John's Billard Balls

It will be easier to ask yourself what are the chances of not getting two of the same numbered ball. Our first ball would not give us two of the same numbered ball (Since only one ball was taken out), so we can put this chance as 1. The second ball would then have a $\frac {5}{6}$ chance of not getting the same numbered ball. The third ball would only have $\frac {4}{6}$ chance of not getting the same numbered ball as picked before, as we would have two different numbers picked out of the six. The fourth ball would have $\frac {3}{6}$ chance of not getting a ball that was not picked before, as there would be three balls that we have already chosen out of the six. So, the chances of not getting the same number at least twice would be: $1 \times \frac {5}{6} \times \frac {4}{6} \times \frac {3}{6} = \frac {60}{216} = \frac {5}{18}$ Thus the chances of getting the same number at least twice would be $\frac {13}{18}$ , where $13 + 18 = 31.$ [LaTeX edits]

The number of ways getting at least 2 equal numbers = The total number of arrangement - The number of ways getting all 4 different numbers

The total number of ways one can get is 6^4 = 1296 The number of ways for one getting all different numbers = 6P4 = 6 x 5 x 4 x 3 = 360 The number of ways getting at least 2 equal numbers = 1296 - 360 = 936

\frac {936}{1296} = \frac {13}{18}

13 + 18 = 31

We can count the probability that John will not draw four balls with at least two numbers that are equal, and subtract this from one to determine the probability that he does . So, we have to find the probability that he draws four different balls. John can draw any ball first, but afterwards cannot match with any previous ones. So the probability is equal to

$(1)(\frac{6-1}{6})(\frac {6-2}{6})(\frac {6-3}{6})=(1)(\frac{5}{6})(\frac {2}{3})(\frac {1}{2})=\frac{5}{18}$

$5+18=23$

The probability that at least two numbers are equal is 1 minus the probability that all the numbers are different. So let's start by finding the probability that all the numbers are different:

We have 6 choices for the first number and one less choice for each draw thereafter because we want the numbers to be different. Thus, our successes total $6 \cdot 5 \cdot 4 \cdot 3 = 360$ . The possible number of draws is $6 \cdot 6 \cdot 6 \cdot 6 = 1296$ because the drawing is done with replacement (meaning each ball is replaced in the pool for drawing). So the probability that we draw all different balls is $\frac {360}{1296} = 5/18$ .

Proceeding with the back door method to obtain the probability that at least two numbers are equal: $1 - \frac{5}{18} = \frac{13}{18}$ .

Summing the numerator and denominator gives 13 + 18 = 31.

You are asked to find the probability that at least two of the numbers on the billiard balls are equal. This probability is just 1 - (the probability that none of the numbers on the billiard balls are equal). This probability is just (6/6) (5/6) (4/6)*(3/6). 6/6 comes from the fact that your first number can be any one of the 6, so 6/6. Then the rest of the numbers come from the fact that you can choose any of the other numbers except the one you just choose. So, 6 choices for the first ball, 5 for the second, and so on, giving you a probability of 5/18. 1 - 5/18 = 13/18. Therefore a + b = 13 + 18 = 31

The complement case of 'at least 2 of the numbers are equal' is 'no two of the numbers are equal.' Of the 6^4 = 1296 possible ways to pick 4 numbers, there are 6 5 4*3 = 360 ways in which the four numbers picked are distinct, hence, the probability that no two of the numbers are equal is 360/1296 = 5/18. Since the two given complementary situations are mutually exclusive, their probabilities add up to 1. Therefore, the probability that at least 2 of the numbers are equal is 1 - 5/18 = 13/18, with a+b=31.

The required condition is atleast 2 of the 4 balls should be of same number. If we were able to calculate the probability that all the balls are of different number and then subtract this result from 1, we will get the probability that at least 2 balls are of same number. Now, for no ball to be of same number, probability is P(6,4)/6^4 = (6 5 4*3)/6^4 = 5/18

1- 5/18 =13/18 so a=13 and b=18 and hence a+b=31

As there are 6 numbers, probability of any number to be chosen in 1/6 We proceed via 4 cases :- 1. Out of 4 balls taken out, two balls are of same number whereas other two are different. ( say event 1)

here, probability of of the event 1 is :- 1/6 * 1/6 * 5/6 * 5/6 = 25/1296

1. Out of 4 ball taken out, three balls are of same number whereas the fourth ball is different ( say event 2)

here, probability of event 2 is : 1/6 * 1/6 * 1/6 * 5/6 = 5/1296

1. Out of four ball taken out, all four balls are of same number. ( say event 3)

here, probability of event 3 is :- 1/6 * 1/6 * 1/6 * 1/6 = 1/1296

1. Out of the four balls taken out, two balls are of same number whereas other two balls are also of same number but this number is different from the previous balls' number. ( say event 4)

here, the probability of event 4 is :-

1/6 * 1/6 * 1/6 * 1/6 = 1/1296

therefore, total probability that out of four balls taken out ATLEAST two are of same number is { 1/1296 + 25/1296 + 5/1296 + 1/1296 } * 6

= 32/1296 * 6 = 32/216 = 4/27 = a/b ( according to the question) now a+b = 4 + 27 = 31

"NOTE - we multiplied by 6 because there are six numbers and all numbers can be proceeded in the same way as above which leads to same probability for each number."

