In a 5 digit ID number, what is the probability of exactly t

5 digit no can be arranged in 9*10*10*10*10 ways ( if repetition allowed )

3 2digit no can be inserted in _ _ _ _ _ 5 digit ID no in 2 ways

(i) ID starting with 2 so thr are 5 ways in whch 3-2's can be placed in 5 slots
222--,22--2,2--22,2-2-2,2-22-
remaining 2 digits can be filled in 9ways ( 0-9 digits excluding 2) each, and they can be arranged among themselves in 2 ways
so total 9*9*5*2 ways

(ii) ID not starting with 2 ,, thr are 4 ways -222-,-22-2,-2-22-,--222
1st digit can be filled by any digit 1-9 excluding 2, so 8 ways
5th slot can be filled by 9 ways
n both themselves can be arranged among themselves in 2 ways
so total 8*9*4*2

that brings us to (9*9*5*2 + 8*9*4*2)/9*10*10*10*10
= 9*2(45+32)/9*10*10*10*10
= 2*77/10000 = 1.52%

I hope the logic works out to be correct

First: there is some problems in calculations, then the answer doesn't match with the choices provided.

Answer to the additional question:

favorable outcomes: \(9C1.9C1.5C3.10C1/=8100\)
Total possible outcomes: \(10^5\)

Probability in %= 8.1%

I could swear the answer is .0081% (which I realize isn't among the answers). Using my 5-step method:

1) Lay out the number of events (5 digits) _ _ _ _ _

2) Write down one specific example of the desired outome (exactly 3 twos): 2 2 2 (not2) (not2)

3) Label each event with its relevant probability and multiply across: 1/10 1/10 1/10 9/10 9/10 = 81/10^5

4)Determine the number of ways in which we could have the desired outcome: 5 events, 3 desired = 5C3 = 10

5) Multiply the result of step 4 by the result of step 3: 10 x 81/10^5 = 81/10^4 = .0081%

Lennon - remember, percentage = divided by 100

You have the right answer, but 81/10^4 = 81/100 (or 81%) * 1/100, thus, .81%

.0081 is the right answer, but .81 is the PERCENTAGE

The answer is simple
Step 1: Total number of ways = 10*10*10*10*10 = 10^5
Step 2: First choose three places for three 2s which can be done in 5C3 ways. (No point arranging these 2s amongst themselves) = 10 ways
Step 3: Now for the remaining two positions there are 9 numbers (as 2 is already used) each to be placed which can be done in 9*9 ways.
Step 4: favorable ways = 10*9*9 = 810

Thus probability = favorable ways / total ways = 810/10^5=0.0081 or 0.81%

Hope the solution helps

Why have you done 5C3 and not 5!/3!. In my opinion the order matters so the 5!/3! should remove the three 2s arrangeemnt out of the equation

Responding to a pm:



5C3 means you are selecting 3 out of 5.
Here, you are selecting 3 spots out of the available 5 spots (1st digit, 2nd digit, 3rd digit, 4th digit, 5th digit). Now you have all such arrangements:

____ 2 ____ 2 2

2 ____ ____ 22

2 ____ 2 ____ 2

... etc

In the two blanks, you can have any of them 9 remaining digits. You will get all ids of the from
23224
23223
32522
etc
i.e. all ids where the remaining two digits are the same or are distinct.
So favourable cases = 5C3 * 9 * 9

Your Method:
In case, you want to specifically arrange, you will need to take two cases:

Case 1: Three 2s and two distinct other digits
No of ways = 9C2 * 5!/3! = 720

Case 2: Three 2s and two another digit
No of ways = 9C1 * 5!/3!*2! = 90

Total no of favourable ways = 720 + 90 = 810

Required probability = 810/100000 = .81%

Step 1:
Choose 3 places where 2 can go
i.e. ways to choose 3 out of 5 places
5C3 = 10

Step 2:
Probability of placing 2's in those 3 places
\(\frac{1}{10}*\frac{1}{10}*\frac{1}{10}\)
(Prob of choosing one digit out of 10 is \(\frac{1}{10}\))

Step 3:
Probability of NOT placing 2's in the remaining 2 places
\(\frac{9}{10}*\frac{9}{10}\)
(Prob of NOT choosing one digit out of 10 is \(1-\frac{1}{10}\))

our answer:
step 1 AND step 2 AND step 3

\(10 * (\frac{1}{10})^3 * (\frac{9}{10})^2\)
=> \(\frac{0.81}{100}\)
=> 0.81%

Option A

Btw, this question is a classic implementation of Binomial theorem in probability.
In general, the questions like 'find out the probability of EXACTLY n successes out of m trials', could be easily solved using Binomial theorem.
Here's the formula:
Probability of 'r' success in 'n' trials where 'p' is the probability of 1 success is given by
\(nCr * p^r * (1-p)^{n-r}\)

Hi bunuel,

we have different combinations of the number 222_ _ using 5C3 but don't we have to look after the arrangement of the remaining last two digits??I am finding it confusing

5C3 means 5 digits in which 3 are of one kind and 2 are of another kind. => we are already considering the arrangement of the other two as well.

If you are still confused look at the solution given by Karishma above. She has mentioned two distinct cases you can split your question into.

What I don't understand about this problem is the possibility of ten digits for the first digit of this five digit number? Doesn't the first digit technically have to be from 1-9 in order to constitute a five digit number?

5-digit ID number can start with 0. For example, you can have an ID number 01234.

chetan2u here arrangements of the digits do not require ?

Hi

Arrangements matters..
So we choose 3 places out of 5 in 5C3 and then we place digit 2 in all 3 places..
Now the remaining 2 places we can place any of two digits so 9*9
Ans 9*9*5C3

Another way..
Case1..
3 are of same type and other two of one type so choose one from 9, so 222aa
They can be arranged in 9C1*5!/(3!2!)=9*10=90
Case 2..
3 are of same type and other 2 from remaining 9 digits but separate, so 222ab
So 9C2*5!/3!=9*4*5*4=720
Combined 90+720=810
Total ways 10*10*10*10*10=10^5

Probability = 810/10^5=0.81/100 or 0.81%

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