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

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14.Probability

hard

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

$\frac{23}{36}$

(JEE MAIN-2022)

Solution

$=$ 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 $=\frac{475}{900}=\frac{19}{36}$

Standard 11

Mathematics

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