(A) Let $N$ be the number appearing on the die. The possible values for $N$ are $\{1, 2, 3, 4, 5, 6\}$.If $N$ is even $(N \in \{2, 4, 6\})$, the numbe

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(A) Let $N$ be the number appearing on the die. The possible values for $N$ are $\{1, 2, 3, 4, 5, 6\}$.If $N$ is even $(N \in \{2, 4, 6\})$, the number of chocolates $X = N + 2$.For $N = 2$, $X = 2 + 2 = 4$.For $N = 4$, $X = 4 + 2 = 6$.For $N = 6$, $X = 6 + 2 = 8$.If $N$ is odd $(N \in \{1, 3, 5\})$, the number of chocolates $X = N + 3$.For $N = 1$, $X = 1 + 3 = 4$.For $N = 3$, $X = 3 + 3 = 6$.For $N = 5$, $X = 5 + 3 = 8$.Thus, the set of all possible values of $X$ is $\{4, 6, 8\}$.Therefore, the range of $X$ is $\{4, 6, 8\}$.

Class 12 Mathematics Probability Probability distribution

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$A$ boy rolls a die once. If an even number appears, the number of chocolates the boy gets is equal to two more than the number that appeared. If an odd number appears on the die, the number of chocolates he gets is equal to three more than the number that appeared. If a random variable $X$ represents the number of chocolates the boy receives, then the range of $X$ is:

TS EAMCET 2019 All TS EAMCET

A
$\{4, 6, 8\}$
B
$\{3, 5, 7\}$
C
$\{3, 4, 7\}$
D
$\{2, 3\}$

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$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

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