Respuesta :
The expected value of game is $ [tex]\frac{-3}{8}[/tex]
Solution:
Given, You toss 4 coins and if you toss all heads you win $9.
When you toss anything other than 4 heads you lose $1
We have to find what is the Expected Value of the game?
The expected value is given as:
[tex]\text { expected value }=\text { probability of winning } \times \text { amount won }+\text { probability of losing } \times \text { amount lost. }[/tex]
since we toss 4 coins, total outcomes [tex]= 2 \times 2 \times 2 \times 2[/tex]
[tex]\text {Expected value}=\frac{1}{2 \times 2 \times 2 \times 2} \times 9+\frac{2 \times 2 \times 2 \times 2-1}{2 \times 2 \times 2 \times 2} \times(-1)[/tex]
[tex]\begin{array}{l}{\text { Expected value }=\frac{1}{16} \times 9+\frac{16-1}{16}(-1)} \\\\ {=\frac{9}{16}+\frac{15}{16}(-1)} \\\\ {=\frac{9}{16}-\frac{15}{16}=\frac{-6}{16}=\frac{-3}{8}}\end{array}[/tex]
Hence, the expected value is [tex]\frac{-3}{8}[/tex]
