really have to do is take into account some sort of probability with this one and that is going to be where this odd can occur, we can get it first, we could get it second, we could get it third, so what we really have to do in this case is multiply this outcome by 3.

Another way you could do that is you want to us combinations, we could multiply this by 3 choose 1, we're rolling it three times, we're choosing where that odd is occuring but either way we took the probability of these three events, we multiplied them together.

In this particular case we have to consider order because a different outcome occurs when you have the odd first or third, so we need to multiply this by 3 and what we're going to end up with is three-eighths chance we roll just one odd.