A conditional probability uses only the marbles allowed by “given that.” The two conditions create different eligible groups, so their fractions need different denominators. Name the unknown red and blue counts, write one equation for each condition, then graph both equations in Desmos. Build each denominator from the marbles that remain, not from the whole bag.
Hints
- Hint 1
A condition limits which marbles can be picked. For “not black,” keep red, blue, and green, but leave black out of the denominator, the eligible total. How would you write that total?
- Hint 2
The condition changes for the blue probability: “not green” includes black. Keep blue on top of the fraction, and put red, blue, and black on the bottom. What equation does that give?
- Hint 3
Both equations use the same red and blue counts. Their intersection makes both probabilities true. When you read that point in Desmos, which coordinate represents blue?
Step-by-step
Graph the two conditional-probability equations
Step 1Build the not-black probability
Let be the red count and be the blue count. The not-black condition leaves red, blue, and the green marbles eligible. Red is the count you want within that group, so its conditional probability gives
- Step 2
Build the not-green probability
Now green is excluded, but the black marbles are eligible. Blue is the count you want within this new group, so .Don't reuse the first denominator: the two conditions leave different groups.
- Step 3
Read the blue count at the intersection
Type both equations in Desmos, one per line. Click their intersection, the point that makes both equations true. Desmos shows . Since represents blue, the second coordinate gives blue marbles, not the red marbles in the first coordinate. Choice C.