A bag contain $30$ tickets numbered from $1, 2, 3, \dots, 30$ of which four are drawn at random and arranged in ascending order $(t_{1} < t_{2} < t_{3} < t_{4})$ . Then the probability of $t_{3}$ being $18$ is
(correct answer + 1, wrong answer - 0.25)

\frac{324}{6289}

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\frac{544}{9135}

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\frac{414}{9139}

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\frac{613}{9372}

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Solution

The correct option is B $\frac{544}{9135}$
Total number of ways of drawning $4$ tickets
$=^{30} C_{4}$

If $t_{3} = 18$ , then $t_{1}$ and $t_{2}$ must be from $1$ to $17$ ,
Number of ways $=^{17} C_{2}$
Also, $t_{4}$ must be from $19$ to $30$ ,
Number of ways $=^{12} C_{1}$

Required probability
$= \frac{^{17} C_{2} \times^{12} C_{1}}{^{30} C_{4}} = \frac{24 \times 17 \times 16 \times 12}{30 \times 29 \times 28 \times 27 \times 2} = \frac{4 \cdot 17 \cdot 4 \cdot 2}{5 \cdot 29 \cdot 7 \cdot 9} = \frac{544}{9135}$

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A bag contain 30 tickets numbered from 1,2,3,…,30 of which four are drawn at random and arranged in ascending order (t1<t2<t3<t4). Then the probability of t3 being 18 is ​​​​​​​(correct answer + 1, wrong answer - 0.25)

A bag contain $30$ tickets numbered from $1, 2, 3, \dots, 30$ of which four are drawn at random and arranged in ascending order $(t_{1} < t_{2} < t_{3} < t_{4})$ . Then the probability of $t_{3}$ being $18$ is
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