Game of Ring Toss

Let be A = "no bottle having more than one ring", i.e. each child throws a ring on a different bottle. $P(A) = 1 \times \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} = \frac{63}{125}$ .

Let's have then B = "at least one bottle having two or more rings". A and B are complementary events, and as such, $P(B) = 1 - P(A)$ Therefore, $\frac{a}{b} = 1 - \frac{63}{125} = \frac{62}{125}$ . Thus, $a + b = 62 + 125 = 187$ .

For this question, we approach it from the other way round, by finding the probability that there does not exist a bottle with two or more rings, e.g) only 1 ring.

With that, after the first ring is being thrown, the probability of the 2nd ring being NOT on the same bottle as the first person is $\frac{9}{10}$

Similarly, after the second ring is being thrown, the probability of the 3rd ring being NOT on the same bottle as the first AND 2nd person is $\frac{8}{10}$ . This applies for the 4th person, with the probability being $\frac{7}{10}$

Thus, the probability that there does NOT exist a bottle with two or more rings, e.g) only 1 ring, is $\frac{9}{10}$ x $\frac{8}{10}$ x $\frac{7}{10}$ $=\frac{63}{125}$

Hence, the probability that there DOES EXIST a bottle with two or more rings, e.g) only 1 ring, is $1- \frac{63}{125}$ $=\frac{62}{125}$

Thus, the value of $a+b = 62+125 = \boxed{187}$

These four children have an equal shot of picking a bottle. So the probability of picking a bottle for each student is 1/10. To find the total possible number of combinations you do 10^4 This shows the total number of combinations of picking 10 bottles from 4 students. This gives us 10000 total combinations. Now we have to find the complement of our question which is the total number of permutations where each student picks a different bottle. This can be represented by 10P4. This comes out to be 5040. This is the opposite of what we are looking for. So we do 10000-5040 = 4960. 4960 is the number of possibilities students picking bottles with two or more rings on them. Now the fraction comes out to be 4960/10000 which can be further simplified to 62/125 which is in the form a/b -- a = 62 and b = 125-- and now can be added. 62 + 125 = 187. The answer is 187 .

The required probability can be obtained by subtracting from 1, the probability of selecting 4 bottles individually.

Total number of ways each of the 4 players can choose 1 of the 10 bottles = $10^{4}$

Total number of ways 4 different bottles can be chosen = $\frac{10!}{(10-4)!}$

P(at least one bottle with two or more rings) = $1-\frac{\frac{10!}{(10-4)!}}{10^{4}} = \frac{62}{125}$

Thus, a+b = 187

We want to find the probability that there is at least one bottle with 2 or more rings on it. This is precisely the complement of the probability that there were no bottles with more than 1 ring on them. The probability that there were no "repeats" was simply $\frac{9}{10}\cdot\frac{8}{10}\cdot\frac{7}{10}$ . Therefore, we consider $1-\frac{9}{10}\cdot\frac{8}{10}\cdot\frac{7}{10}=\frac{496}{1000}=\frac{62}{125}$ . Thus, we get $62+125=\boxed{187}$ .

We first look at the total number of ways that the children can hit the ten bottles. The first child can throw onto 10 bottles the next can also throw onto 10 bottles and so can the other two giving the total number of possibilities $10^4=10000$ .

We now look at the ways in which the children can throw onto bottles so that no bottle has more than one ring in the end. The first child can throw in 10 ways, the next in 9, then 8 and finally 7 (Here I thought at first that the order of children didn't matter so I got a different answer at first but then I tried this, so maybe the problem is equivocal or I am missing something). The number of ways this can happen is then $10 \times 9 \times 8 \times 7 = 5040$ .

The number of ways the children can then throw so there exists a bottle with two or more rings on it is $10000-5040=4960$ . The probability of this happening is then $\frac{4960}{10000}=\frac{62}{125}$ , hence the answer is $62+125=187$ .

The question asks for the probability that there is at least one bottle with more than ring. Note that the opposite of this case would be that there are no bottles with more than one ring. Together, these cases make up all possible outcomes; these are called complementary. We choose to find the probability of the complementary case because it is much simpler to find. This method of finding the complement is called Complementary Counting .

We can call the probability of the original case $P_{1}$ and the complementary probability $P_{2}$ . Since they are complementary, $P_{1}+P_{2}=1$ .

To find $P_{2}$ : To have no bottles with duplicate rings, each player must choose a different bottle. To find the number of ways this is possible, we consider the choices each player has. Player 1 can choose any of the 10 bottles. Player 2 can choose any but the one Player 1 choose; hence, Player 2 has 9 choices. Player 3 can't choose any of the previous bottles, so 8 choices, and with the same reasoning Player 4 has 7 choices.

We multiply the number of choices to find the number of possible favorable outcomes, so $10 \cdot 9 \cdot 8 \cdot 7 = 5040$ . If we were to disregard the rules for the favorable outcomes, each player would have 10 choices. Hence the number of total outcomes is $10 \cdot 10 \cdot 10 \cdot 10 = 10000$ . To find probability, we take $\frac{favorable outcomes}{total outcomes}$ . Using this, $P_{2}=\frac{5040}{10000}=\frac{63}{125}$ .

Using the rule of complementary probabilities outlined at the beginning: $P_{1}+\frac{63}{125}=1$ . And so the desired probability of having a bottle with at least two rings is $P_{1}=\frac{62}{125}$ .

Step 1:

Every children has 10 options to through their rings. There are four children so total number of permutation is

10^{4}.

step 2:

We will now calculate the number of permutation in which no bottle will have more than 1 ring.

so the first children has * 10 options. *

The second children has 9 options because 1 bottle already has a ring.

the third and fourth children has 8 and 7 options.

So,total number of permutation is * 10 * 9 * 8 * 7 or 5040. *

So the number of permutations in which at least a bottle contains * at least 2 rings is 10^{4}-5040. *

So, the probability of given circumstance is * [10^{4}-5040]/10^{4} or 62/125 *.

so the answer is 62+125=187.

total possibilities = $10*10*10*10$

possibilities that there are no 2 bottles with two rings = $10P4$

therefore probability that there exits bottle with two rings = $1- \frac {10P4 }{10^4}$ =[farc{62}{125}]

therefor $a + b= 62 + 125=\boxed{187}$

the total number of events = 10^4

the number of events that every player play in a distinct bottle=10!/4!=5040

so the number of events that there exists a bottle with two or more rings = 10000-5040=4960

its probability=4960/10000=0.496=62/125

answer a+b=62+125=187

Let's assign each bottle a number from 0 to 9. A ring landing on one of these numbered bottles will be in our sequence. We need to find the number of 4 digit numbers with no repeating digits, which is 10 9 8*7 or 5040. Take that from 10,000, the total number of four digit combinations. So, 10,000-5,040=4,960. We now have a fraction 4960/10000. The HCF of both sides is 80, giving us 62/125, which cannot be simplified further. So our answer is 62+125, which is 187.

$P($ at least one bottle has two or more rings )

$= 1-P($ no bottle have more than one ring $)$

$= 1 - \frac{10}{10} \frac{9}{10} \frac{8}{10} \frac{7}{10}$

$= \frac{62}{125}$

$62 + 125 = \boxed{187}$