Golden rule : Probability = (Desired Possibility)/(Total Possibilities)

Since John draws the ball four times, the total possibilities in any case would be 6x6x6x6 i.e. 6^4 because each and every time he draws the ball, it can be any of the six balls.

Now in order to find "At least 2 balls equal", it would be easy to find the negation of it first (i.e. "None of 2 are equal") and then subtract the probability from 1.

So p' is probability for "None of 2 balls are equal" And p is probability for "At least 2 balls are equal"

Thus p = 1 - p'

Now Required possibility for p' is 6x5x4x3 since none of the ball should be repeated

p' = (6x5x4x3)/(6x6x6x6) = 5/18

Thus p = 1 - p' = 1 - 5/18 = 13/18

Comparing with the question i.e. p = a/b We get a=13 and b=18 Hence a+b = 13+18 = 31 is the answer ['x' indicates multiplication]

Since the opposite of the event that "at least 2 of the numbers are equal" is "all the numbers picked are different" which is easier to compute, we calculate the way the latter event can be done instead.

The number of ways we can pick 4 distinct numbers from 6 distinct numbers (with regard of sequence) is $6\times5\times4\times3=360$ .

The number of ways we can pick 4 numbers from 6 distinct numbers (with regard of sequence) is $6\times6\times6\times6=1296$ .

The probability that all numbers are different is $\frac{360}{1296}=\frac{5}{18}$ .

Therefore, the probability that at least 2 of the numbers are equal is $1-\frac{5}{18}=\frac{13}{18}$ and the answer is $13+18=31$

Rather than directly calculating the probability of at least two of the numbers drawn out being equal, we can calculate the probability of its complementary event happening. In a sample space $S$ , the complementary event of an event $E$ (denoted by $E'$ ) is the set of possible outcomes in $S$ that are not in $E$ . In other words, the complementary event of $E$ is the set of events of $E$ not occurring.

$E$ and $E'$ are related by $1 - P(E) = P(E')$ . (Recall: $P(E) = \frac {n(E)}{n(S)}$ , where $n(S)$ is the total number of possible outcomes in sample space $S$ , and $n(E)$ is the number of outcomes in $S$ for $E$ to occur.)

In this scenario, let the event of at least two of the numbers drawn out being equal be $X$ . $X'$ would then be the event of all numbers drawn out being different . Let us proceed to calculate $P(X')$ now.

In order to calculate $P(X')$ , we use the multiplication rule of probability. For independent events $E_1, E_2, E_3, \ldots E_n$ , the probability of all events $E_1, E_2, E_3, \ldots E_n$ occurring (or $P(E_1\cap E_2 \cap E_3 \cap \ldots \cap E_n)$ ) can be found by $P(E_1) \times P(E_2) \times P(E_3) \times \ldots \times P(E_n)$ . The event of John drawing out each ball can be considered as independent of the rest, since John replaces the ball after taking it out. So the result of any trial would not affect the other results. To ensure that all the numbers on the four balls are different from each other, whenever each ball is drawn, we need to check that the number on the ball is different from the numbers on the other balls.

When the second ball is taken out, there is a probability of $\frac {5}{6}$ that the number on the ball is not equal to that on the first ball. (Since there is an equal chance of drawing out any of the six numbers, but we do not want to draw out the number that is equal to that on the first ball.) Likewise, for the third ball, there is a probability of $\frac {4}{6} = \frac {2}{3}$ that the number on the ball is not equal to that on the first or second ball. For the final ball, the probability that the number is different from that on the previous balls is $\frac {3}{6} = \frac {1}{2}$ .

Following this, using the multiplication rule, $P(X') = \frac {5}{6} \times \frac {2}{3} \times \frac {1}{2} = \frac {5}{18}$ . Using the relation of $1 - P(E) = P(E')$ , $P(X) = 1 - P(X') = 1 - \frac {5}{18} = \frac {13}{18}$ . To answer the question, $a + b = 13 + 18 = 31$ .

