Out of 21 tickets marked with numbers from 1 | vedclass

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(A) Total number of ways to draw $3$ tickets out of $21$ is given by ${}^{21}C_3 = \frac{21 	imes 20 	imes 19}{3 	imes 2 	imes 1} = 1330$. For the

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$\frac{10}{133}$

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(A) Total number of ways to draw $3$ tickets out of $21$ is given by ${}^{21}C_3 = \frac{21 	imes 20 	imes 19}{3 	imes 2 	imes 1} = 1330$. For the numbers to be in $A.P.$,let the numbers be $a, a+d, a+2d$. Since the numbers are between $1$ and $21$,we have $1 \le a This implies $a+2d \le 21$. If $d=1$,$a+2 \le 21 \implies a \le 19$. There are $19$ such sets. If $d=2$,$a+4 \le 21 \implies a \le 17$. There are $17$ such sets. If $d=k$,$a+2k \le 21 \implies a \le 21-2k$. The maximum value of $d$ is $10$ (since $a+20 \le 21 \implies a=1$). Total number of favourable cases $= 19 + 17 + 15 + \dots + 1$. This is an arithmetic progression with $10$ terms,where the first term is $19$ and the last term is $1$. Sum $= \frac{10}{2}(19+1) = 5 	imes 20 = 100$. Required probability $= \frac{100}{1330} = \frac{10}{133}$.

Out of $21$ tickets marked with numbers from $1$ to $21$,three are drawn at random. The chance that the numbers on them are in $A.P.$ is

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