Question

Cards numbered 10, 11, 12, ..., 30 are kept in a box and shuffled thoroughly. Rahit draws a card at random from the box. The probability that the number on the card is a multiple of 6 or 5 is:

• A. \( \dfrac{9}{20} \)
• B. \( \dfrac{9}{21} \)
• C. \( \dfrac{10}{20} \)
• D. \( \dfrac{10}{21} \)

Hint:

Always use inclusion-exclusion principle when asked about "multiples of A or B".

Answer 1

The Correct Option is D

Total numbers: \[ 30 - 10 + 1 = 21 \]
Multiples of 5: 10, 15, 20, 25, 30 \( \Rightarrow 5 \text{ numbers} \)
Multiples of 6: 12, 18, 24, 30 \( \Rightarrow 4 \text{ numbers} \)
Common multiples (both 5 and 6): 30 \( \Rightarrow 1 \text{ number} \)
By inclusion-exclusion:
\[ n(5 \cup 6) = n(5) + n(6) - n(5 \cap 6) = 5 + 4 - 1 = 8 \]
\[ \Rightarrow \text{Probability} = \frac{8}{21} \]
But none of the options say \( \frac{8}{21} \), so let's recheck.
Verification:
Multiples of 5 → 5 numbers
Multiples of 6 → 4 numbers
Common → 1 number
So total = \( 5 + 4 - 1 = 8 \)
\[ \therefore \text{Probability} = \frac{8}{21} \]
Since that’s not in options, likely correct option is missing or misprinted.
Best fit answer: \( \frac{8}{21} \)