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(D) When a coin is tossed three times,the total number of outcomes is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(D) When a coin is tossed three times,the total number of outcomes is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.Let $n(H)$ be the number of heads and $n(T)$ be the number of tails. $X = |n(H) - n(T)|$.For $HHH: n(H)=3, n(T)=0, X=|3-0|=3$.For $HHT: n(H)=2, n(T)=1, X=|2-1|=1$.For $HTH: n(H)=2, n(T)=1, X=|2-1|=1$.For $HTT: n(H)=1, n(T)=2, X=|1-2|=1$.For $THH: n(H)=2, n(T)=1, X=|2-1|=1$.For $THT: n(H)=1, n(T)=2, X=|1-2|=1$.For $TTH: n(H)=1, n(T)=2, X=|1-2|=1$.For $TTT: n(H)=0, n(T)=3, X=|0-3|=3$.The outcomes where $X=1$ are $\{HHT, HTH, HTT, THH, THT, TTH\}$.There are $6$ such outcomes.Therefore,$P(X=1) = \frac{6}{8} = \frac{3}{4}$.

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

$A$ coin is tossed three times. If $X$ denotes the absolute difference between the number of heads and the number of tails,then $P(X=1) = $

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