In this video, we’re gonna look at how to solve some probability questions. Now, we’re gonna be using probability trees and we’re gonna be using some algebra.
A bag contains five green and three red balls. A ball is taken from the bag at random and then replaced. Another ball is then taken from the bag at random. Complete the tree diagram.
And then we gotta answer the question “what is the probability that both balls are red?” So we’ve got a bag that’s got five green and three red balls; let’s do a quick sketch of that. Now, we’re told that we’re gonna take a ball at random from the bag. Now for this to work, it’s important that we can actually see them when we’re selecting them that they’ve all gone equally likely chance of being selected. So they all need to look and feel the same apart from their colour. So they need to be the same size, the same texture, the same weight, and so on.
And in this experiment, we’re gonna take a ball and see what color it is, put it back in the bag, shuffle them up again, and then take a second one at random again. Now, this part of the probability tree here is telling us what’s happening when we select the first ball and these are the possible outcomes when we select that second ball. So let’s go ahead and try to complete the probability tree.
Well, if the bag has got five green and three red balls, in total we got eight balls. And the random selection means that all of them are equally likely to be selected. So five out of eight of the balls are green; the other three out of eight are red. Those are the proportions of the balls of those two colors. Now, those proportions represent the probabilities. So if the property of green is five-eighths, the probability of red must be the other three out of eight or three-eighths.
Now, onto the selection of the second ball, we put the first one back whether it’s red or green. So we’re starting off with five green and three red again. So there are eight balls in total in the bag and they’re all equally likely to be selected. So the probability of green or red is exactly the same as it was before regardless of whether we got green or red first time. So the probability is five-eighths for the green and three-eighths for red if we picked the green to start off with and the same again even if we picked a red first time round.
So now we’re going to work out what the probability is that both balls are red. Well, we got four different options. We could have picked a green first and a green second, a green first and a red second, a red first and a green second, or a red and a red. Now, the probability that we picked green both times will be five-eighths for the first green and five-eighths for the second green, so that’s five-eighths times five-eighths, which is 25 over 64.
The probability that we picked a green first and then a red is five-eighths times three-eighths, so we’re just multiplying the probabilities along that path of the tree together and that makes 15 over 64. So the probability of getting a red and a green, that’s gonna be three-eighths times five-eighths, again multiplying the probabilities along that path of the probability tree. And again, that’s 15 over 64.
Now, the probability that we’re really interested in — the probability that we get a red first time and a red second time — well, looking along that route through the tree, we got three-eighths times three-eighths. Now as a quick check, we can add up all of those different probabilities. Those four outcomes are the only possible outcomes, so we know all of the probabilities must sum to one. So 25 over 64 plus 15 over 64 plus 15 over 64 plus nine over 64, well that does give us 64 over 64, which is of course one.
So it looks like we haven’t made any simple arithmetic errors along the way. So all what we need to do now is just read of what was the probability of getting red first time and red second time and that was nine over 64. So our answer is the probability that both balls are red is nine over 64.
Now, the second part of the question goes like this. Some more green balls are added to the five green and three red balls in the bag. A ball is taken at random and replaced. Another ball is then taken from the bag at random. The probability that both balls are red this time is one 25th. How many green balls were added to the bag?
So some more green balls were added to the bag. Now, we don’t know how many green balls were added. So let’s define a variable called 𝑥, which represents the number of green balls that were added to the bag. So we’re now going to repeat the experiment that we did in the first part of the question, but with a different number of red and green balls. And then our question boils down to find the value of 𝑥, the variable that we just defined.
Okay, so in the bag, we’ve now got three red balls and some green balls. Well, how many green balls have we got? We have the five that we started off with and we’ve added 𝑥. So we’ve got five plus 𝑥 green balls in total. So how many balls altogether? Well, three plus five plus 𝑥 is eight plus 𝑥 balls in total. And the important thing is that these are all equally likely to be picked when we choose a ball at random from the bag.
Now, because we’re doing replacement in the bag, whatever ball we take out we’re putting back, those probabilities are staying the same for the first draw and the second draw. So for one draw, whether it’s first or second, the probability that we get a red ball is three because there are three red balls out of eight plus 𝑥 because there were eight plus 𝑥 balls in total. So three over eight plus 𝑥 is the proportion of those in the bag, which are red.
So to work out the probability that we’re gonna get red both times again, so going along that bottom line of the probability tree, it’s gonna be three over eight plus 𝑥 times three over eight plus 𝑥. And remember the question told us that the probability that both balls are red this time is one over 25. So we can say that three over eight plus 𝑥 times three over eight plus 𝑥 is equal to one over 25.
So we just gotta solve this equation using algebra and find the value of 𝑥. Well, just to make it clear, I’m going to put parentheses around both of these denominators to make sure that we know that the eight and the 𝑥 terms go together. So we gotta multiply those two fractions together on the left-hand side. And when we do that, we multiply the numerators together and the denominators. So three times three is nine and eight plus 𝑥 times eight plus 𝑥 is eight plus 𝑥 all squared. And that’s equal to one over 25.
So now I’ve got an equation involving 𝑥 on the denominator and I’ve got to try to get that on the numerator. I’ve got to simplify this whole expression. So I think the first thing that I’m gonna do is to try to get rid of the fraction from the right-hand side by multiplying everything through by 25. So multiplying both sides by 25, well in fact we’re dealing with fractions here, so I’m going to say 25 over one just to make that a bit clearer.
On the right-hand side, if I divide the top by 25, I get one; if I divide the bottom by 25, I get one. So I’ve now got one times one over one times one, which is one. And on the left-hand side, I’ve got 25 times nine which is 225 over eight plus 𝑥 squared. So let’s just write that down. Now to get eight plus 𝑥 squared off the denominator, I’m going to multiply both sides of my equation by 𝑎 plus 𝑥 squared. And of course, on the left-hand side, that means that these things cancel out leaving me with just 225. And on the right-hand side, I’ve got one times eight plus 𝑥 all squared.
So just writing that out to make it a bit clearer, I’ve got 225 is equal to eight plus 𝑥 all squared. Now, if I just take square roots of both sides, I’ve got the square root of 225 is equal to the square root of eight plus 𝑥 all squared. Now, the square root of 225 is 15 and the square root of eight plus 𝑥 all squared is just eight plus 𝑥. Now remember 15 times 15 is 225, but also negative 15 times negative 15 is 225. So since we were trying to solve this equation, we’ve got two possible values are on the left-hand side that we need to be aware of.
So positive or negative 15 is equal to eight plus 𝑥. And if I take away eight from both sides of that on the right-hand side, I’ve got eight take away eight is nothing; that just leaves me with 𝑥. And on the left-hand side, I’ve got negative eight add 15 or negative eight take away 15. And negative eight plus 15 is equal to positive seven and negative eight take away another 15 is equal to negative 23.
Now looking at these two possible mathematical solutions, that second one there, adding negative 23 green balls to the bag, would actually means we’re taking away 23 green balls. Now, given that there are only five in there to start off with, this answer can clearly be ignored; it’s wrong. So the other answer must be the real answer; 𝑥 is equal to seven. We added seven green balls to the bag. So the answer is we added seven green balls to the bag.