To solve this problem, we need to find the probability that John either hits the Bull's eye or scores 10 points. Let's denote the following probabilities:
1. The probability that John hits the Bull's eye: [tex]\( P(B) = \frac{1}{10} \)[/tex].
2. The probability that John scores 10 points: [tex]\( P(T) = \frac{3}{10} \)[/tex].
Since hitting the Bull's eye and hitting the 10-point ring are mutually exclusive events (John can't hit both with a single throw), we can use the addition rule for probabilities of mutually exclusive events. This rule states that if two events are mutually exclusive, the probability of either event occurring is the sum of their individual probabilities.
The formula for finding the probability of either of two mutually exclusive events occurring is:
[tex]\[ P(B \cup T) = P(B) + P(T) \][/tex]
Now, we can substitute the probabilities into the formula:
[tex]\[ P(B \cup T) = \frac{1}{10} + \frac{3}{10} \][/tex]
Adding these fractions:
[tex]\[ P(B \cup T) = \frac{1 + 3}{10} = \frac{4}{10} = 0.4 \][/tex]
Therefore, the probability that John either hits a Bull's eye or scores 10 points is 0.4.
Now, let's convert this probability into a fraction to match the given options:
0.4 is equivalent to [tex]\(\frac{4}{10}\)[/tex], which simplifies to [tex]\(\frac{2}{5}\)[/tex].
Given the options:
A. [tex]\(\frac{1}{3}\)[/tex]
B. [tex]\(\frac{2}{3}\)[/tex]
Neither [tex]\(\frac{1}{3}\)[/tex] nor [tex]\(\frac{2}{3}\)[/tex] is equivalent to 0.4 or [tex]\(\frac{2}{5}\)[/tex].
Thus, the correct answer is not listed in the options provided.
