Two letters are chosen at random from the word delete and two are chosen at random from the word enter. what is the probability that the selection contains 4 e's or no e's?

Answer 1

Let's first find the number of possible combinations that we can find:

Let's first find the number of possibilities of letters we can choose from the word delete. Given there are 6 letters, we can choose one of these 6. However, once we choose one of these 6, we have 5 letters left to choose from. So, the number of possibilities for delete is 6*5=30.

For enter, we have 5 letters to choose from. However, once we pick one letter, we have 4 left. So, the number of pairs of letters we can pick from enter is, 5*4=20.

Now, we can multiply 30*20 to obtain the total groupings of letters we can possibly pick, which is 600.

Now, let's find the number of possibilities that a potential selection of four letters contains four e's. In delete, there are three possible combinations of two e's we can combine, the first two e's, the last two e's, and the first and last e's.

Now, let's go to enter. For enter, there is only one combination of e's we can pick as there are only two e's in the word.

To find the total groupings that we can get of 4 e's. We must multiply the numbers we found for each word. 3*1=3 possibilities.

Now, let's find the number of possibilities that a potential selection can contain no e's. For the word delete, there are three possible combinations in which we do not pick e as a letter, dl, dt, and lt. Same with enter. They are rn, nt, and rt. Since there are three possible combinations for both letters, we multiply by 3*3=9.

Now, let's add 3, which is the number of possibilities in which we can get 4 e's, with 9, the number of possibilities we can get no e's. This is 12. We divide this by 600, our total possible groupings of letters we can choose.

12/600=1/50

There is a 1 in 50 chance that we will get 4 e's or no e's.