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by nikhilkatira » Sun Jun 27, 2010 10:06 pm
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

Can this problem be solved using Combination formula ?
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Nikhil H. Katira
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by tpr-becky » Sun Jun 27, 2010 10:10 pm
It is best using basic probability - decide how many items you are choosing - here we have 5 digits. Then write a want/total formula for each 1 - then if you want the probability of one thing AND another multiply

There are 5 numbers that fulfil the criteria and so the want/total is 5/10 for each one (assuming we can repeat numbers) we want them all to be prime or zero so we multiply

5/10 * 5/10 *5/10 * 5/10 * 5/10 = 1/32 answer is A.
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by nikhilkatira » Sun Jun 27, 2010 10:17 pm
tpr-becky wrote:It is best using basic probability - decide how many items you are choosing - here we have 5 digits. Then write a want/total formula for each 1 - then if you want the probability of one thing AND another multiply

There are 5 numbers that fulfil the criteria and so the want/total is 5/10 for each one (assuming we can repeat numbers) we want them all to be prime or zero so we multiply

5/10 * 5/10 *5/10 * 5/10 * 5/10 = 1/32 answer is A.
thanks Becky...

But I want to know whether this problem can be solved using formula nCr ( combination) ?
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by rockeyb » Mon Jun 28, 2010 12:26 am
nikhilkatira wrote:
tpr-becky wrote:It is best using basic probability - decide how many items you are choosing - here we have 5 digits. Then write a want/total formula for each 1 - then if you want the probability of one thing AND another multiply

There are 5 numbers that fulfil the criteria and so the want/total is 5/10 for each one (assuming we can repeat numbers) we want them all to be prime or zero so we multiply

5/10 * 5/10 *5/10 * 5/10 * 5/10 = 1/32 answer is A.
thanks Becky...

But I want to know whether this problem can be solved using formula nCr ( combination) ?
Number available 0 to 9 . total = 10 numbers .

The code is of 5 numbers _ _ _ _ _ .

first number can be selected in 10c1 ways . Since the numbers are not unique and repetition is allowed we can select the second number in 10c1 ways and so on .

So total possible ways in which this code can be made is (10c1)(10c1)(10c1)(10c1)(10c1)

Now the possibility of finding the code with only odd numbers and zero .

So total numbers to select from 0,2,3,5,7 . Total numbers 5 .

The first number can be selected in 5c1 ways again repetition allowed so second number can be selected in 5c1 ways and so on .

So numbers of ways in which the code with odd numbers is selected is

(5c1)(5c1)(5c1)(5c1)(5c1) .

Probability = (number of required outcomes )/ (total possible outcomes)

= (5c1)(5c1)(5c1)(5c1)(5c1) / (10c1)(10c1)(10c1)(10c1)(10c1)

= 1/32 .
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by tpr-becky » Mon Jun 28, 2010 6:32 am
Rock correctly used the combination route but he essentially did the exact same thing I did. So yes, this problem can be solved using combination but it would be unnecessarily complicated in the thinking and when doing problems you should certainly look for the simplest route to the solution.
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by globalcitizen » Mon Jun 28, 2010 6:59 am
So I got the wrong answer for this....

Can someone please explain to me why the calculation shouldn't be:

(1/10) - because there is only one 0
(4/10) - because there are only four prime numbers (2,3,5,7)

and thus,

(1/10)(4/10)(4/10)(4/10)(4/10)

When I look at the correct answers, I understand why 5/10 was used...and though I now know that my calculation is wrong, I still don't understand why it is wrong to breakdown prime number probability vs. likelihood of picking 0. If someone can be so kind as to explain why, I would be most appreciative. I.e. What does my incorrect calculation mean in words?

Thanks!
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by kvcpk » Mon Jun 28, 2010 7:13 am
globalcitizen wrote:So I got the wrong answer for this....

Can someone please explain to me why the calculation shouldn't be:

(1/10) - because there is only one 0
(4/10) - because there are only four prime numbers (2,3,5,7)

and thus,

(1/10)(4/10)(4/10)(4/10)(4/10)

When I look at the correct answers, I understand why 5/10 was used...and though I now know that my calculation is wrong, I still don't understand why it is wrong to breakdown prime number probability vs. likelihood of picking 0. If someone can be so kind as to explain why, I would be most appreciative. I.e. What does my incorrect calculation mean in words?

Thanks!
Let me try to explain. The first step of your approach is incorrect.
The problem asks to have either prime numbers or zero as password.
So, between 0 to 9, which digits will satisfy the above condition?
{0,2,3,5,7} - let me call this as PWD set

So we have 5 digits from which should appear in the password.

Now we have 5 digit password.
probability that first digit comes form PWD set is 5/10 [because we can pick any from PWD set out of the whole 10 digits available]
Similar is the case for other 4 digits of the password also.

Also, important thing to remember here is that the question is not asking for 5 digit numbers. So 0 can be in any position.

Hope this helps. Let me know in case you have any doubt.
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by globalcitizen » Mon Jun 28, 2010 5:20 pm
Ok...just to double check....

The answer is (5/10)^5 and not my approach because we are picking from set {0,2,3,5,7} whereby it doesn't matter which is picked first (prime# vs. 0). While, my incorrect answer would be correct in the situation where we had to choose 0 separately and then the prime #s.

Did I understand that correctly? I just want to know...when would my approach ever be used...or would it never be because it is flat out incorrect?

Sorry for tiring you with this again!

Many thanks for your help, I really appreciate it!
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by tpr-becky » Mon Jun 28, 2010 8:13 pm
yes, you are correct.
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by kvcpk » Mon Jun 28, 2010 8:16 pm
globalcitizen wrote:Ok...just to double check....

The answer is (5/10)^5 and not my approach because we are picking from set {0,2,3,5,7} whereby it doesn't matter which is picked first (prime# vs. 0). While, my incorrect answer would be correct in the situation where we had to choose 0 separately and then the prime #s.

Did I understand that correctly? I just want to know...when would my approach ever be used...or would it never be because it is flat out incorrect?

Sorry for tiring you with this again!

Many thanks for your help, I really appreciate it!
I believe your method would be correct, if the question says that the password should contain only one zero. and the rest of them should be prime.
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by tpr-becky » Mon Jun 28, 2010 8:41 pm
The passage does not say only 1 zero - it says soley prime numbers and zero - the meaning of this is that there are only prime numbers and zero in the password - you can have more than one zero.
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by globalcitizen » Tue Jun 29, 2010 7:19 am
Thanks so much for making everything crystal clear! You guys are great :)
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by missrochelle » Tue Jun 29, 2010 9:33 am
hi can someone pls explain to me how to QUICKLY multiply those fractions to get 1/32? how do u simplify that math?

i tried 5^5 / 10^5 and that gives me 1/2^5 which is the same as 1^1 / 2^5.

is this the most efficient way to do it?
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by tpr-becky » Tue Jun 29, 2010 11:55 am
5/10 reduces to 1/2 then if you want 1/2^5 - you have to do the 2^5 part - which is just multiplying by 2's - 2, 4, 8, 16, 32 - the 1^5 will be one becuase 1 multiplied by itself is always 1.

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