This solution details the steps to calculate the probability of drawing exactly one white ball when two balls are drawn simultaneously from a bowl containing balls of different colors.
The total number of ways to choose 2 balls from the 21 available balls is determined using combinations ($C(n, k)$).
The formula for combinations is $C(n, k) = \frac{n!}{k!(n-k)!}$.
For this problem, $n=21$ (total balls) and $k=2$ (balls drawn).
Total ways = $C(21, 2) = \frac{21!}{2!(21-2)!} = \frac{21!}{2!19!} = \frac{21 \times 20}{2 \times 1} = 210$.
There are 210 possible ways to draw 2 balls from the bowl.
Favorable outcomes involve drawing exactly one white ball and one non-white ball.
The total number of favorable outcomes is the product of these two combinations:
Favorable outcomes = $C(7, 1) \times C(14, 1) = 7 \times 14 = 98$.
Probability is calculated as the ratio of favorable outcomes to total possible outcomes.
Probability = $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{98}{210}$
To simplify the fraction $\frac{98}{210}$, we can divide both the numerator and the denominator by their greatest common divisor, which is 14.
$\frac{98 \div 14}{210 \div 14} = \frac{7}{15}$
The probability of drawing exactly one white ball is $\frac{7}{15}$.
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