Video: Determining the Probability of Intersection of Two Independent Events

A bag contains 7 blue marbles and 42 red marbles. A marble is drawn from the bag, recorded, and then replaced. Another marble is then drawn. What is the probability that the first marble is blue and the second is red?

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Video Transcript

A bag contains seven blue marbles and 42 red marbles. A marble is drawn from the bag, recorded, and then replaced. Another marble is then drawn. What is the probability that the first marble is blue and the second is red?

Well, to help us solve this problem what I’ve drawn is a tree diagram. And in our tree diagram, we’re gonna have the first branch is our first marble and the second branch is our second marble. So, next, what we do is we label the branches. And in this case, what we’re gonna label them is B and R for blue and red because it says that we’ve got blue marbles and red marbles. And now, what we need to do is add our probabilities.

So, what is the probability on the first pick that we get a blue marble? Well, we can see that there’re seven blue marbles. So, we can say that the probability of blue marble is gonna be equal to seven. Then, it’s gonna be over the total number of marbles, which is seven plus 42. But be careful here cause a common mistake is just to put the 42, but that’s only the number of red marbles. So, this will give us 49. So, we get seven over 49.

We can put this into our tree diagram, but what we wanna do first is simplify. And we can do that by dividing both the numerator and denominator by seven because it’s common factor. And if we do that, we get one over seven, or one-seventh. So, that means we can add that to our first branch. So, That’s the probability of getting a blue marble.

Well, what’s the probability of picking a red marble? Well, we could work it out in the same way. So, we could say the probability of being red would be 42 out of 49, or 42 over 49. And again, we could divide the numerator and denominator by seven cause it’s common factor. And that would give us six-sevenths. However, there would’ve been a quicker way of working it out.

And that is to see what was needed to add to one-seventh to make a whole, or one. And that’s because the branches sum to one because they’re exhaustive. So, that means that they’re all the probability that anything could occur. So therefore, if blue is gonna be one-seventh, then red is gonna be six-sevenths because one-seventh plus six-sevenths make seven-sevenths, which is one whole one.

Well, now, what we need to do is look at the second branch and think, well, what are our probabilities going to be here? Well, the key to working this out is the word replaced cause what we’re told is that the marble is drawn from the bag, recorded, and then replaced. And because it’s replaced, it doesn’t affect the probabilities because they remain the same. Because we’ve got the same number of blue and red marbles on the second pick. So therefore, we can fill in our probabilities accordingly.

It’s worth noting that if it wasn’t replaced, then this’d be something called conditional probability. And there’d have to be a slight difference with the second lot of probabilities. Okay, great, we’ve completed our tree diagram. What’s the next step? Then, it’s good practice to work out each of the probabilities for the different possible combinations or scenarios arising from our tree diagram. We might not necessarily needed for this question because we know what we’re looking for. However, it is good practice cause it can help with more complex problems.

So, first of all, we’ve got the probability of picking a blue then picking a blue. Then, we’ve got the probability of picking a blue then picking a red, then picking a red, picking a blue, and then picking two reds. Well, before we work out our probabilities, it’s worth noting that what we’re gonna do is apply the AND rule. So, for instance, blue and blue, blue and red, etcetera. And what that means is we’re gonna multiply along our branches.

So, the probability of a blue then a blue is gonna be equal to one-seventh, because that was the probability of picking a blue in the first pick, multiplied by one-seventh, and that’s because that was the probability of picking a blue on the second pick. And this would give us an answer of one over 49. And that’s because when we’re multiplying fractions, we multiply the numerators then multiply the denominators.

Then, for blue red, we get one-seventh multiplied by six-sevenths, which is six over 49. Then, for red blue, we have six-sevenths multiplied by one-seventh, which is six over 49 again. Then, finally, we have six over seven multiplied by six over seven, which is equal to 36 over 49. So, that’s the red red. So, now, to check that we’ve got the right probabilities, all we do is we add them together. So, we’ve got one over 49 add six over 49, which will be seven over 49. Add another six over 49 gives us 13 over 49. Add 36 over 49 gives us 49 over 49, which is one, which is what we’d be looking for.

Okay, now, we’ve done this and completed our tree diagram, let’s solve the problem. Well, the question wants us to find the probability that the first marble is blue and the second is red. So, if we follow the path I’ve drawn in pink, it’s the first one is being blue, the second one is being red, we can see that we’ve already calculate the probability. And that’s six over 49. Therefore, we can say that in answer to the question, the probability of the first marble is blue and the second marble is red is six over 49.

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