1. **Problem statement:** We have three people: Alice, Bob, and Charlotte, each independently finding a butterfly with probabilities 0.17, 0.25, and 0.45 respectively. Define random variables: - $X$: total number of butterflies caught by all three. - $Y$: number of people who do not catch a butterfly. We want to find the joint probability mass function (pmf) $p_{X,Y}(x,y) = P(X=x, Y=y)$. 2. **Understanding the variables:** - Each person either catches 1 butterfly or 0. - Since there are 3 people, $X$ can be 0,1,2,3. - $Y$ counts how many people catch 0 butterflies, so $Y = 3 - X$ because each person either catches or does not catch a butterfly. 3. **Relation between $X$ and $Y$:** Since $Y = 3 - X$, the joint pmf $p_{X,Y}(x,y)$ is zero unless $y = 3 - x$. 4. **Calculating $p_X(x)$:** Since the people are independent, the pmf of $X$ is the convolution of three Bernoulli variables with parameters $p_A=0.17$, $p_B=0.25$, $p_C=0.45$. 5. **Enumerate all possible outcomes:** Each person can be 0 or 1 butterfly: - Let $A,B,C$ be indicators for Alice, Bob, Charlotte catching a butterfly. - $P(A=1)=0.17$, $P(A=0)=0.83$; similarly for $B$ and $C$. 6. **Calculate probabilities for each $X=x$:** - $X=0$: all fail $$P(X=0) = (1-0.17)(1-0.25)(1-0.45) = 0.83 \times 0.75 \times 0.55 = 0.341625$$ - $X=1$: exactly one succeeds $$P(X=1) = P(A=1,B=0,C=0) + P(A=0,B=1,C=0) + P(A=0,B=0,C=1)$$ $$= 0.17 \times 0.75 \times 0.55 + 0.83 \times 0.25 \times 0.55 + 0.83 \times 0.75 \times 0.45$$ $$= 0.070125 + 0.114125 + 0.279375 = 0.463625$$ - $X=2$: exactly two succeed $$P(X=2) = P(A=1,B=1,C=0) + P(A=1,B=0,C=1) + P(A=0,B=1,C=1)$$ $$= 0.17 \times 0.25 \times 0.55 + 0.17 \times 0.75 \times 0.45 + 0.83 \times 0.25 \times 0.45$$ $$= 0.023375 + 0.057375 + 0.093375 = 0.174125$$ - $X=3$: all succeed $$P(X=3) = 0.17 \times 0.25 \times 0.45 = 0.019125$$ 7. **Joint pmf $p_{X,Y}(x,y)$:** Since $Y=3-X$, $$p_{X,Y}(x,y) = \begin{cases} P(X=x) & \text{if } y=3-x \\ 0 & \text{otherwise} \end{cases}$$ 8. **Summary table:** | $x$ | $y$ | $p_{X,Y}(x,y)$ | |-----|-----|----------------| | 0 | 3 | 0.341625 | | 1 | 2 | 0.463625 | | 2 | 1 | 0.174125 | | 3 | 0 | 0.019125 | This completes the joint pmf. **Final answer:** $$p_{X,Y}(x,y) = \begin{cases} 0.341625 & (x,y)=(0,3) \\ 0.463625 & (x,y)=(1,2) \\ 0.174125 & (x,y)=(2,1) \\ 0.019125 & (x,y)=(3,0) \\ 0 & \text{otherwise} \end{cases}$$