The probability that our school will host soccer and rugby tournament this year is 0.8. If we host the the probability of winning soccer is 0.7. if we do not host the probability of winning soccer is 0.4. if we win soccer the probability of winning rugby is 0.8, otherwise if we lose soccer the probability of winning rugby is 0.3.
a)draw a tree diagram to represent this information.
b)use the tree diagram to find:-
i) the probability that we will lose both games.
ii) the probability that we will win only one game.
iii) the probability that we will host and lose both games.
first branch:
host (h) = .8
not host (nh) = .2
second level:
h, win soccer(ws) = .7
h, lose soccer(ls) = .3
nh, ws = .4
nh, ls = .6
third level:
h, ws, win rugby(wr) = .8
h, ws, lose rugby(lr) = .2
h, ls, wr = .3
h, ls, lr = .7
nh, ws, wr = .8
nh, ws, lr = .2
nh, ls, wr = .3
nh, ls, lr = .7
lose both games: h,ls, lr + nh, ls, lr
= (.8)(.3)(.7) + (.2)(.6)(.7)
= .168 + .084
= .252
follow the paths in your tree to answer the other two questions
(btw, I tested my tree, the prob of all 8 end points add up to 1)
It is hard do it for me
It is hard do it for me
a) To draw a tree diagram, we start by representing the initial event: hosting the tournament. Then, we branch out to represent the two possible outcomes - hosting or not hosting.
```
Hosting
/ \
Soccer No Soccer
/ \ / \
Win Lose Win Lose
| | | |
Rugby Rugby Rugby Rugby
```
b) i) The probability of losing both games is the multiplication of the probabilities along the path of losing both soccer and rugby. From the diagram, we can see that both soccer and rugby are lost if we do not host the tournament.
```
P(Lose both games) = P(No Soccer) * P(Lose Rugby | No Soccer)
= (1 - P(Soccer)) * P(Lose Rugby | No Soccer)
= (1 - 0.8) * 0.3
= 0.2 * 0.3
= 0.06
```
Therefore, the probability of losing both games is 0.06 or 6%.
ii) The probability of winning only one game is the sum of the probabilities along the paths where we win one game and lose the other.
```
P(Win only one game) = P(Soccer) * P(Lose Rugby | Soccer) + P(Lose Soccer | Soccer) * P(Rugby | Lose Soccer)
= 0.8 * 0.3 + 0.2 * 0.8
= 0.24 + 0.16
= 0.4
```
Therefore, the probability of winning only one game is 0.4 or 40%.
iii) The probability of hosting and losing both games is the multiplication of the probabilities along the path of hosting, losing soccer, and losing rugby.
```
P(Host and Lose both games) = P(Soccer) * P(Lose Rugby | Soccer)
= 0.8 * 0.3
= 0.24
```
Therefore, the probability of hosting and losing both games is 0.24 or 24%.