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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