In a 5 digit ID number probability of 3 digits + tough

[cramya]

In a 5 digit ID number, what is the probability of 3 digits are the digit 2(question from beat the gmat forum)?


Can one of the experts please weigh in on this?

My questions are:

1) How will we generalize or come up with a perm/comb formula for taking numbers like 22255 or 44222 also in to account?

2) Also 55555 etc... For example are the total number of 5 digit numbers that can be formed
9*10*10*10*10 since we cant have 0 as the first digit but it 50000 or 200000 or 21111 are all valid 5 digit numbers)

[Guest]

What is the answer for this problem?? I was just thinking about it, but how i would have approached the problem is to have 5 blank digits and fill them in with the probablilities. To have exactly 3 digits in the 5 digit id be 2, I think you can just multiple the probability of each probability of getting the nbr 2 in 3 digits and NOT 2 in the other 2 digits. Example,

1/10 x 1/10 x 1/10 x 9/10 x 9/10
_____ _____ _____ _____ _____

81/10^5 or 81/100000 or .081%

I could be way wrong.....

[RonPurewal]

i'm not sure whether this problem requires that EXACTLY 3 digits be "2" or that AT LEAST 3 digits be "2".
i'll solve the former. you can generalize the following method to be able to solve the latter; post back if you want details on that.

--

probability that EXACTLY three digits are "2"s:

you need to account separately for each different way in which the "2"s can be distributed among the digits.

if you know and love combinatorial formulas, you can figure out that there are 5c3 = (5!)/(3!2!) = 10 such arrangements.

if you don't know or love combinatorial formulas, you can just list them (in an organized fashion, of course):
222__
22_2_
22__2
2_22_
2_2_2
2__22
_222_
_22_2
_2_22
__222
there are 10 arrangements.

for each of the arrangements, the probability is the product of three 0.1's or (1/10)'s, and two 0.9's or (9/10)'s. this is the case because the probability of each of the "2"s is 1/10, and each of the non-"2"s is 9/10. we know that products will work out the same way no matter in what order they're multiplied, so we can just find one of these products and multiply it by 10:
10 x (1/10)^3 x (9/10)^2 = 81/10,000
or, 10 x (0.1)^3 x (0.9)^2 = 0.0081
or, 0.81%.

--

the previous post contains a "solution" that is actually 1/10 of the correct solution, because the poster didn't account for the fact that there are 10 different ways in which to arrange the the three "2"s.
this poster's solution represents just the probability of "222__", or just the probability of "2_2_2", or just the individual probability of any other one of the individual arrangements only.

[hema.jce]

Hi Ron,

In the above method, we are assuming that having a zero in the first digit is valid ?

[Ben Ku]

I think the more we look at this question, the more nuances we see.

Guest started us off by looking at all the ways we can have three 2's and two of another number:
(1/10)^3(9/10)^2

Ron then added and said that there are 10 different arrangements the 2's can be placed in this ID number.

I'd like to add that we also need to take into account what those two digits are, specifically whether they repeat. For example, suppose we have 2, 2, 2, 3, and 4. One possible arrangement would be 22234. Another would be 22243. Those would be different IDs. However, if we have 2, 2, 2, 3, and 3, one possible arrangement would be 22233. Here, if I switched the threes (i.e. 22233) I would not get a different ID. Basically the product that we've been using with Ron and Guest does NOT take into account that last two digits could be the same.

I'd like to prose we find the probability this way:

Probability = P(three 2s + two different digits) + P (three 2s + two of same digit)

Let's take the first probability first: P(three 2s + two different digits). As Ron said, we need to first find ALL the different arrangements possible, then multiply that by the probability of one of those arrangements happening.

Possible arrangements: This is like making anagrams from the word AAAXY. In this case, it would be 5!/3!, or 20.
Probability of one arrangement: There's a 1/10 chance of selecting a two, 9/10 chance of selecting the next digit (since 2 cannot be used), and 8/10 chance of selecting the last digit (since it cannot be either of the last two digits). So it would be (1/10)(1/10)(1/10)(9/10)(8/10) = 72/100,000.

Therefore, the first probability is:
P(three 2s + two different digits) = (20)(72/100,000) = 1440/100,000

Let's look at the second probability: P(three 2s + two of same digit).

Possible arrangements: This is like making anagrams from the word AAAXX. In this case, it would be 5!/3!2!, or 10.
Probability of this arrangement: There's a 1/10 chance of selecting a two, 9/10 chance of selecting the next digit (since 2 cannot be used), and 1/10 chance of selecting the last digit (since it must be the same as the previous digit). So it would be (1/10)(1/10)(1/10)(9/10)(1/10) = 9/100,000.

Therefore, the second probability is:
P(three 2s + two of same digit) = (10)(9/100,000) = 90/100,000

The total probability would be the sum of the two probabilities:
1440/100,000 + 90/100,000 = 1530/100,000 = 154/10,000

As a decimal, this would be 0.0152, or 1.52%

-----------------------------------------------
hema.jce said:

Hi Ron,

In the above method, we are assuming that having a zero in the first digit is valid ?

In our calculations, we are assuming that 0 can be the first digit in the ID. Since it's an ID, I'm not worrying about it looking like a real number. On a real GMAT question, it would be explicitly tell you whether having 0s at the front would count or not.

Hope this helps.

[jerly_vivek]

How about the following solution:
Total number of ways in which exactly 3 digits have value 2: 5C3 = 10.
Total number with 5 digits = 9*10*10*10*10 = 90000
Required probability = 10/90,000 = 1/9000

[esledge]

How about the following solution:
Total number of ways in which exactly 3 digits have value 2: 5C3 = 10.

5C3 only considers the number of ways the repeated 2s can be placed in the 5 digit series:
x y 2 2 2
x 2 y 2 2
x 2 2 y 2
etc.

But you aren't considering the many possibilities for what could fill in for xy as the non-2 digits. For example, 13222 and 34222 are distinct id numbers, but your method counts them as one.

Those xy possibilities increase our cases by a factor of 9*9 = 81, assuming 0 can be the first digit. (Another way to think of the xy's: there are 100 two-digit numbers between 00 and 99, inclusive. Subtract out the 10 two-digit numbers from 20-29, then subtract the 9 other two-digit numbers that end in a 2, i.e. 02,12,32,42,52,62,72,82,92).

Total number with 5 digits = 9*10*10*10*10 = 90000

This is accurate if we can't have 0 as the first digit. But since the question didn't rule out 0 as the first digit, the total number of 5 digit ids must be 10^5 = 100,000.

In sum, the probability is 81*10/10^5 = 81/10^4 = .81/100 = 0.81%

Note that this replicates Ron's answer, and also proves that it takes repeated digits (Ben's concern) into account. For xy, the non-2 digits, both repeated digits (e.g. 44 or 33) and non-repeated digits (e.g. 63 or 49) were included in this probability.