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Let $\mathrm{X}$ represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of $X$ ?
Solution:
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Verified Answer
For one coin, $\mathrm{S}=\{\mathrm{H}, \mathrm{T}\}$
$\mathrm{n}(\mathrm{S})=2$, Let $\mathrm{A}$ represent $\mathrm{Head}$
$$
\begin{gathered}
\mathrm{A}=\{\mathrm{H}\}, \mathrm{n}(\mathrm{A})=1 \\
\mathrm{P}(\mathrm{A})=\frac{1}{2}, P(\bar{A})=\frac{1}{2} \\
\mathrm{n}=6, \mathrm{r}=0,1,2,3,4,5,6 \\
\mathrm{P}(\mathrm{X}=0)=[P(\bar{X})]^6=\left(\frac{1}{2}\right)^6=\frac{1}{64} \\
\mathrm{P}(\mathrm{X}=1)=6 \mathrm{P}(\mathrm{X})[P(\bar{X})]^5=6 \times\left(\frac{1}{2}\right)^6=\frac{6}{64} \\
\mathrm{P}(\mathrm{X}=2)=15[\mathrm{P}(\mathrm{X})]^2[P(\bar{X})]^4=15\left(\frac{1}{2}\right)^6=\frac{15}{64} \\
\mathrm{P}(\mathrm{X}=6)=[\mathrm{P}(\mathrm{X})]^6=\left(\frac{1}{2}\right)^6=\frac{1}{64} \\
\mathrm{P}=3)=20[\mathrm{P}(\mathrm{X})]^3[P(\bar{X})]^3=20\left(\frac{1}{2}\right)^6=\frac{20}{64} \\
\mathrm{P}(\mathrm{X}=4)=15[\mathrm{P}(\mathrm{X})]^4[P(\bar{X})]^2=15\left(\frac{1}{2}\right)^6=\frac{15}{64} \\
\mathrm{P}=6[\mathrm{P}(\mathrm{X})]^5[P(\bar{X})]=6\left(\frac{1}{2}\right)^6=\frac{6}{64} \\
\mathrm{~F}
\end{gathered}
$$
$\mathrm{n}(\mathrm{S})=2$, Let $\mathrm{A}$ represent $\mathrm{Head}$
$$
\begin{gathered}
\mathrm{A}=\{\mathrm{H}\}, \mathrm{n}(\mathrm{A})=1 \\
\mathrm{P}(\mathrm{A})=\frac{1}{2}, P(\bar{A})=\frac{1}{2} \\
\mathrm{n}=6, \mathrm{r}=0,1,2,3,4,5,6 \\
\mathrm{P}(\mathrm{X}=0)=[P(\bar{X})]^6=\left(\frac{1}{2}\right)^6=\frac{1}{64} \\
\mathrm{P}(\mathrm{X}=1)=6 \mathrm{P}(\mathrm{X})[P(\bar{X})]^5=6 \times\left(\frac{1}{2}\right)^6=\frac{6}{64} \\
\mathrm{P}(\mathrm{X}=2)=15[\mathrm{P}(\mathrm{X})]^2[P(\bar{X})]^4=15\left(\frac{1}{2}\right)^6=\frac{15}{64} \\
\mathrm{P}(\mathrm{X}=6)=[\mathrm{P}(\mathrm{X})]^6=\left(\frac{1}{2}\right)^6=\frac{1}{64} \\
\mathrm{P}=3)=20[\mathrm{P}(\mathrm{X})]^3[P(\bar{X})]^3=20\left(\frac{1}{2}\right)^6=\frac{20}{64} \\
\mathrm{P}(\mathrm{X}=4)=15[\mathrm{P}(\mathrm{X})]^4[P(\bar{X})]^2=15\left(\frac{1}{2}\right)^6=\frac{15}{64} \\
\mathrm{P}=6[\mathrm{P}(\mathrm{X})]^5[P(\bar{X})]=6\left(\frac{1}{2}\right)^6=\frac{6}{64} \\
\mathrm{~F}
\end{gathered}
$$
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