Question

The probability, that in a randomly selected 3-digit number at least two digits are odd, is

• A. $\frac {19}{36}$
• B. $\frac {15}{36}$
• C. $\frac {13}{36}$
• D. $\frac {23}{36}$

Answer 1

Answer: (A)

$\frac {19}{36}$

Answer 2

Required cases = Total cases – all digits even – exactly one digit even
Total Cases $= 900$ ways

The required Probability = $\frac {900 - 425}{ 900}$ = $\frac {19}{36}$

So, the correct option is (A): $\frac {19}{36}$.