Ok, so in this example, following on from

the first example, we’re gonna look at events where something is taken out and isn’t replaced.

So for example, this could be taking counters out of a bag and not replacing the counters.

Taking cards out of a deck of cards and not replacing them. There’s lots of examples where

this exists. And, in this sense, the two events are dependent on each other in the way that

the second occurrence is dependent on what happens in the first event. So, if we look

at this example, we know that there are 7 red and 5 blue counters in a bag. Will takes

out 1 counter and does NOT replace it. He then takes out a second counter. When we’re

putting our probabilities onto our probability tree, we do have to take into consideration

that Will does NOT replace the counter. So let’s start with the first outcome of the

event. We know that there are 7 red counters and 5 blue counters. So in total there are

12 counters in the bag. For examples like this, I usually like to present my probabilities

as fractions as it gives more of a visual representation of what’s going on. So red

counters, the probability of picking a red is 7/12. And then, if we do 12/12 minus 7/12,

we get 5/12, so we know this is the probability of achieving a blue counter. So now, when

we move on to what happens next, when Will picks out another counter, we know that he’s

picked out either a red or a blue counter, we don’t know which colour counter he’s picked

out. We know that he’s picked out one of these counters. So overall, the amount of counters

has been reduced now to 11, because one counter has been taken out and isn’t replaced. So

the total number of counters now is 11. So the bottom of our fraction for each of these

occurrences is now gonna be 11. Let’s just write the denominator of the fraction onto

each of these branches. So now, we have to look at what the colour of the counter was

before in the first occurrence. So we know if we’re travelling along this branch, it’s

the first counter was red. So if 1 red counter has been taken out of the bag, we know now

that the amount of red counters is going to reduce by 1. So previously there were 7 counters

and now there are only 6 counters that are red. So now we can put our numerator onto

the fraction as we know there are now only 6 red counters because a red counter has been

taken out of the bag. Let’s do this again. So a red counter has been taken out of the

bag and now we’re gonna pick a blue counter. No blue counters have been taken out of the

bag so the same number of blue counters still applies. So we can just put a 5 onto that

fraction. This time, we’re travelling along the blue branch, so we know a blue counter

has been taken out of the bag and not replaced. So we can apply the same logic. No red counters

have been taken out so the same number of red counters remain in the bag. So we get

7/11. And then, we know a blue counter has already been taken out of the bag and now

another blue counter is gonna be taken out of the bag. So this number reduces by 1, so

we have 4/11. Now, to find the probability of all possible outcomes, we do the same thing

we did in the first example and we multiply the two branches together using the multiplication

rule. So when we multiply fractions, we know that we simply times the tops of the fractions

and the bottoms of the fractions together. All of the possible denominators are going

to be 12 times 11 and the answer to that is 132 so we can write in our denominators of

the fractions. And therefore, this gives us an indication that all of the numerators of

our fractions need to add together to equal 132. So our fraction will be 132 over 132

which equals 1 which therefore is a certain outcome. So now we know that 7 times 6 is

42, so we can write that at the top of our fraction. 7 times 5 is 35. 5 times 7 again

is 35. And 5 times 4 is 20. So when we add all these fractions together, because we can

because it’s a common denominator, 35 add 35 is 70, add 20 is 90, add 42 is 132. So

we get 132 over 132 which equals 1.