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?  Pick one:
A) 9/16 B) 1/2 C) 1/3 D) 1/4 E) 1/6
Remaining number of digits after three '0' digits= 4   Constraint- no more '0' and no more '1'   Well, these four digits can be odd or even   1/2*1/2*1/2*1/2= 1/16   Ways to choose 3 odd digits out of total 4 digits= 4C3= 4   4* 1/16= 1/4   D is the answer   Hi! Since we are talking about remaining 4 digits. Each digit can either be odd or even (Total ways to select = 2. desired outcome= 1)   so let's say we want to choose first 3 digits as odd digits and last digit as even. The probability will be :-   1/2 *1/2*1/2*1/2= 1/16   But out of remaining 4 digit , any 3 can be odd and one even as there are no constraints in the question. Therefore, total ways of choosing 3 odd digits out of 4 total digits= 4C3   So, the probability = 4*1/16= 1/4   answer is 1/4
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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?