How to directly calculate $P(A)^{\prime}$?

Question

Six different coloured balls are placed in a box. Kendra and Abdul each select a ball without replacement. What is the probability that Kendra does not select the green ball and Abdul does not select the red ball?

The indirect method of solving the aforementioned question is realizing that the scenario described is one of the thirty possible scenarios; so the answer is $$1-\frac{1}{30}=\frac{29}{30}.$$

For a direct method for solving this question, I thought to multiply the probabilities of Kendra not selecting a green ball and Abdul not selecting a red ball together. After all, $P(A\cap B) = P(A)\times P(B).$ However, when we substitute each probability into the aforementioned equation, the answer is $$\frac{5}{6}\times\frac{4}{5}=\frac{2}{3}.$$ Why isn't the direct method giving us the correct answer?

Answer 1

Six different coloured balls are placed in a box. Kendra and Abdul each select a ball without replacement. What is the probability that Kendra does not select the green ball and Abdul does not select the red ball?

Without loss of generality, let's assign trials $1$ and $2$ to Kendra and Abdul, respectively.

The indirect method of solving the aforementioned question is realizing that the scenario described is one of the thirty possible scenarios; so the answer is $$1-\frac{1}{30}=\frac{29}{30}.$$

There are indeed $30$ possible outcomes, but the described event's complement contains $9$ outcomes, including $(r, b)$ and $(y, b),$ rather than just one.

For a direct method for solving this question, I thought to multiply the probabilities of Kendra not selecting a green ball and Abdul not selecting a red ball together. After all, $P(A\cap B) = P(A)\times P(B).$

This formula doesn't apply, because Kendra selecting Green and Abdul selecting Red are not independent: knowing that Kendra selects Green increases the original $\frac16$ probability of Abdul selecting Red. The correct answer: $$P(R_1)\,P(R_2'\mid R_1)+P(G_1'\cap R_1')\,P(R_2'\mid G_1'\cap R_1')=\frac16\times\frac55+\frac46\times\frac45=\frac{7}{10}.$$

Answer 2

Another approach, which agrees with ryang's answer, is as follows.

Let $X$ be the event that we are interested in, and let $Y$ be the event that neither the red ball nor the green ball are selected at all.

Now clearly $P(X\mid Y)=1$. Also we have $P(X\mid Y')=1/2$, since (provided at least one ball is red or green) swapping balls will change whether $X$ occurs. So $$P(X)=P(X\mid Y)P(Y)+P(X\mid Y')P(Y')=\frac{P(Y)+1}{2}.$$

Now $P(Y)=\frac 46\times \frac35=\frac4{10}$ giving $P(X)=\frac7{10}$.