Out of 40 consecutive natural numbers, two ar | vedclass

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Question

(b) Required probability $ = \frac{{{}^{20}{C_1} 	imes {}^{20}{C_1}}}{{{}^{40}{C_2}}} = \frac{{20 	imes 20 	imes 2}}{{40 	imes 39}} = \frac{{20}}{

Vedclass · Accepted Answer

$\frac{{20}}{{39}}$

Vedclass · Answer

(b) Required probability $ = \frac{{{}^{20}{C_1} 	imes {}^{20}{C_1}}}{{{}^{40}{C_2}}} = \frac{{20 	imes 20 	imes 2}}{{40 	imes 39}} = \frac{{20}}{{39}}.$

Out of $40$ consecutive natural numbers, two are chosen at random. Probability that the sum of the numbers is odd, is

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