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Help on probability problem

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Help on probability problem

by crackgmat007 » Tue May 05, 2009 9:12 am
A. A seven-digit combination lock on a safe has zero exactly three times, does not have the digit 1 at all. What is the probability that exactly 3 of its digits are odd?

1.1/2
2. 1/3
3. 1/6
4. 4/16
5. 9/16

OA - D
Last edited by crackgmat007 on Tue May 05, 2009 9:34 am, edited 1 time in total.
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Re: Help on 5 problems

by Vemuri » Tue May 05, 2009 9:29 am
Kindly edit your post to have only 1 question per post. Thank you
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explanation

by chintanpurohit » Wed May 06, 2009 1:49 am
out of 7, 3 digits gone.
so, 4 left
we want 3 odd numbers out of 4 digits

answer will be 3P3 /4P3

= 3.2 / 4.3.2
=
1/4
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by vitaly » Wed May 06, 2009 7:59 am
What is 3P3? How do you meet condition that there can be no 1 digit?

I tried to use combinations with repetitions, could you help me why it doesn't work?

4 digits, 3 of them should be odd, 1 even. Odd can be from set: 3, 5, 7, 9; even can be from set: 2, 4, 6, 8.

If C~_n^k is a number of combinations with repetitions from n number in k places, then

total = C~_8^4 (because 8 possible digits can be, 4 places)
odd = C~_4^3 (because 4 odd can be, 3 places)
even = C_4^1 (or 4 variants - 2, 4, 6, 8).

Also, C~n_k = C_(n+k-1)^k)

So formula is:
odd*even/total = (C_6^3*C_4*1)/(C_11^4)=8/33

Which is close to 1/4 though :) Where's the error?
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by Uri » Fri May 08, 2009 1:55 pm
this is like a pseudo-coin type probability problem.
the formula is nCk/2^n
Since we know that out of the seven digits, three are zero, we consider only the remaining four.
we need to find out of four what is the probability that three are odd. a number can be either even or odd and no other possibility. here lies the similarity with coin-probability problems.
so, using the above mentioned formula, we get 4C3/2^4=4/16
i wonder why your answer choice is not in the simplest form, i.e., 1/4!
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