Cards with numbers 2 to 101 are placed in a b | vedclass

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(A) Total number of outcomes with numbers $2$ to $101$ is $n(S) = 101 - 2 + 1 = 100$.$(i)$ Let $E_1$ be the event of selecting a card which is an even

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$\frac{1}{2} , \frac{9}{100}$

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(A) Total number of outcomes with numbers $2$ to $101$ is $n(S) = 101 - 2 + 1 = 100$.$(i)$ Let $E_1$ be the event of selecting a card which is an even number. The even numbers are $\{2, 4, 6, \dots, 100\}$.This is an arithmetic progression where $a = 2$,$d = 2$,and $l = 100$.Using $l = a + (n - 1)d$,we get $100 = 2 + (n - 1)2$,which implies $98 = 2(n - 1)$,so $n - 1 = 49$,and $n = 50$.Thus,$n(E_1) = 50$.The probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{50}{100} = \frac{1}{2}$.$(ii)$ Let $E_2$ be the event of selecting a card which is a square number.The square numbers between $2$ and $101$ are $\{4, 9, 16, 25, 36, 49, 64, 81, 100\}$.These correspond to $\{2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2, 10^2\}$.Thus,$n(E_2) = 9$.The probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{9}{100}$.

Cards with numbers $2$ to $101$ are placed in a box. $A$ card is selected at random. Find the probability that the card has
$(i)$ an even number
$(ii)$ a square number

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Cards with numbers $2$ to $101$ are placed in a box. $A$ card is selected at random. Find the probability that the card has$(i)$ an even number$(ii)$ a square number