Answer:
0.1935 or 19.35%
Step-by-step explanation:
If only one shot hit the target, two other people missed the target. The probability that only one person hit the target is:
[tex]P(X=1) =P(Only\ A)+P(Only\ B)+P(Only\ C)\\P(X=1) =\frac{1}{6}*\frac{3}{4}* \frac{2}{3} +\frac{5}{6}*\frac{1}{4}* \frac{2}{3} +\frac{5}{6}*\frac{3}{4}* \frac{1}{3} \\P(X=1) = 0.4306[/tex]
The probability that only Alice hit the shot is:
[tex]P(Only\ A)=\frac{1}{6}*\frac{3}{4}* \frac{2}{3}\\P(Only\ A) = 0.08333[/tex]
Therefore, the probability that Alice's shot hit the target, given that only one shot hit the target:
[tex]P(Only\ A|X=1) = \frac{P(Only\ A)}{P(X=1)} \\P(Only\ A|X=1) =\frac{0.08333}{0.4306}\\P(Only\ A|X=1) =0.1935=19.35\%[/tex]
The probability is 0.1935 or 19.35%.
