1. **Problem statement:** We have three independent people (Alice, Bob, Charlotte) each catching a butterfly with given probabilities. Define $X$ as the total number of butterflies caught and $Y$ as the number of people who do not catch a butterfly. We want the joint mass function $p_{X,Y}(x,y) = P(X=x, Y=y)$. 2. **Given probabilities:** - $P(Alice=1) = 0.17$, $P(Alice=0) = 0.83$ - $P(Bob=1) = 0.25$, $P(Bob=0) = 0.75$ - $P(Charlotte=1) = 0.45$, $P(Charlotte=0) = 0.55$ 3. **Important rules:** - Since catches are independent, joint probabilities multiply. - $X$ can be $0,1,2,3$ (total butterflies caught). - $Y$ can be $0,1,2,3$ (people who do not catch butterflies). - Note that $X + Y = 3$ always, since each person either catches or does not catch exactly one butterfly. 4. **Calculate joint probabilities:** We consider all 8 possible outcomes (each person catches or not): | Alice | Bob | Charlotte | $X$ | $Y$ | Probability | |-------|-----|-----------|-----|-----|-------------| | 0 | 0 | 0 | 0 | 3 | $0.83 \times 0.75 \times 0.55 = 0.341625$ | | 0 | 0 | 1 | 1 | 2 | $0.83 \times 0.75 \times 0.45 = 0.279375$ | | 0 | 1 | 0 | 1 | 2 | $0.83 \times 0.25 \times 0.55 = 0.114125$ | | 0 | 1 | 1 | 2 | 1 | $0.83 \times 0.25 \times 0.45 = 0.093375$ | | 1 | 0 | 0 | 1 | 2 | $0.17 \times 0.75 \times 0.55 = 0.070125$ | | 1 | 0 | 1 | 2 | 1 | $0.17 \times 0.75 \times 0.45 = 0.057375$ | | 1 | 1 | 0 | 2 | 1 | $0.17 \times 0.25 \times 0.55 = 0.023375$ | | 1 | 1 | 1 | 3 | 0 | $0.17 \times 0.25 \times 0.45 = 0.019125$ | 5. **Group probabilities by $(X,Y)$:** - $p_{X,Y}(0,3) = 0.341625$ - $p_{X,Y}(1,2) = 0.279375 + 0.114125 + 0.070125 = 0.463625$ - $p_{X,Y}(2,1) = 0.093375 + 0.057375 + 0.023375 = 0.174125$ - $p_{X,Y}(3,0) = 0.019125$ 6. **Final joint mass function:** $$ p_{X,Y}(x,y) = \begin{cases} 0.341625 & (x,y) = (0,3) \\ 0.463625 & (x,y) = (1,2) \\ 0.174125 & (x,y) = (2,1) \\ 0.019125 & (x,y) = (3,0) \\ 0 & \text{otherwise} \end{cases} $$ This completes the solution.