To solve this problem, we begin by understanding the given probabilities:
1. The probability of drawing a blue card, [tex]\( P(\text{blue}) \)[/tex], is given by:
[tex]\[
P(\text{blue}) = \frac{5}{9}
\][/tex]
2. The probability of drawing a white card, [tex]\( P(\text{white}) \)[/tex], is given by:
[tex]\[
P(\text{white}) = \frac{1}{18}
\][/tex]
Next, we need to calculate the probability of drawing either a blue card or a white card. This is done by summing the probabilities of the two events since they are mutually exclusive (a card cannot be both blue and white at the same time):
3. The probability of drawing either a blue or a white card, [tex]\( P(\text{blue or white}) \)[/tex], is:
[tex]\[
P(\text{blue or white}) = P(\text{blue}) + P(\text{white}) = \frac{5}{9} + \frac{1}{18}
\][/tex]
To add these fractions, we need a common denominator. The least common multiple of 9 and 18 is 18:
[tex]\[
\frac{5}{9} = \frac{5 \times 2}{9 \times 2} = \frac{10}{18}
\][/tex]
Thus,
[tex]\[
P(\text{blue or white}) = \frac{10}{18} + \frac{1}{18} = \frac{10 + 1}{18} = \frac{11}{18}
\][/tex]
Finally, we calculate the probability of drawing a card that is neither blue nor white. This is the complement of drawing either a blue or a white card:
4. The probability of neither blue nor white is:
[tex]\[
P(\text{neither blue nor white}) = 1 - P(\text{blue or white})
\][/tex]
Substituting the value we found for [tex]\( P(\text{blue or white}) \)[/tex]:
[tex]\[
P(\text{neither blue nor white}) = 1 - \frac{11}{18}
\][/tex]
Writing [tex]\( 1 \)[/tex] as a fraction with a denominator of 18:
[tex]\[
1 = \frac{18}{18}
\][/tex]
Thus,
[tex]\[
P(\text{neither blue nor white}) = \frac{18}{18} - \frac{11}{18} = \frac{18 - 11}{18} = \frac{7}{18}
\][/tex]
Therefore, the probability of picking a card that is neither blue nor white is:
[tex]\[
P(\text{neither blue nor white}) = \frac{7}{18}
\][/tex]
Thus, [tex]\( P(\text{neither blue nor white}) \)[/tex] is [tex]\( \boxed{\frac{7}{18}} \)[/tex].
