Dom is taking part in a card
tournament. He plays three games. And the probability that he wins
each game is 0.3. Assume each game is
independent. Part a) Complete the tree
As each game is independent, the
result of the first game has no impact on the results of the second game. Likewise, the result of the second
game has no impact on the third game. The probability of winning any game
will always be 0.3.
As there are only two possible
outcomes, winning and losing, if the probability of Dom winning the game is 0.3,
then the probability of him losing the game is one minus 0.3 as the probabilities
must sum to one.
One minus 0.3 is equal to 0.7. Therefore, the probability of Dom
losing any game is 0.7. This means that any branch in our
tree diagram that says win will have a probability of 0.3 and any branch that says
loss will have a probability of 0.7.
If Dom wins the first game, he can
have a win or a loss on the second game. The probabilities on these branches
will be 0.3 and 0.7, respectively. Likewise, if he lost the first
game, he could also win or lose the second game. The probability of a win is 0.3 and
the probability of a loss is 0.7.
We then repeat this process for the
third game. Once again, the probability of
winning any of these games is 0.3 and the probability of losing them is 0.7.
We now have our completed tree
diagram, showing the eight possible outcomes over the three games: win, win, win,
win, win, loss, win, loss, win, and so on all the way down to loss, loss, loss.
The second part of the question
says the following.
To qualify for the next stage of
the tournament, Dom needs to win at least two games. Part b) Calculate the probability
that after the third game, Dom qualifies for the next stage by winning exactly two
We want to work out the probability
that Dom wins exactly two out of his three games. This could happen in three
ways. He could win the first game, win
the second game, and lose the third game: win, win, lose. Or he could win the first game,
lose the second game, and win the third game. Or he could lose the first game and
win the second and third games.
In probability, the word “and”
means multiply. In this case, Dom has to win the
first game and the second game. And then, he has to lose the third
game. Or he has to win the first game and
lose the second game and win the third game. Or finally, he needs to lose the
first game and win the second game and win the third game.
We need to multiply the
probabilities of each event. As the probability of winning is
0.3 and the probability of losing is 0.7, we need to multiply 0.3 by 0.3 by 0.7.
To work out the probability of win,
lose, win, we need to multiply 0.3 by 0.7 by 0.3. Finally, to work out the
probability of lose, win, win, we need to multiply 0.7 by 0.3 by 0.3.
You will notice that all three of
these calculations are the same as multiplication is commutative. It doesn’t matter which order we
multiply the numbers.
As this is a non-calculator paper,
we might wish to write these decimals as fractions. 0.3 is equal to three-tenths or
three over 10 and 0.7 is equal to seven-tenths. We can then solve this calculation
by multiplying the numerators three, three, and seven and multiplying the
denominators 10, 10, and 10.
Three multiplied by three is equal
to nine. And nine multiplied by seven is
equal to 63. 10 multiplied by 10 is equal to
100. And 100 multiplied by 10 equals
This means that the probability of
Dom winning the first game and winning the second game and losing the third game is
63 out of 1000. This is the same as 63 divided by
1000, which is equal to 0.063.
As mentioned previously,
multiplication is commutative. Therefore, the probability of win,
lose, win is also 0.063. Likewise, the probability of lose,
win, win is 0.063.
There are three possible ways that
Dom could win exactly two out of his three games. The word “or“ in probability means
add. Therefore, we need to add these
As the three numbers are identical,
we could also do this by multiplying 0.063 by three. 63 multiplied by three is equal to
189. Therefore, our probability is
We can say that the probability of
Dom qualifying by winning exactly two games is 0.189.