Probability tree diagrams

1. In this question we must consider two events. The first event R is the event ‘it rains’, B is the event ‘there is a big turnout’ We use branches to represent these events

   Given that it does rain, in the question we’re told the probability of a big turnout is 0.4 written as;

   P(B|R) = 0.4

   …and the probability that it rains is 0.75 written as;

   P(R) = 0.75

   Therefore we can put 0.75 and 0.4 on the first pair of branches respectively.

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   There must be tw branches at each point; one showing the probability of the outcome being true and the other showing the probability of the outcome not being true. We know that in general;

   P(A) = 1 – P(A’)

   We must put a branch below the probability of B

   P(A) = 1 – P(A’)

   We must put a branch below the probability of B.

   P(B’) = 1 – 0.4 = 0.6

   Updating the tree diagram gives;

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   We have to add the outcome that it does not rain and there is a big turnout. We know the probability it does not rain is 0.25 and the probability there is a big turnout is 0.9

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   The multiplication rule can also be used to work out the probabiities.

2. We must work out the probability that there is a big turn out and it rains. To find this probability we multiply the top branches. The probability that there is a big turn out and it rains is written as

   P(BnR) = P(B|R)PR)
   P(BnR) = 0.4 × 0.75 = 0.3

3. Now we must find the probability there is a big turnout. There are two probabilities; It can either rain and there is a big turnout or it does not rain and there is a big turn out.

   We can use the previous answer for ‘there is a big turnout and it rains. So the answer for this becomes;

   P(B) = 0.3 + P(A|R’)P(R’)

   0.3 + 0.25 × 0.9

   = 0.525

   Notice: When you move along the branches of a tree diagram you multiply the probabilities and when you move between the branches you add the probabilities.

Using formulae and venn diagrams

1. P(AnB) = P(A|B) × P(B)

   = 0.1 × 0.3

2. P(AnB’) = P(A|B’) × P(B’)

   = 0.6 × 0.7

   = 0.42

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   P(A) = 0.42 + 0.03

   = 0.45

4. P(B|A) = P(BnA)/P(A)

   = 0.03/0.45 = 0.06

5. P(B|A’) = P(BnA’)/P(A’)

   = 0.27/0.55 = 0.490

Using a tree diagram

We can use a tree diagram to answer the question. Lets draw a tree diagram first.

The first branches are for P(B); P(B)=0.3 and P(B’)=0.7 thus;

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The second branch is the conditional probability P(A|B)=0.1 and the corresponding branch is the P(A’|B) which is 1-P(B|A)=1-0.1 = 0.9

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We do the same for the rest of the branches; The P(A’|B’)=0.6 and therefore P(A’|B’)=1-0.6=0.4. The tree diagram becomes;

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We can now use the branches to work out the questions;

1. P(AnB) = 0.3 × 0.1 = 0.3

   Notice when we move along the branches we multiply.

2. Again we multiply along the appropriate branches;

   P(AnB’) = P(B’nA) = 0.7 × 0.6

   = 0.42

3. We can simply add the previous two exnswers to answer this question;

   P(A) = P(AnB) + P(AnB’) = 0.45