Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let $E_1$ be the event that the ball chosen belonged to Box I and let $E_2$ be the event that the ball chosen belonged to Box II. Let $F_1$ be the event that the ball chosen is red and let $F_2$ be the event that the ball chosen is green. Then which of the following statements is (are) TRUE ?

$\text{Box I: } 6R,\ 9G \qquad \text{Box II: } 8R,\ 12G$ $\text{All mixed, one ball drawn at random}$ $E_1=\{\text{Box I}\},\ E_2=\{\text{Box II}\},\ F_1=\{\text{red}\},\ F_2=\{\text{green}\}$ $\text{To check: which of (1)-(4) are TRUE}$ $\text{independence / dependence of } E_i, F_j \text{ and conditional probabilities}$ $1.\ 2\times2 \text{ table} \Rightarrow \text{marginals}$ $2.\ \text{independence}: P(A\cap B)=P(A)P(B)$ $3.\ \text{conditional probabilities and compare}$ $\text{Box I: } 15,\quad \text{Box II: } 20,\quad \text{Total } N=35$ $\text{Red}=14,\qquad \text{Green}=21$ $\begin{array}{|c|c|c|c|}\hline & R & G & \text{Tot}\\ \hline \text{Box I} & 6 & 9 & 15\\ \hline \text{Box II} & 8 & 12 & 20\\ \hline \text{Tot} & 14 & 21 & 35\\ \hline \end{array}$ $P(E_1)=\dfrac{15}{35}=\dfrac{3}{7},\qquad P(E_2)=\dfrac{20}{35}=\dfrac{4}{7}$ $P(F_1)=\dfrac{14}{35}=\dfrac{2}{5},\qquad P(F_2)=\dfrac{21}{35}=\dfrac{3}{5}$ $A,\ B \text{ independent} \iff P(A\cap B)=P(A)\,P(B)$ $P(E_1\cap F_1)=P(\text{red from Box I})=\dfrac{6}{35}$ $P(E_1)P(F_1)=\dfrac{3}{7}\cdot\dfrac{2}{5}=\dfrac{6}{35}\ \Rightarrow\ (1)\ \checkmark$ $P(E_2\cap F_2)=P(\text{green from Box II})=\dfrac{12}{35}$ $P(E_2)P(F_2)=\dfrac{4}{7}\cdot\dfrac{3}{5}=\dfrac{12}{35} \Rightarrow \text{independent} \Rightarrow (2)\ \times$ $P(F_1\mid E_1)=\dfrac{P(F_1\cap E_1)}{P(E_1)}=\dfrac{\text{red in Box I}}{\text{total in Box I}}$ $P(F_1\mid E_1)=\dfrac{6}{15}=\dfrac{2}{5}$ $P(F_1\mid E_2)=\dfrac{8}{20}=\dfrac{2}{5}\ \Rightarrow\ (3)\ \checkmark$ $P(F_2\mid E_2)=\dfrac{12}{20}=\dfrac{3}{5}$ $\dfrac{2}{5} > \dfrac{3}{5}\ ?\ \text{No} \Rightarrow (4)\ \times$ $\boxed{(1)\ \text{and}\ (3)\ \text{are TRUE}}$