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An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" () and "tails" () which we write , , etc. For each outcome, let be the random variable counting the number of tails in each outcome. For example, if the outcome is , then . Suppose that the random variable is defined in terms of as follows: . The values of are given in the table below.

An ordinary fair coin is tossed 3 times Outcomes are thus triples of heads and tails which we write etc For each outcome let be the random variable counting the class=

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SOLUTION

Consider the table given below.

From the table above, the random value x of x are

[tex]X(x)=-4,\text{ -2 and 4}[/tex]

To obtain the probability distribution for each value of x, we use the probability formula where,

[tex]\text{Total outcome=8}[/tex]

For X(x)= -4, we have

[tex]\begin{gathered} -4\text{ occurs thr}ee\text{ times, } \\ P_X(x)=\frac{\text{ number of occurence}}{Total\text{ outcome }} \end{gathered}[/tex]

Then

[tex]P_X(-4)=\frac{3}{8}[/tex]

Also, for X(x)= -2, from the table we have

[tex]\begin{gathered} -2\text{ occurs four times } \\ \text{Hence } \\ P_X(-2)=\frac{\text{ Number of occurence}}{total\text{ number}} \\ \\ \end{gathered}[/tex]

Then

[tex]P_X(-2)=\frac{4}{8}=\frac{1}{2}[/tex]

Similarly, value X(x)=4

4 occur once, hence

[tex]\begin{gathered} P_x(x)=\text{ }\frac{\text{number of occurence}}{total\text{ outcome }} \\ \text{Then} \\ P_x(x)=\frac{1}{8} \end{gathered}[/tex]

The solution is given in the image below:

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