 # 3 Probability Tree Diagrams and Dependent Events

Inside a bag, there are three green balls,
two red balls, and four yellow balls. Two balls are randomly drawn without replacement. Calculate the probability of drawing one red
ball and one yellow ball. So, let’s consider, : what’s What’s the probability
of drawing one red ball and one yellow ball? So, the probability of drawing one red and
one yellow. Well, there are two different ways in which
we could draw a red ball and a yellow ball. We could draw a red ball first, and then on
the second ball we could draw a yellow ball. Or, we could draw a yellow ball first and
then a red ball. So, in other words, this probability is going
to be the sum of drawing red then yellow and drawing yellow then red. And, in order to find out these two probabilities,
let’s draw a probability tree diagram. So, there are actually three probabilities
here. There are three possible outcomes that we
have to consider. So, on the first draw we could get a green
ball or we could get a red ball or we could get a yellow ball. So, because of that we’re going to have three
branches coming out of our probability tree. And here we’re told there are three green
balls, and how many balls we’ve got all up? We’ve got three, two, four, so that’s nine
balls all up. So, the probability of drawing a green ball
on the first draw is going to be 3 over 9. Now, there are two red balls, so the probability
of drawing a red ball on the first draw is going to be 2 over 9. And the probability of drawing a yellow ball,
there are four yellow balls, so that’s gonna be 4 over 9. OK, once we’ve drawn the green ball, there
are still some green balls left in the bag and still some red balls and still some yellow
balls. So, from this green ball, we now have another
three branches coming out. We could draw a green ball, a red ball, or
yellow ball. Now, what’s the probability of drawing a green
ball if we’ve already drawn a green ball in our first draw? Let’s quickly draw a bag and imagine that
we have a bunch of balls here. Imagine that we have three green balls and
we have two red balls and we also have four yellow balls. Well, if we on the first draw have already
drawn out a green ball, on our second draw we’re going to have one fewer green ball than
we had before. Which means that the probability of now drawing
a green ball is going to be—there only two green balls left in this bag and all up there
are only eight balls—so the probability of drawing a green ball considering that these
are drawn without replacement is going to be 2 over 8. What about the probability of drawing a red
ball? Well, there are two red balls here and eight
balls all up, so that’s also going to be a probability of 2 over 8. So, the probability that we draw a yellow
ball, given that we’ve drawn a green ball on our first draw, is going to be one, two,
three, four—there are four yellow balls and eight balls all up, so that’s going to
be 4 over 8. Now, if we are to repeat that process, we
have to repeat that process a couple more times in order to fill out this probability
tree. So, here for red we’ve got three possibilities. We could have green, red, and yellow. So, remember this is the outcome that we’ve
got a red ball on the first draw. So, if we had a red ball on the first draw,
let’s quickly draw our balls again, and have three green and four yellow, so here we’re
imagining the first draw we reach in and we draw a red ball. So, what’s the probability from this bag now
of drawing a green ball? Well, there are three green balls and eight
balls all up, so the probability is 3 over 8. What about the probability of drawing a red
ball? That’s going to be 1 over 8. And a yellow ball? That’s going to be 4 over 8. OK, this is going well. We’ve just got three more branches to go. So, here imagine that we draw a yellow ball
on our first draw. Well, we could draw a green, a red, or a yellow
on a second. Again, if we’ve got our bag, imagine we’ve
still got three green, got two red, and we’ve got a few yellow balls here. We’ve got four yellow. So, imagine that we draw out a yellow ball
on the first draw. So, we’ve got one fewer yellow ball than
the first draw, . so So here what’s the probability of getting a green ball? Well, there are eight balls, and three of
them are green, so this is going to be 3 over 8. Here, the probability of getting a red ball,
that’s going to be 2 over 8. And the probability of getting a yellow ball,
that’s going to be 3 over 8. So, we’re interested in two outcomes, red
yellow and yellow red. So, red yellow, that corresponds to red on
the first ball, yellow on the second ball. So, red yellow is here. So, the probability of getting red yellow. S, so, red on the first, yellow on the second,
is going to be 2 over 9 times 4 over 8. So, the probability associated with first
branch multiplied by the probability associated with the second branch. So, 2 over 9 times 4 over 8. Well, 4 over 8 is the same as a half, so that’s
going to be 2 over 9 times a half. We’ve got a common factor of 2, so this is
going to be 1 over 9 multiplied by 1 over 1. That’s just going to be 1 over 9. So, that’s the probability of drawing a red
on the first draw and a yellow ball on the second draw. What about yellow red? So, it’s yellow on the first draw, red on
the second draw. So, the probability here of yellow red, we’re
going to do something very similar. Going toThen you get 4 over 9 times 2 over
8. Four over 9 times 2 over 8. Two over 8 is a quarter, so it’s going to
be 4 over 9 times a quarter. And here we’ve got a common factor again of
4, so this is going to be 1 over 9. That’s 1 over 1. So, this is going to be 1 over 9 as well. So, here, we’ve got the probability of one
red and one yellow is going to be 1 over 9, the probability of red and yellow. Plus the probability of yellow red, that’s
going to be 1 over 9 again, so that’s going to be 2 over 9.

1. DhamaalHD says:

Thanks, you've helped a lot.

2. Angelo Acanto says:

struggled for tree diagrams, it was great to see how everything made more sense how you drew out the bag and crossed out the red green and yellow ball. Keep up the great work

3. Knewton DTI says:

Hi Mark! Interested in adding transcripts to your videos. Would you please email me at [email protected]? Thanks!

4. Shamekh Ali says:

But it is written that 2 balls r drawn randomly
Wouldnt this solution be if one is drawn randomly