The probability that all numbers are different is equal to $\frac{{^6 P_4 }}{{6^4 }}$ therefore required probability is $1 - \frac{{^6 P_4 }}{{6^4 }} = \frac{{13}}{{18}}$

Begin by computing the total number of possibilities. Which is $6^{4}$ in this case because John picks from a total of 6 balls 4 times. We then take $6^{4}$ (a repeating permutation) and subtract by the number of possibilities in which not a single number is repeating, this equals $6\times5\times4\times3$ (a none repeating permutation). $6^{4} - 6\times5\times4\times3$ or $1296-360$ if you will. Divide that answer by the total number of possibilities $6^{4}$ . You'll have $\frac{936}{1296}$ . Finally Simplify to make them coprime ( $\frac{13}{18}$ ), add ( $13+18$ ) to get 31 .

The total possibilities for sequences drawn are $6 \cdot 6 \cdot 6 \cdot 6$ . The possibilities that no number is drawn twice is $\frac {6!}{2!}$ . That is because we have a ball less to be possible for drawing. So, the number of possibilities that at least two of the numbers John rote down are equal is: $1296 - \frac {6!}{2!}$ . Then: $\frac {936}{1296}$ = $\frac {13}{18}$ .

The total possible outcomes, $N=6^4=1296$ .

The number of outcomes with no same number, $N_1=6\times 5\times 4\times 3=360$ .

The number of outcomes with at least two same numbers, $N_{2-4}=N-N_1=1296-360=936$ .

And its probability, $P=\frac{N_{2-4}}{N}=\frac{936}{1296}=\frac{13}{18}=\frac{a}{b}$

Therefore, $a+b=13+18=\boxed{31}$ .

a/b=1-(combinações onde todos são diferentes/todas as combinaçōes possiveis)=1-(10/36)=26/36=13/18 logo a resposta (a+b) é igual a 31

To calculate the probability that at least 2 of the numbers are equal, we calculate the probability that none of the numbers are equal to every other numbers.

When John take the first ball, it can be any numbers of the 6 given numbers. To make it sure that none of the numbers are equal, the second ball's number must be one of remaining 5 numbers. Similarity, the third ball must be one of 4 remaining balls, and the final ball must be one of 3 remaining balls.

So, the total amouth of cases that none of numbers are equal to every other numbers is $6 \times 5 \times 4 \times 3 = 360$

The total amouth of possible cases is $6^4 = 1296$ Take the ratio: $\frac{360}{1296} = \frac{5}{18}$

Now we have the probility that none of the numbers are equal to every other numbers. Now we can calculate the probility that at least 2 of the numbers are equal. It is

$1 - \frac{5}{18} = \frac{13}{18}$

Therefore, a = 5 and b = 13. We have the solution of the problem is 18.

It is easier to think about the probability that no 2 numbers are equal In the first draw, the probability that no 2 numbers are equal is 6/6

In the first draw, the probability that no 2 numbers are equal is 5/6

In the first draw, the probability that no 2 numbers are equal is 4/6

In the first draw, the probability that no 2 numbers are equal is 3/6

Multiplying the values we get $6/6*5/6*4/6*3/6=5*4*3/6*6*6=60/6*6*6=10/36=5/18$

Subtracting this from 1, $1-5/18=13/18$ Our final answer=13+18=31

the probability that at least 2 of the numbers are equal = 1 - the probability 2 of the numbers are different . so we get 1 - \frac{6P4}{6x6x6x6} = 1 - \frac{5}{18} = \frac{13}{18} so a+b = 13 + 18 = 31

The probability that "at least 2 of the numbers are equal" is the complement to the probability of "none of the numbers are the same". The latter is calculated as $\frac {6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} = \frac{5}{18}$ . Therefore the probability asked in the question is obtained as $1-\frac{5}{18}=\frac{13}{18}$ and the final answer would be $13+18=31$ .

The probabillity that all of 4 numbers are different is $\dfrac{6.5.4.3}{6.6.6.6}=\dfrac{5}{18}$ , So the probability that at least 2 of the numbers are equal is $1-\dfrac{5}{18}=\dfrac{13}{18}$ , and we get $13+18=31$

To make it easier, we can compute the probability that they are all distinct: After we choose our first number, there's a $\frac{5}{6}$ chance that the second number is different; then there's a $\frac{4}{6}$ chance the third number is different; then finally there's a $\frac{3}{6}$ chance that the fourth number will be different.

We then multiply $\frac{5}{6}*\frac{2}{3}*\frac{1}{2}=\frac{5}{18}$ .

To get the probability that at least two numbers are equal, we do $1-\frac{5}{18}=\frac{13}{18}$ .

$13+18=\boxed{31}$ .

The number of cases when picking four balls in the bag - \­(2^{34}\­). The number of cases when all the numbers on the ball aren't equal - \­(3 \times 4 \times 5 \times 6\­) Therefore, the probability is \­(1-\frac{3 \times 4 \times 5 \times 6}{18}\­)=\frac{13}{18} ∴a+b=31