7

In a race, the probabilities of A and B winning the race are $\frac{1}{3}$ and $\frac{1}{6}$ respectively. Find the probability of neither of them winning the race.

I solved the question in the following manner-

Since A and B are running in a race, probability of neither of them winning is $$1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}$$

However, all the solution books that I refer to are solving it in the following manner

$$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)=\frac{5}{9}$$

Now this does not make any sense to me since the events of A winning and B winning are not independent. I was pretty confident of my answer but even the official answer key of the test has given $\frac{5}{9}$ as the answer.

Where exactly am I wrong? How can the winning of A and B be independent of each other since given that one does not win, the winning chances of the other increases?

lulu
  • 70,402
Anurag Saha
  • 679
  • 5
  • 16
  • +1. I vote for your solution. The solution of the book seems wrong to me. – the_candyman Mar 02 '20 at 19:39
  • 2
    You have found the same problem in multiple "solution books"? I'd say your interpretation is correct. There is only one race here and only one winner is permitted. – lulu Mar 02 '20 at 19:39
  • I must also add, that my calculations are based on the fact that a tie does not occur. I believe the events will be independent if ties are also taken into consideration(ie A and B can win together.) – Anurag Saha Mar 02 '20 at 19:39
  • they can't both win... – Saketh Malyala Mar 02 '20 at 19:41
  • 2
    Ties aren't the issue. To defend the second approach you need to interpret the problem as speaking of two separate and independent races, with $A$ in one and $B$ in the other. If you have quoted the problem accurately, I'd say that was an eccentric reading. – lulu Mar 02 '20 at 19:41
  • @lulu I believe running two races at the same time has the same effect as considering ties. Since A and B can win together, P(A wins) and P(B wins) become independent of each other. That is to say, even if B wins, probability of A winning remains same. (Also, the question is quoted properly.) – Anurag Saha Mar 02 '20 at 19:46
  • Separating the races ensures independence. If this is a single race, we need information on the independence. If the race is required to produce a single winner, that settles the point..."$A$ winning" and "$B$ winning " are mutually exclusive (as in your interpretation). I think assuming independence within a single race is a very strange assumption, and it would need to be spelled out. – lulu Mar 02 '20 at 19:48
  • "Separating the races ensures independence" Not necessarily. Independence and mutual exclusivity in real world problems might occasionally be assumed to make the problems easier, but are in several instances invalid assumptions. Take as extreme example both racers come from the same impoverished family who can afford to only feed one of them a day. If $A$ was the racer who was fed, $A$ goes on to have a more successful race while $B$ has less success, while the opposite occurs if $B$ was the one who was fed. This is regardless of the fact that they participated in separate races. – JMoravitz Mar 02 '20 at 19:52
  • If two events are mutually exclusive and their probabilities are nonzero, then they can't be independent. – Robert Israel Mar 02 '20 at 19:53
  • @JMoravitz True, but I do think that in a problem involving separate races, independence is the most natural assumption. And that in a problem involving a single race, mutual exclusivity is the most natural assumption. – lulu Mar 02 '20 at 19:58
  • @JMoravitz, if the question is rephrased by adding an extra line-"consider ties are also possible and more than one participants can win at the same time" then would $\frac{5}{9}$ be a valid answer? In this case, the race becomes more like an archery contest, where the participant's chances of success(hitting the target) is independent of others and more than 2 participants can hit the target simultaneously. – Anurag Saha Mar 02 '20 at 20:11
  • If the question were rephrased and included an extra line "consider ties are also possible and more than one participant can win at the same time ---and $A$ and $B$'s chances of winning are independent of one another---" the explicit mention of independence is what would make the problem unambiguous. Alternatively the explicit mention of mutual exclusivity would make the problem unambiguous. Without either, either assumption is possible to make but would require an assumption which may turn out to be incorrect and the answer you get depends on which assumption you made. – JMoravitz Mar 02 '20 at 20:14
  • If $A$ and $B$ are competing in two separate races whose outcomes are independent of each other, then the books' answers are right. If they're both in the same race and only one can win, then your solution is right. This seems like a good example for illustrating what "independent" and "mutually exlcusive" mean when teaching beginning probability. – Michael Hardy Mar 02 '20 at 20:14

2 Answers2

5

As has been discussed in the comments, your interpretation of the question makes much more sense than one that might lead to these events being viewed as independent. Even if this were a race that can be won by more than one participant, it would be a strange race indeed if the chance of winning it were independent of someone else’s chance of winning it.

I did find the wrong answer that you found online, but your correct answer is also online here.

joriki
  • 238,052
  • 1
    My thoughts exactly. Also, thank you for finding a source that gives the right answer. Now I have something to show my teacher. – Anurag Saha Mar 02 '20 at 19:58
-1

Important

Since $A$ and $B$ are not opponents (this race is not zero-sum),then the probabilities of winnings do not sum up to $1$ and it is possible for both $A$ and $B$ to win the race simultaneously.

Your mistake in summing up the probabilities is that you have considered $$1-[P(A)+P(B)]$$instead of $$P(A'\cap B')=1-[P(A\cup B)]$$i.e. you have assumed that it is impossible for both $A$ and $B$ to win the race.

Remark

The case of not winning one of them, does not increase the chance for the other one's winning. You can equivalently assume that neither of them is aware of the state of the other one and the race is simply a math competition during limited time.

Community
  • 1
Mostafa Ayaz
  • 31,924
  • The probability of rolling a $1$ and the probability of rolling a multiple of $3$ on a standard fair six-sided die don't add up to $1$ either... yet it is clearly impossible for us to simultaneously roll a $1$ which is a multiple of $3$ on a fair six-sided die. – JMoravitz Mar 02 '20 at 19:53
  • So basically ties must also be considered. Since in the 2nd interpretation, P(A wins)+P(B wins)+P(neither A nor B wins) adds to less than 1, the remaining probability must be of P(A and B together wins). Adding all 4 probability terms gives us 1. – Anurag Saha Mar 02 '20 at 19:54
  • @JMoravitz, you are correct. My intuition was that $A$ and $B$ are not opponents and hence $A\cap B\ne \emptyset$. A correction is applied on the answer. – Mostafa Ayaz Mar 02 '20 at 19:57