A bag contains $12$ tickets numbered $1, 2, 3, . . ., 12$ .Five of them are drawn at random and arranged in the ascending order of their numbers . What is the probability that the third in the order is $5$ ?

\frac{7}{44}

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\frac{5}{12}

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\frac{2}{3}

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\frac{1}{2}

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Solution

The correct option is A $\frac{7}{44}$
Total number of cases $=^{12} C_{5}$
If five tickets are arranged in ascending order and third ticket is $5$ ,
then tickets are $a_{1}, a_{2}, 5, a_{4}, a_{5}$ where $a_{1}, a_{2} \in {1, 2, 3, 4}$ and $a_{4}, a_{5} \in {6, 7, 8, . . ., 12}$
Number of favorable cases $=^{4} C_{2} \times^{7} C_{2}$ .
$∴$ Required probability $= \frac{^{4} C_{2} \times^{7} C_{2}}{^{12} C_{5}} = \frac{7}{44}$

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