A box contains 100 balls,numbered from 1 to 1 | vedclass

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(A) Let $O$ denote an odd number and $E$ denote an even number. In the set $\{1, 2, \dots, 100\}$,there are $50$ odd and $50$ even numbers. Thus,$P(O)

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $O$ denote an odd number and $E$ denote an even number. In the set $\{1, 2, \dots, 100\}$,there are $50$ odd and $50$ even numbers. Thus,$P(O) = \frac{50}{100} = \frac{1}{2}$ and $P(E) = \frac{50}{100} = \frac{1}{2}$.For the sum of three numbers to be odd,we must have an odd number of odd-numbered balls. The possible cases are:$1$. Three odd balls: $P(O, O, O) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$$2$. One odd and two even balls: The odd ball can be in any of the $3$ positions $(OEE, EOE, EEO)$.$P(O, E, E) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$$P(E, O, E) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$$P(E, E, O) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$Total probability $= \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$.

$A$ box contains $100$ balls,numbered from $1$ to $100$. If $3$ balls are selected one after the other at random with replacement from the box,then the probability that the sum of the three numbers on the balls selected is an odd number,is

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