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Second exercise of combinations

The first example of combinations is this.

Remember our formula of combinations

C(n,r)=\left( \begin{array}{c}n\\r \end{array} \right) = \cfrac{n!}{r!\left(n - r\right)!}

Let’s start with the paragraph of our second example of combinations:

From an urn, with 6 numbered white balls of 1 up to 6 and 7 black balls from 1 up to 7, at the same time, 4 are taken out. What is the probability of obtaining:

Balls with numbers whose product is even?

To ensure that the product is even, the following must be fulfilled as shown in the following table:

\begin{array}{| c | c | c | c |} \hline \text{PAIR} & \text{PAIR} & \text{PAIR} & \text{PAIR} \\ \hline \text{PAIR} & \text{PAIR} & \text{PAIR} & \text{ODD} \\ \hline \text{PAIR} & \text{PAIR} & \text{ODD} & \text{ODD} \\ \hline \text{PAIR} & \text{ODD} & \text{ODD} & \text{ODD} \\ \hline \end{array}

Let’s think a little bit to know if the result will be even.

  • If we multiply 4 even numbers, the result will always be an even number.
  • If we multiply 3 even numbers and 1 odd number, the result will always be an even number.
  • If we multiply 2 even and odd numbers, the result will always be an even number.
  • Finally, if we multiply 1 even number and 3 odd numbers, the result will always be an even number.

If you need to make a table to better visualize the balls, draw a table. We made the following table:

\begin{array}{|c|c|} \hline \text{White balls} & \text{Black balls}\\ \hline 1 \ 2 \ 3 \ 4 \ 5 \ 6 & 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7\\ \hline \end{array}

And as we have only 4 ways to make the product result in an even number, we have to put the 4 cases in the following way with the operations already done:

\begin{array}{|c|c|c|} \hline \text{Case number} &\text{Case with letters} & \text{Caso with numbers}\\ \hline \hline 1 &\text{All pairs} & 6C4= 15 \\ \hline 2 &\text{Three pairs and one odd} & 6C3*7C1= 140\\ \hline 3 &\text{Two pairs and two odd} & 6C2* 7C2 = 315\\ \hline 4 &\text{A pair and three odd} & 6C1*7C3 = 210\\ \hline \end{array}

Consider that of the total of 13 balls, we have 6 with even number and 7 with odd number.

As in case 1 all the numbers of the balls have to be even and in total we only have 6 balls that have even numbers, the operation was made 6\text{C}4 that gives us as result 15 combinations.

In case 2 you need 3 pairs and 1 odd, then what you have to do is combinations of 3 even balls and combinations of 1 odd ball, in mathematical terms is 6\text{C}3*7\text{C}1 which results in 140.

In case 3 you need 2 pairs and 2 odd, so you can easily apply combinations of 2 even balls and combinations of 2 odd balls, you would mathematically do 6\text{C}2*7\text{C}2 which results in 315.

In the last case you have to leave 1 ball with even number and 3 with odd number, we simply apply 6\text{C}1*7\text{C}3 which results in 210.

We have to sum all the results that we just obtained, to have a total of 680. Then what you have to do is apply combinations to know all the possible cases that there are to get 4 balls from the 13 there is, so applying 13\text{C}4 results in 715 combinations.

Finally what you have to do is divide the sum of all the combinations calculated between the number of combinations that can be obtained from 4 balls of the 13:

\cfrac{680}{715} = 0.951

So the probability of getting 4 balls whose product is even is 0.951

Balls with numbers less than 2?

The answer is the following:

In this case the probability is zero because we only have two balls that are smaller than 2, that is, the balls with the number 1.

Balls with numbers greater than 2?

This paragraph is quite simple because all you have to do is discard the 4 balls that have numbers less than or equal to 2. Instead of having 13 balls, now we will have 9 balls and apply the following combination:

9\text{C}4=126

And the result obtained has to be divided between 715 which is the total of combinations that there are to grab 4 balls of the 13:

\cfrac{126}{715}=0.1762

So the probability of taking balls with numbers greater than 2 is 0.1762

Balls with different numbers?

Of all the balls, we must bear in mind that there are 6 repeated numbers, so it is necessary to know the number of possible combinations that can be made by excluding the ball with the 7 number:

6\text{C}4 = 15

The number of combinations that can be alternated with white and black balls is calculated:

(2\text{C}1)^{4}

The calculation that was made was because we have 6 repeated balls, that means that for the number 1 we have 2 combinations: that is white or black. For number 2 we have 2 combinations: that is white or black. And so with all the others. Why is the exponent 4 and not 6? Because we’re only going to extract 4 balls, not 6.

Then proceed to multiply the number of possible combinations that can be done by excluding the ball with the number 7 with the number of combinations that can be alternated with the white and black balls:

6C4*(2C1)^{4}=240

Now consider the 7 that had been discarded. We identify he number of combinations that can be made with 6 pairs and the 7:

6\text{C}3

We wrote a 3 because a place of those 4 we have already occupied because we are including 7.

Then calculate the number of combinations that can be alternated with white and black balls, of the 4 possible, 3 are repeated and 1 is not:

6\text{C}3*1\text{C}1*(2\text{C}1)^{3}=160

Of the two events calculated (when it is considered and when it is not considered the number 7), we have to sum them:

6\text{C}4*(2\text{C}1)^{4}+6\text{C}3*1\text{C}1*(2\text{C}1)^{3}=400

The 400 obtained has to be divided among the total of combinations that there are to remove 4 balls from the 13 (we already calculated it, remember that 13\text{C}4 equals 715):

\cfrac{400}{715}=0.5594

So the probability that different balls can be extracted will be 0.5594

Thank you for being at this moment with us:)

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