We can also construct a probability tree diagram to help us solve some probability problems.
A probability tree diagram shows all the possible events. The first event is represented by a dot. From the dot, branches are drawn to represent all possible outcomes of the event. The probability of each outcome is written on its branch
Example:
A bag contains 3 black balls and 5 white balls. Paul picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.
a) Construct a probability tree of the problem.
b) Calculate the probability that Paul picks:
i) two black balls
ii) a black ball in his second draw
Solution:
a)
Check that the probabilities in the last column add up to 1.
b) i) To find the probability of getting two black balls, first locate the B branch and then follow the second B branch. Since these are independent events we can multiply the probability of each branch.
ii) There are two outcomes where the second ball can be black.
Either (B, B) or (W, B)
From the probability tree diagram, we get:
P(second ball black)
= P(B, B) or P(W, B)
= P(B, B) + P(W, B)
Example:
Bag A contains 10 marbles of which 2 are red and 8 are black. Bag B contains 12 marbles of which 4 are red and 8 are black. A ball is drawn at random from each bag.
a) Draw a probability tree diagram to show all the outcomes the experiment.
b) Find the probability that:
(i) both are red.
(ii) both are black.
(iii) one black and one red.
(iv) at least one red.
Solution:
a) A probability tree diagram that shows all the outcomes of the experiment.
b) The probability that:
(i) both are red.
P(R, R) =
(ii) both are black.
P(B, B) =
(iii) one black and one red.
P(R, B) or P(B, R) =
(iv) at least one red.
1- P(B, B) =
Example:
A box contains 4 red and 2 blue chips. A chip is drawn at random and then replaced. A second chip is then drawn at random.
a)
Show all the possible outcomes using a probability tree diagram.
b)
Calculate the probability of getting:
(i)
at least one blue.
(ii)
one red and one blue.
(iii)
two of the same colour.
Solution:
a) A probability tree diagram to show all the possible outcomes.
b) The probability of getting:
(i) at least one blue.
P(R, B) or P(B, R) or P(B, B) =
(ii) one red and one blue.
P(R, B) or P(B, R) =
(iii) two of the same colour.
P(R, R) or P(B, B) =
The following video gives more examples of solving probability problems using probability tree diagrams.
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