MCQ
The probability, that in a randomly selected $3-$digit number at least two digits are odd, is
- ✓$\frac{19}{36}$
- B$\frac{15}{36}$
- C$\frac{13}{36}$
- D$\frac{23}{36}$
✓
Answer
Correct option: A.
$\frac{19}{36}$
a
$=$ exactly two digits are odd $+$ exactly there 3 digits are odd
$=$ exactly two digits are odd $+$ exactly there 3 digits are odd
For exactly three digits are odd
For exactly two digits odd :
If $0$ is used then $: 2 \times 5 \times 5=50$
If $0$ is not used then : ${ }^{3} C _{1} \times 4 \times 5 \times 5=300$
Required Probability $=\frac{475}{900}=\frac{19}{36}$
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