A box contains 10 white balls and 15 red ones. If two balls are drawn in succession, without replacement, find the probability that i) the two balls are red, ii) the second ball is red if the first one is white, iii) the second ball is white if the first one is red.
Now, the original question actually says that we only need to answer two of the three parts of this question. But I’m going to go through how to answer all three parts in this video. Let’s just remind ourselves of the situation first of all. This box contains 10 white and 15 red balls. We’re then told that two balls are drawn in succession, that means one after the other, without replacement. And the key here is that this means the first ball is not put back in the bag before the second ball is chosen. We’re then asked to work out various probabilities, the first of which is that the two balls are both red. So we can use a tree diagram to help with this.
A tree diagram is just a way of representing all of the possible outcomes for two or more events and also their associated probabilities. For the first ball, we have two options for its colour. It could be either white or red. In each case, we also have two options for the colour of the second ball. If the first ball is white, then the second ball can be white or red. And if the first ball was red, the second ball can still be white or red.
We now need to add the probabilities onto each branches. First, we note that if there are 10 white and 15 red balls in this box, then the total number of balls in the box is 10 plus 15, which is 25. This means that the probabilities for the first set of branches on our tree diagram can be represented as fractions with denominators of 25. Now, there are 10 white balls in the box to begin with. So the numerator for the first fraction is 10. And there are 15 red balls in the box to begin with. So the numerator for the second fraction is 15.
So we filled in the probabilities on the first set of branches on our tree diagram. But what about the second? We need to be really careful here because, remember, the balls are chosen without replacement, which means that, at this point, one ball has been removed from the box and not put back in. So there are no longer 25 balls available to be chosen. Instead, there are 24, which means that the probabilities for the second set of branches on our tree diagram will be fractions with denominators of 24. We also need to be very careful when considering the numerators of these fractions.
Let’s look at the first probability, the probability that the second ball is white if the first ball was also white. Well, if the first ball chosen was white, then this means there is one less white ball available in the box to be chosen at this point. So instead of 10 white balls, there are actually nine, which means that the probability that the second ball is white if the first was white is nine over 24. If the first ball was white, then there are still all 15 red balls in the bag, which means that the numerator for the second fraction will be 15. Notice that the probabilities on this pair of branches on the tree diagram will sum to one. And this must be the case for every pair of branches on our tree diagram.
Now let’s consider the probabilities if the first ball removed from the box was actually red. Well, if the ball removed was red, then we still have the same number of white balls as we started off with, that is, 10. So the probability that the second ball would be white if the first was red is 10 over 24. There will now only be 14 red balls in the bag because one red ball has been removed. So the probability that the second ball is red if the first was red is 14 over 24. Again, notice that these probabilities sum to one. Now, we can use our tree diagram to answer the three parts of the question. And it’s actually much more straightforward to answer parts two and three than it is to answer part one.
Part two asked us for the probability that the second ball is red if the first one is white. And we’ve actually already found this probability when we filled in our tree diagram. It is this probability here. Remember, we said that if one ball has already been removed, then there are 24 left. And if that first ball was white, then there are still 15 red balls remaining in the bag. So the probability that the second ball is red if the first was white is 15 over 24. This fraction can be simplified by dividing both the numerator and denominator by three to give five over eight.
We can answer part three of the question in the same way. This time, we’re looking for the probability that the second ball is white if the first one was red. Well, that is this probability here. There are now 24 balls in the bag. And if the first was red, then there will still all 10 white balls remaining. So the probability is 10 over 24. This probability can be simplified by dividing both the numerator and denominator by two to give the simplified fraction five over 12.
So we’ve answered parts two and three of the question. Now, let’s think about part one, which asked us for the probability that both balls are red. That’s the probability here at the end of the bottom branches of our tree diagram. To work this probability out, we need to multiply together the probabilities along the route that we need to take. So we multiply 15 over 25, that’s the probability that the first ball is red, by 14 over 24. That’s the probability that the second ball is red if the first ball is red.
Before we multiply, we can simplify. Within the first fraction, we have 15 over 25. And both the numerator and denominator can be divided by five to give the simplified fraction three-fifths. In the second fraction, both the numerator and denominator can be divided by two to give the simplified fraction seven over 12. We can also cross-cancel a factor of three from the three in the numerator of the first fraction and the 12 in the denominator of the second, giving one over five or one-fifth multiplied by seven over four. Now, we multiply. Multiplying the numerators together, one multiplied by seven gives seven. And multiplying the denominators, five multiplied by four gives 20. So we find that the probability that the two balls are red is seven over 20.
And so we have our answers to the three parts of this problem. Part one is seven over 20. Part two is five-eighths. And part three is five twelfths.