hi this is a probability question that we're going to solve using a tree diagram but let's just read the question first jenny has a bag with seven blue sweets and three red sweets in it she picks a sweet at random from the bag replaces it and then picks again at random draw a tree diagram to represent this situation and use it to calculate the probabilities that she picks a two red sweets be no red sweets see at least one blue sweet d1 sweet of each color well let's start off by drawing our tree diagram would look at the first pick first there are two possibilities that she has so we do draw a little double branch like that one represents her picking a blue sweet the other one a red sweet I'm using B and R to mean blue and red now one of the probabilities of that well because there are ten in all and seven of them are blue probability of a blue is 7/10 and in the same reasoning probability of red is 3/10 now what about the second tick well she has replaced whatever she picked first time so we got exactly the same situation in the bag as we started with so she picks a blue first then she still got the same probabilities of clicking picking a blue or a red second time and if she picks red first it's exactly the same again so those are the probabilities for her tree diagram just got two more columns to do first of all the outcome and secondly the probability calculations well the outcome of this first branch is blue and then blue so I've labeled it be B and the probability is 7/10 times 7/10 because it's blue and blue if you remember when you're anding two things you multiply them together that gives us 49 over 100 the next branch is first blue then red B R for short 7/10 times 3 tenths 2100 this branch here is first red then blue RB 3/10 time 7/10 2121 hundredths and the fourth one is two Reds first red second red three tenths times three tenths nine hundredths now it's always a good idea to do a check here because all of these four probabilities are add up to one because they're the only four possibilities in the situation and they do add up to 100 over 100 which is one so we've done our check now let's look back at the question we need to work out the probability that she picks two red sweets well we've actually done that one it's this number here nine hundredths there we are the answer to that first part second part no red sweets will the only thing which has no red sweets in it is the blue blue combination and we've already calculated that that's 49 hundredths and says that now we're getting it more complicated we need to find at least one blue so this has got a blue this is going to blue this is going to blue there are three possibilities and we get these by adding because it's this or this or this and if you remember or in probability means add them together so when we add those three together we'll get our answer it's blue red red blue or blue blue adding together we get that which totals 291 over 100 so that's our answer but there is another slightly quicker way of doing it let's do that as well now the only one which doesn't have a blue in it is this one and singles we got all our possibilities adding up to one if we take this possibility away from one and then everything else must have a blue in it at least one blue so our working is going to be 1 minus the possibility of two Reds which is one minus nine hundredths which is ninety one hundredths same answer no surprise it's the same answer good it's just a different way of getting to it now what about the last one one sweet of each color well that's just the two middle branches first blue then red or first red then blue so it's this or this so we add those two probabilities together blue red plus red blue 21 hundreds plus twenty one hundred forty two hundredths which we may choose to simplify to twenty one fiftieth just finishing off let's look at our tree diagram again remember that as you go along a branch you're doing one thing and another thing so you multiply the probabilities each branch has multiplied probabilities but when you're doing answering questions at the end you might have to add these because you're doing this or this this or this etc etc so remember multiply along the branches but add the results of those multiplications to get your answers