Problem Solving

The question is from the 300 GMAT problems with best solutions from our forum.

A password of a computer used five digits where they are from 0 and 9. What is the probability that the password solely consists of prime numbers and zero?
A 1/32
B 1/16
C 1/8
D 2/5
E ½

My approach:

We often use permutation and combinations in probability to arrive at no of outcomes.

Total number of outcomes = 10C5 - Choosing 5 digits out of 10 digits.
Total number of favorable outcomes = 5C5 x 5! ((Choosing 5 digits - 4 prime and zero out of the same) x (the number of ways each digit can be arranged))

5C5 = 1
so P(Password being prime and zero) = (1 x 5!)/10C5

The solution given in the document is different from the way i solved it. There are many problems which i have solved it using combinations and this one is confusing as even the answer choices won't fit in if i solve it using the above approach.

Need your help on why the above approach is wrong.


OA is A

Solution given in the document:
There are 10 possible options (0,1,2,3,4,5,6,7,8,9) for each digit.
5 of the options (0,2,3,5,7) are zero or prime.
So, P(a given digit is zero or prime) = 5/10 = 1/2

A quick way is to look at this as an AND probability.
P(all five digits are zero or prime) = P(1st digit is zero or prime AND 2nd digit is zero or prime AND 3rd digit is zero or prime AND 4th digit is zero or prime AND 5th digit is zero or prime)

This is equal to P(1st digit is zero or prime) x P(2nd digit is zero or prime) x P(3rd digit is zero or prime) x P(4th digit is zero or prime) x P(5th digit is zero or prime)
So, we get 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32

Thanks in Advance.

sam2304 wrote: The solution given in the document is different from the way i solved it. There are many problems which i have solved it using combinations and this one is confusing as even the answer choices won't fit in if i solve it using the above approach.

Hi,
The question never said five distinct digits. So, digits can be repeated.
So, total number of cases will be 10^5 instead of 10C5.
Btw, even in your case(assuming all 5 are distinct), the number of cases should be 10P5, not 10C5.
As, the password should be chosen from 5digits(0,2,3,5,7), number of ways = 5^5
So, probability is 5^5/10^5 = (1/2)^5 = 1/32.

Total number of outcomes = 10C5 - Choosing 5 digits out of 10 digits. This is wrong

Apply basic counting methods on these types of problems, rather than jumping straight on to formulas.

Total number of outcomes = (10)^5 (you can repeat the digits , because the question does not restrict you.)

Favorable outcomes= 5*5*5*5*5 = 5^5 (again no restriction on choosing same digits out of 0,2,3,5,7)

So, P(E) = (5/10)^5=(1/2)^5 = 1/32

So A

Thanks a lot for the tip. I forgot about repetition. Very true to your tagline

sam2304 wrote:Thanks a lot for the tip. I forgot about repetition. Very true to your tagline

sorry for nitpicking, but the method detailed above would not have worked even had the question said "without repetition". This is because you are assuming that the order matters for the top of the fraction (by multiplying by 5!), while at the same time you're assuming that the order does not matter for the bottom of the fraction (10c5 means "without order). In order to correctly find the number of wanted outcomes out of the total number of outcomes, you need to be consistent - either the order matters for both, or does not matter for both. For probability, it doesn't really matter whether the order matters or not, as long as you're consistent in both top and bottom.

the correct calculation (assuming that the question uses without repetition) would be

total number of outcomes: 10C5 *5! (first choose which 5 digits, then multiply by internal order)
Wanted number of outcomes 5C5 * 5! (internal order of the 5 numbers - 4 primes and zero)

Once you reduce the 5!, you get 1 / 10C5 - Out of all the different ways of choosing 5 digits out of 10 (regardless of order), only one choice will include all 4 primes and zero 2,3,5,7,0. If you then want to find all the permutations in the wanted outcomes (0,2,3,5,7), that's fine, as long as you remember that the total number of outcomes should also include the permutations of all the different choices in 10C5.

probability = (5+1)^5/ 10^5 = 1/32