mohshu wrote:

Divyadisha wrote:

Bunuel wrote:

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?

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= 1/16Ways to choose 3 odd digits out of total 4 digits= 4C3= 4

4* 1/16= 1/4

D is the answer

Hi, divyadisha u have multipled the probability for 4 digits but number of ways for 3 digits

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

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