Can anyone please explain this probability question to me, i have a test tomorow and i dont get this question:
Suppose the odds of the Toronto Maple Leafs winning the Stanley cup are 1:5, while the odds of the Montreal Canadiens winning the Stanley cup are 2:13. What are the odds in favour of either Toronto or Montreal winning the Stanley Cup?
Thanks to anyone who explains it to me
2007-01-14 03:14:30 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
probability = odds/(1+odds)
odds = probability/(1-probability)
odds 1:5 is the same as probability = (1/5)/[1+1/5] = 1/6
odds 2:13 is the same as probability = (2/13)/[1+2/13] = 2/15
Since either Toronto or Montreal winning the Stanley Cup is independent event,
Probability(either Toronto or Montreal winning the Stanley Cup)
= Probability(Toronto winning the Stanley Cup) + Probability(Montreal winning the Stanley Cup)
= 1/6 + 2/15 = 3/10
The odds in favour of either Toronto or Montreal winning the Stanley Cup = 0.3/(1-0.3) = 3:7
2007-01-14 04:56:14 · answer #1 · answered by sahsjing 7 · 1⤊ 1⤋
First beware of the difference between odds and probability.
The odds being 1:5 means the probability is 1/6 and for 2:13, it is 2/15. Since those two events are mutually exclusive, you can sum the probabilities which gives you 5/30 + 4 / 30 that is 3/10. So the odds are 3:7.
2007-01-14 03:38:28 · answer #2 · answered by gianlino 7 · 3⤊ 0⤋
The probability is the sum of the 2 probabilities which is:
1/5+2/13 = 23/65.
This is like saying the probability of flipping a coin "heads" is 1/2 and the probability of flipping a coin tails is 1/2. Oviously the chances that the coin will come up either heds or talis is 1/2 +1/2 = 1.
2007-01-14 03:27:57 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋
Easy.
A has a 1 in 5 chance.
B has a 2 in 13 chance.
So, A has a 13 in 65 chance and B has a 10 in 65 chance.
So, the chance of A or B winning is 23 in 65.
2007-01-14 03:21:04 · answer #4 · answered by RS 4 · 0⤊ 1⤋
since the probabilities of those two winning the stanley cup are "independent" of each other, you can simply add the odds... T.M.L : 1/5 M.C: 2/13 so either one of them: 1/5 + 2/13 = 23/65
2007-01-14 03:21:10 · answer #5 · answered by chirkhin 2 · 0⤊ 1⤋
The probability of TMF winning is 1/6 and that of MC winning is 2/15. Both of these events and disjoint,i.e both teams can't win.
Therefore P(A \intersection B) = 0
Therfore
P(A \union B) = P(A) + P(B) - P(A \intersection B)
= 1/6 + 2/15.
2007-01-14 03:22:38 · answer #6 · answered by Vijay Krishnan 1 · 1⤊ 0⤋
first make it equal, 2: 10 and 2:13, now toronto has better chances of winning
2007-01-14 03:22:13 · answer #7 · answered by Blue Rain 6 · 0⤊ 1⤋
Yes. I'm sure someone can solve this question.
2007-01-14 03:21:08 · answer #8 · answered by BigBrain 2 · 0⤊ 1⤋
yes the answer is 1.34
2007-01-14 03:23:54 · answer #9 · answered by Bee.x 1 · 0⤊ 1⤋