Answer
The probability that someone will be eliminated on the first toss is \[\frac{3}{4}\].
Work Step by Step
When we toss three coins simultaneously, then the possible sample space is
\[S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}\]
Therefore, total numbers of outcome are \[{{2}^{3}}=8\]and the probability of each outcome is \[\frac{1}{8}\]
Let event A denote someone will be eliminated on the first toss; that means there is a possibility that either two heads and one tail or two tails and one head.
Therefore, the 6 possible outcomes of event A are:
\[S=\{HHT,HTH,HTT,THH,THT,TTH\}\]
The probability that someone will be eliminated on the first toss is
\[\begin{align}
& P(A)=P(HHT)+P(HTH)+P(HTT)+PH(THH)+P(HTH)+P(TT) \\
& P(A)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8} \\
& P(A)=\frac{6}{8} \\
\end{align}\]
The required probability is \[\frac{6}{8}\].
