Bunuel
A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?
A. \(\frac{860}{90,000}\)
B. \(\frac{810}{100,000}\)
C. \(\frac{858}{100,000}\)
D. \(\frac{860}{100,000}\)
E. \(\frac{1530}{100,000}\)
We're going to apply the MISSISSIPPI rule at the end of my solution, so here's what it says:
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....] So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are
11 letters in total
There are
4 identical I's
There are
4 identical S's
There are
2 identical P's
So, the total number of possible arrangements =
11!/[(
4!)(
4!)(
2!)]
-----NOW ONTO THE QUESTION------------------------------
We'll consider two cases:
case i: We have three 6's and two DIFFERENT digits (e.g., 66612)
case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)
case i: We have three 6's and two DIFFERENT digits (e.g., 66612)We already have three 6's. So, we must select 2 different digits from (0,1,2,3,4,5,7,8 and 9)
We can do this in 9C2 ways (=
36 ways)
Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule.
We can arrange 3 identical digits and 2 different digits in 5!/3! ways =
20 ways
So, we can have three 6's and two DIFFERENT digits in (
36)(
20) ways (=
720 ways)
case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)We already have three 6's. So, we must select 1 digit (which we'll duplicate)
Since we're selecting 1 digit from (0,1,2,3,4,5,7,8,9) we can do so in
9 ways
Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule.
We can arrange 3 identical 6's and 2 other identical digits in 5!/3!2! ways =
10 ways
So, we can have three 6's and two IDENTICAL digits in (
9)(
10) ways (=
90 ways)
So, TOTAL number of ways to have three 6's =
720 +
90 =
810Since there are 100,000 possible 5-digit codes, P(having exactly three 6's) = 810/100,000
Answer: B
Cheers,
Brent