A bag contains 19 tickets numbered from 1 to | vedclass

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(D) The total number of tickets is $19$ (numbered $1$ to $19$).The even numbers between $1$ and $19$ are $2, 4, 6, 8, 10, 12, 14, 16, 18$. There are $

Vedclass · Accepted Answer

$\frac{4}{19}$

Vedclass · Answer

(D) The total number of tickets is $19$ (numbered $1$ to $19$).The even numbers between $1$ and $19$ are $2, 4, 6, 8, 10, 12, 14, 16, 18$. There are $9$ even numbers.The probability of drawing an even number in the first draw is $P(E_1) = \frac{9}{19}$.Since the ticket is drawn without replacement,there are now $18$ tickets left in the bag,and $8$ of them are even.The probability of drawing an even number in the second draw,given that the first was even,is $P(E_2|E_1) = \frac{8}{18}$.The probability that both tickets show even numbers is $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2|E_1) = \frac{9}{19} 	imes \frac{8}{18} = \frac{9}{19} 	imes \frac{4}{9} = \frac{4}{19}$.

$A$ bag contains $19$ tickets numbered from $1$ to $19$. $A$ ticket is drawn and then another ticket is drawn without replacement. The probability that both the tickets will show even numbers is:

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