Problem
Let X be the set of all five digit numbers formed using 1, 2, 2, 2, 4, 4, 0. For example, 22240 is in X while 02244 and 44422 are not in X. Suppose that each element of X has an equal chance of being chosen. Let p be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5. Then find the value of 38p.
Solution
To find the conditional probability, we’ll first count the total number of 5-digit numbers formed using the given digits, that are divisible by 5.
Then, from this set, we’ll count the numbers that are divisible by 20. Finally, we’ll divide the latter by the former to get the value of p. Let’s begin!
For a number to be divisible by 5, its last digit must be 5 or 0. Thankfully, there’s no 5 in the given set, making our counting easier.
So, our number must look like XXXX0, where the first four digits are chosen from the set {1, 2, 2, 2, 4, 4}.
Since there are multiple 2s and 4s, and we’re allowed to select multiple digits of each type, we’ll have to list down individual cases for the possible selections, and then use the standard formulas to count their arrangements.
Case 1: {2, 2, 2, 4}
Here, the total number of arrangements will be 4!/3! or 4.
Case 2: {2, 2, 2, 1}
Same here. The total number of arrangements will be 4!/3! or 4.
Case 3: {2, 2, 4, 4}
Here, the total number of arrangements will be 4!/2!2! or 6.
Case 4: {2, 2, 4, 1}
Here, the total number of arrangements will be 4!/2! or 12.
Case 5: {2, 4, 4, 1}
Same here. The total number of arrangements will be 4!/2! or 12.
Finally, the total number of 5-digit numbers that are divisible by 5 is 4 + 4 + 6 + 12 + 12 or 38.
That 38 gives some hope, since it’ll be the denominator, and we’re required to find 38p.
Let’s move on to the favorable cases now, i.e., the numbers that are divisble by 20.
To be divisble by 20, the number must be of the form XXX20, XXX40, XXX60, XXX80, or XXX00. Only the first two matter here, since there’s no 6 or 8 in the set, and only one 0.
Now, we’ll go back to each case above, and count the required cases.
But wait, we can same some effort here! What if we remove the cases where the number is not divisible by 20? That is, numbers of the form XXX10 (ignoring XXX30, XXX50, XXX70, XXX90, since there’s no 3, 5, 7, or 9)
That’s much easier to count. Let’s do that for each case.
Case 1: {2, 2, 2, 4}
Here, there’s no 1. So, each number formed will be divisble by 20. Hence 0 numbers that are not divisible by 20.
Case 2: {2, 2, 2, 1}
Here, the only possible number is 22210. So, 1 possible number.
Case 3: {2, 2, 4, 4}
Again, no 1 here. So, 0 numbers that are not divisble by 20.
Case 4: {2, 2, 4, 1}
Here, if we fix the last two digits as 10, then the remaining digits can be arranged in 3!/2! ways. So, 3 possible numbers here.
Case 5: {2, 4, 4, 1}
Same here. The total number of arrangements will be 3!/2! or 3, giving us 3 numbers again.
Finally, the number of 5-digit numbers that are not divisible by 20 is 0 + 1 + 0 + 3 + 3 or 7. That gives a total of 38 – 7 or 31 numbers that are divisible by 20.
Therefore, the required conditional probability, or p, equals 31/38. The value of 38p is left as an exercise to the reader.
Comments
A nice, balanced problem based on permutations and probability. A few things to be careful about:
- Avoiding counting the entire set, and only those relevant for the probability condition.
- Carefully listing out the all the favourable and total cases.
- Counting the complementary set of the favorable cases.
This could have been posed as a paragraph style problem as well, where the second problem asked to count the total number of elements in X. That would’ve made the student’s effort worthwhile, given the considerable pain they had already taken in counting the cases.