A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contai...
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A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered \( 2,4,6,7,8,9 \). One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4 , is equal to
(A) \( \frac{5}{12} \)
(B) \( \frac{5}{24} \)
(C) \( \frac{7}{12} \)
(D) \( \frac{19}{24} \)
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