Question

Write the probability distribution when three coins are tossed.

• A. $X:\begin{array}{cccc}0& 1& 2& 3\end{array}P\left(X\right):\begin{array}{cccc}\frac{1}{8}& \frac{3}{8}& \frac{3}{8}& \frac{1}{8}\end{array}$
• B. $X:\begin{array}{cccc}0& 1& 2& 3\end{array}P\left(X\right):\begin{array}{cccc}\frac{1}{8}& \frac{3}{8}& \frac{5}{8}& \frac{7}{8}\end{array}$
• C. $X:\begin{array}{cccc}0& 1& 2& 3\end{array}P\left(X\right):\begin{array}{cccc}\frac{7}{8}& \frac{5}{8}& \frac{3}{8}& \frac{1}{8}\end{array}$
• D. $X:\begin{array}{cccc}0& 1& 2& 3\end{array}P\left(X\right):\begin{array}{cccc}\frac{1}{8}& \frac{3}{8}& \frac{5}{8}& \frac{1}{8}\end{array}$

Answer 1

The correct option is A $X:\begin{array}{cccc}0& 1& 2& 3\end{array}P\left(X\right):\begin{array}{cccc}\frac{1}{8}& \frac{3}{8}& \frac{3}{8}& \frac{1}{8}\end{array}$

Assume $X=$number of heads for the coin.
Sample space$=\{HHH,HHT,HTT,TTT,HTH,THT,THH,TTH\}$
Hence,$n\left(S\right)=8$
For no heads,we have $n(X=0)=1\Rightarrow P(X=0)=\frac{1}{8}$

For one heads,we have $n(X=1)=3\Rightarrow P(X=1)=\frac{3}{8}$

For two heads,we have $n(X=2)=3\Rightarrow P(X=2)=\frac{3}{8}$
similarly,for three heads,we get $n(X=3)=1\Rightarrow P(X=3)=\frac{1}{8}$