A box contains 9 tickets numbered 1 to 9 both | vedclass

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(D) Total tickets = $9$. Odd numbers are ${1, 3, 5, 7, 9}$ ($5$ tickets) and even numbers are ${2, 4, 6, 8}$ ($4$ tickets).Case $1$: The sequence is {

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$\frac{5}{18}$

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(D) Total tickets = $9$. Odd numbers are ${1, 3, 5, 7, 9}$ ($5$ tickets) and even numbers are ${2, 4, 6, 8}$ ($4$ tickets).Case $1$: The sequence is {odd,even,odd}.The probability is $P(O_1 \cap E_2 \cap O_3) = \frac{5}{9} 	imes \frac{4}{8} 	imes \frac{4}{7} = \frac{80}{504} = \frac{10}{63}$.Case $2$: The sequence is {even,odd,even}.The probability is $P(E_1 \cap O_2 \cap E_3) = \frac{4}{9} 	imes \frac{5}{8} 	imes \frac{3}{7} = \frac{60}{504} = \frac{7.5}{63} = \frac{5}{42}$.Total probability = $\frac{10}{63} + \frac{5}{42} = \frac{20 + 15}{126} = \frac{35}{126} = \frac{5}{18}$.

$A$ box contains $9$ tickets numbered $1$ to $9$ both inclusive. If $3$ tickets are drawn from the box one at a time without replacement,then the probability that they are alternatively either {odd,even,odd} or {even,odd,even} is

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