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(A) Let $H$ be the number of heads and $T$ be the number of tails in $6$ tosses. Since the total number of tosses is $6$,we have $H + T = 6$,which imp

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$0, 2, 4, 6$

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(A) Let $H$ be the number of heads and $T$ be the number of tails in $6$ tosses. Since the total number of tosses is $6$,we have $H + T = 6$,which implies $T = 6 - H$.The random variable $X$ is defined as the absolute difference between the number of heads and tails: $X = |H - T|$.Substituting $T = 6 - H$,we get $X = |H - (6 - H)| = |2H - 6|$.Since $H$ can take values from $0$ to $6$,we calculate $X$ for each case:If $H = 0, T = 6$,then $X = |0 - 6| = 6$.If $H = 1, T = 5$,then $X = |1 - 5| = 4$.If $H = 2, T = 4$,then $X = |2 - 4| = 2$.If $H = 3, T = 3$,then $X = |3 - 3| = 0$.If $H = 4, T = 2$,then $X = |4 - 2| = 2$.If $H = 5, T = 1$,then $X = |5 - 1| = 4$.If $H = 6, T = 0$,then $X = |6 - 0| = 6$.Thus,the set of possible values for $X$ is $\{0, 2, 4, 6\}$.

Let $X$ represent the difference between the number of heads and the number of tails obtained when a coin is tossed $6$ times. What are the possible values of $X$?

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