A bag contains 17 tickets numbered from 1 | vedclass

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Question

(A) The total number of tickets is $ 17 $. The even-numbered tickets are $ 2, 4, 6, 8, 10, 12, 14, 16 $. Thus,the total number of even-numbered ticket

Vedclass · Accepted Answer

$ \frac{7}{34} $

Vedclass · Answer

(A) The total number of tickets is $ 17 $. The even-numbered tickets are $ 2, 4, 6, 8, 10, 12, 14, 16 $. Thus,the total number of even-numbered tickets is $ 8 $. Let $ E_1 $ be the event that the first ticket drawn is even,and $ E_2 $ be the event that the second ticket drawn is even. The probability of drawing an even ticket first is $ P(E_1) = \frac{8}{17} $. Since the ticket is not replaced,the remaining number of tickets is $ 16 $,and the number of remaining even tickets is $ 7 $. The conditional probability of drawing an even ticket second is $ P(E_2|E_1) = \frac{7}{16} $. The probability that both tickets show even numbers is $ P(E_1 \cap E_2) = P(E_1) 	imes P(E_2|E_1) = \frac{8}{17} 	imes \frac{7}{16} = \frac{7}{34} $.

$A$ bag contains $ 17 $ tickets numbered from $ 1 $ to $ 17 $. $A$ ticket is drawn at random,then another ticket is drawn without replacing the first one. The probability that both the tickets show even numbers is

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