KCET · Maths · Probability
A bag contains \( 17 \) tickets numbered from \( 1 \) to \( 17 \). A ticket is drawn at random, then another
ticket isdrawn without replacing the first one. The probability that both the tickets may show
even numbers is
- A \( \frac{7}{34} \)
- B \( \frac{8}{17} \)
- C \( \frac{7}{16} \)
- D \( \frac{7}{17} \)
Answer & Solution
Correct Answer
(A) \( \frac{7}{34} \)
Step-by-step Solution
Detailed explanation
Number of tickets are \( 17 \)
Even number balls are \( 2,4,6,8,10,12,14,16 \). So, number of even balls are \( 8 \).
\( P \) (both tickets show even number)
\( =\frac{8}{17} \times \frac{7}{16}=\frac{7}{34} \)
Even number balls are \( 2,4,6,8,10,12,14,16 \). So, number of even balls are \( 8 \).
\( P \) (both tickets show even number)
\( =\frac{8}{17} \times \frac{7}{16}=\frac{7}{34} \)
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