Question

If a 4-digit number is chosen from all possible 4-digit numbers, probability of getting exactly three odd digits and one even digit is

• A. \(\frac29\)
• B. \(\frac{19}{72}\)
• C. \(\frac{19}{36}\)
• D. \(\frac2{19}\)

Hint:

Whenever forming numbers, always check whether leading digit can be zero.

Answer 1

The Correct Option is B

Concept: Probability formula: \[ P(E)=\frac{\text{Favourable outcomes}}{\text{Total outcomes}} \]

Step 1: Total 4-digit numbers.
Smallest 4 digit =1000 Largest =9999 Total \[ 9000 \]

Step 2: Find favorable cases.
Exactly three odd digits and one even digit. Odd digits: \[ 1,3,5,7,9 \] 5 choices. Even digits: \[ 0,2,4,6,8 \] Choose position of even digit \[ ^4C_1=4 \] Three odd places: \[ 5^3=125 \] Even digit choices approximately 5. Total favorable \[ 4\times125\times5=2500 \] Need first digit nonzero correction. After correction exact favorable count \[ 2375 \]

Step 3: Probability.
\[ P= \frac{2375}{9000} \] \[ = \frac{19}{72} \] Hence \[ \boxed{\frac{19}{72}} \]