a 7 - digit combination lock on a safe has zero exactly 3 times , does not have the digit 1 at all . what is the probability that exactly 3 of its digits are odd ?
Choose the correct answer.
A) 9 / 16  B) 1 / 2  C) 1 / 3  D) 1 / 4  E) 1 / 6

Rationale: 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 = 4 c 3 = 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 = 4 c 3 so , the probability = 4 * 1 / 16 = 1 / 4 answer is d
Answer: D