Consider a Stanley Cup playoff series in which the Toronto Maple Leafs hockey team faces the Ottawa Senators. Toronto hosts the first, second, and if needed, fifth and seventh games in this best-of-seven contest. The Leafs have a 65% chance of beating the Senators at home in he first game. After that, they have a 60%chance of a win at home if they won the previous game, but a 70% chance if they are bouncing back from a loss. Similarly, the Leaf's chances of victory in Ottawa are 40%after a win and 45% after a loss.
Construct a tree diagram to illustrate all the possible outcomes of the first three games
A) consider the following events:
A={Leafs lose the first game but go on to win the series in the fifth game}
B={Leafs win the series n the fifth game}
C={Leafs lose the series in the fifth game}
Identify all the outcomes that make up each event, using strings of letters, such as LLSLL. Are any pairs from these three events mutually exclusive?
What is the probability of event A in part a)
What is the chance of the Leafs winning in exactly five games?
Explain how you found your answer to the previous questions?
To construct a tree diagram for the first three games, we can start with the first game and then branch out to the possible outcomes for the subsequent games.
First, we begin with the first game:
- Leafs win (W)
- Leafs lose (L)
Now, based on the given information, we can determine the probabilities for the second game:
- If Leafs won the previous game:
- Leafs win with a 60% chance (W)
- Leafs lose with a 40% chance (L)
- If Leafs lost the previous game:
- Leafs win with a 70% chance (W)
- Leafs lose with a 30% chance (L)
For the third game, we consider the outcomes in Ottawa:
- If Leafs won the previous game:
- Leafs win with a 40% chance (W)
- Leafs lose with a 60% chance (L)
- If Leafs lost the previous game:
- Leafs win with a 45% chance (W)
- Leafs lose with a 55% chance (L)
Combining these possibilities, we can create a tree diagram showing all the possible outcomes of the first three games.
Event A (Leafs lose the first game but win the series in the fifth game) is defined as LLLW.
Event B (Leafs win the series in the fifth game) includes all the outcomes where the Leafs win the series in the fifth game. These can be WLLW, LWLW, and LLWW.
Event C (Leafs lose the series in the fifth game) includes all the outcomes where the Leafs lose the series in the fifth game. These can be LLLWW, LLWLW, and LWLLW.
None of the pairs in events A, B, and C are mutually exclusive, as there are no overlapping outcomes between them.
To find the probability of event A, we need to calculate the probability of the specific outcome LLLW. Based on the given information, the Leafs have a 65% chance of losing the first game (L), followed by a 70% chance of winning the second game (L to W transition), a 45% chance of winning the third game (W in Ottawa), and a 40% chance of winning the fifth game (W to L transition).
The probability of event A is calculated as follows:
P(A) = P(LLLW) = 0.65 * 0.7 * 0.45 * 0.4 = 0.0657 or 6.57%
To determine the chance of the Leafs winning in exactly five games (event B), we need to add up the individual probabilities of the outcomes WLLW, LWLW, and LLWW. Since these outcomes are mutually exclusive, we can sum up the probabilities:
P(B) = P(WLLW) + P(LWLW) + P(LLWW)
= (0.35 * 0.6 * 0.4 * 0.4) + (0.65 * 0.4 * 0.45 * 0.4) + (0.65 * 0.7 * 0.6 * 0.4)
= 0.0674 + 0.0468 + 0.1092
= 0.2234 or 22.34%
To find these probabilities, we used the information provided in the question about the chance of winning after a win/loss, and the probability of winning/losing in Toronto and Ottawa. We applied the given probabilities to the specific outcomes related to the events of interest and calculated the joint probabilities using multiplication.