Question

Judith's bakery has a promotion where they print three letters at the bottom of their receipts. The three letters are chosen randomly from J, U, D, I, T, H, and S, and they don't repeat any of the letters on a receipt. For instance, a receipt may have DJI printed on it, or HSU, but not JJT. If the three letters spell a word, you get a free doughnut. The bakery manager has come up with a list of acceptable answers: HID, HIS, HIT, HUT, IDS, JUT, and SIT. What is the probability that a customer will win a free doughnut after one purchase?

Answer 1

Type your answer as a fraction reduced to the simplest form, like this: 7/11

Answer 2

First you would want to identify which kind of probability formula you will want to use to find out the number of 3 letter words without any of the 7 seven letters repeating that the name JUDITHS would make and then since you can only have 7 correct words from that name you will want to put that the probability of getting those 7 words will be out of the number of 3 letter word combinations that there

Answer 3

So first figure out how many 3 letter word combinations JUDITHS makes without a letter repeating and then put that number under 7 to find the probability of getting 1 of the 7 words.

Answer 4

I see..wait so the answer is 1/7 ? not 1/3 ??? @Johnbc

Answer 5

I dont believe either is the answer? I have not performed the calculations to find the number of possible 3 letter word combinations the word JUDITHS will have

Answer 6

hmmm alright

Answer 7

This is a permutation problem without repetition with a restricted only 3 words use out of the 7 so we can use the formula \[\frac{ n! }{ (n-r)! }\] Where N is the number of letters to choose from (7) and R is the number of letters we can only choose from those letters (3)

Answer 8

Plugging in what we know then we will have:
n = 7 letters
r = 3 letters per word
so the possible combination of 3 letter words is\[\frac{ 7! }{ (7-3)!}\] Which then equals\[\frac{ 7! }{ 4! }\]