The problem describes drawing two balls without replacement from a bag containing 7 red (R), 4 white (W), and 6 blue (B) balls, for a total of 7+4+6=17 balls.

The problem describes drawing two balls without replacement from a bag containing 7 red (R), 4 white (W), and 6 blue (B) balls, for a total of $7+4+6=17$ balls.
The tree diagram shows the probabilities for the first draw and the conditional probabilities for the second draw.

First Draw:

• $P(R_1) = \frac{7}{17}$
• $P(W_1) = \frac{4}{17}$
• $P(B_1) = \frac{6}{17}$

Second Draw (without replacement):

• If $R_1$: $P(R_2|R_1) = \frac{6}{16}$, $P(W_2|R_1) = \frac{4}{16}$, $P(B_2|R_1) = \frac{6}{16}$
• If $W_1$: $P(R_2|W_1) = \frac{7}{16}$, $P(W_2|W_1) = \frac{3}{16}$, $P(B_2|W_1) = \frac{6}{16}$
• If $B_1$: $P(R_2|B_1) = \frac{7}{16}$, $P(W_2|B_1) = \frac{4}{16}$, $P(B_2|B_1) = \frac{5}{16}$

The tree diagram in the image correctly illustrates these probabilities.

Use the tree diagram to find the probability that

This occurs if the sequence of draws is (Red, Red), (White, Red), or (Blue, Red).

This occurs if the sequence of draws is (Red, Red), (White, White), or (Blue, Blue).

This means both balls drawn are either red or blue. The possible sequences are (Red, Red), (Red, Blue), (Blue, Red), or (Blue, Blue).

It is easier to calculate the complement: $1 - P(\text{no blue ball})$.
"No blue ball" means both balls drawn are either red or white. The possible sequences are (Red, Red), (Red, White), (White, Red), or (White, White).