SL 4.11—Conditional and independent probabilities, test for independence

Let $A$ = 'second ball is blue' and $B$ = 'first ball is red'. We want $P(A|B)$. The bag starts with 4 red, 3 blue, 3 green = 10 balls total.

Given that the first ball drawn was red, there are now $10-1=9$ balls remaining. The composition is 3 red, 3 blue, and 3 green — the red count decreased by 1.

$P(A|B)=\frac{\text{blue balls remaining}}{\text{total balls remaining}}=\frac{3}{9}=\frac{1}{3}$

The probability that the second ball is blue, given the first was red, is $\frac{1}{3}$.

We need $P(\text{2nd blue}|\text{1st red})$. After removing one red ball, 9 balls remain: 3 red, 3 blue, 3 green.

The option $\frac{3}{10}$ uses the original total of 10 balls, ignoring that one ball has already been removed. This does not account for drawing without replacement.

The option $\frac{3}{30}$ appears to be a joint probability calculation rather than a conditional probability. This is not what is being asked.

Both $\frac{3}{9}$ and $\frac{1}{3}$ are equivalent. However, $\frac{1}{3}$ is the fully simplified form. The answer written as $\frac{3}{9}$ is the same value but unsimplified — the correct answer in simplified form is $\frac{1}{3}$.

After removing one red ball, 3 out of 9 remaining balls are blue, giving $P=\frac{3}{9}=\frac{1}{3}$.

Event $B$ = 'at least one die shows a 3'. Using complementary counting: total outcomes = 36, outcomes with no 3 = $5\times 5=25$. So $|B|=36-25=11$.

Event $A$ = 'sum equals 9'. Pairs summing to 9: $(3,6),(4,5),(5,4),(6,3)$. Of these, those containing at least one 3: $(3,6)$ and $(6,3)$. So $|A\cap B|=2$... wait, also check $(3,6)$: sum = 9 ✓, has a 3 ✓. And $(6,3)$: sum = 9 ✓, has a 3 ✓. Also consider if any pair with a 3 sums to 9: $(3,6)$ and $(6,3)$. That gives $|A\cap B|=2$. But also $(3,6),(6,3)$ — let me recount: pairs with at least one 3 summing to 9 are $(3,6)$ and $(6,3)$, so $|A\cap B|=2$. Hmm, but let me check: does $(3,6)$ sum to 9? Yes. Does $(6,3)$ sum to 9? Yes. Any other? No. So $|A\cap B|=2$. Wait — I should also include if 3 appears on the second die: that's already covered. Actually, there are exactly 3 outcomes: $(3,6),(6,3)$... no those are 2. Let me recount the sum-9 outcomes with a 3 more carefully: $(3,6)$: yes; $(6,3)$: yes; $(4,5)$: no; $(5,4)$: no. So $|A\cap B|=2$. Hmm, but the answer should give $3/11$, so let me recount $|B|$. $|B|=11$, so we need $|A\cap B|=3$. Re-examine: pairs with at least one 3 that sum to 9 — also $(3,6),(6,3)$ are 2 pairs. But wait, we also need to check if $(3+6=9)$ yes. What about $(3,6)$ first die 3, second die 6: sum 9 ✓. First die 6, second die 3: sum 9 ✓. What about first die 3, second die 6 already counted... I get 2 outcomes. Let me reconsider: perhaps the answer is $2/11$.

Outcomes in $B$ (at least one die shows 3): There are 11 such outcomes. Outcomes in $A\cap B$ (sum = 9 AND at least one die is 3): $(3,6)$ and $(6,3)$. So $|A\cap B|=2$. Wait — but also we could have both dice showing values that sum to 9 with a 3: first die 3, second die 6 ✓; first die 6, second die 3 ✓. That's only 2. But $3+3=6\ne 9$. So $P(A|B)=\frac{2}{11}$.

$P(A|B)=\frac{|A\cap B|}{|B|}=\frac{2}{11}$ Wait — I need to recount $|B|$ to verify: outcomes with at least one 3: $(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(1,3),(2,3),(4,3),(5,3),(6,3)$ = 11 outcomes. Outcomes with sum 9 among these: $(3,6)$ and $(6,3)$, giving $|A\cap B|=2$. However, checking $(3,6)$: first die=3, second die=6, sum=9 ✓ and $(6,3)$: first die=6, second die=3, sum=9 ✓. Also $(3+?)$ — if first=3, second must=6 for sum 9. If second=3, first must=6 for sum 9. So indeed $P(A|B)=\frac{2}{11}$.

Event $B$ = at least one die shows 3. Listing outcomes: $(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(1,3),(2,3),(4,3),(5,3),(6,3)$ = 11 outcomes. The denominator must be 11.

The denominator 9 doesn't correspond to any natural count here. It could come from confusing this with a simpler problem (9 outcomes after fixing one die), but $|B|=11$, not 9.

There are not 4 outcomes in $A\cap B$. Pairs summing to 9 are $(3,6),(4,5),(5,4),(6,3)$ — only $(3,6)$ and $(6,3)$ contain a 3, giving 2 outcomes, not 4.

There is no third outcome in $A\cap B$; $(3,6)$ and $(6,3)$ are the only ones. The option $\frac{3}{11}$ overcounts the intersection.

$|A\cap B|=2$ (the pairs $(3,6)$ and $(6,3)$), and $|B|=11$, so $P(A|B)=\frac{2}{11}$.