Where did I go wrong with my answer?

Question

Question: 10 unique balls are distributed randomly among 8 ordered baskets, what is the probability that the first basket is empty and the second basket has at least 1 ball in it?

My approach: first we put one ball inside the second basket, there are $10$ ways to do this, for each ball we put in the second basket we have $7^9$ ways to distribute the remaining 9 balls into 7 baskets. Now since it's obvious that our sample space is of size $8^{10}$ we get that the answer is $\frac{10*7^9}{8^{10}}$

however that answer makes no sense since $10*7^9 > 7^{10}$ which is the number of ways to distribute 10 balls into 7 baskets(without the limitation that the second basket must have atleast 1 ball). Can someone explain to me where I went wrong?

Also please note that I'm not interested in getting the right answer using a different approach I'm specifically trying to understand what is wrong with my reasoning

Answer 1

Let $A_i$, for each $i=1,2,\ldots, 10$, be the event

$$A_i = \{\text{Ball $i$ is in the second basket,}$$ $$\text{and the first basket is empty}\}.$$ Then indeed, ${\bf{P}}[A_i]$ is, $\frac{7^9}{8^{10}}$ for each $i=1,2,\ldots, 10$. You however, want to calculate ${\bf{P}}[A_1 \cup A_2 \cup \ldots A_{10}]$, and to this end you [incorrectly] claimed that ${\bf{P}}[A_1 \cup A_2 \cup \ldots A_{10}]$ is $\sum_{i=1}^{10}{\bf{P}}[A_i]$ $=$ $\frac{10 \times 7^9}{8^{10}}$. This would be correct iff the $A_i$s were disjoint.

Where this is wrong however, is that the $A_i$s are not disjoint, and so ${\bf{P}}[A_1 \cup A_2 \cup \ldots A_{10}]$ is strictly less than $\sum_{i=1}^{10}{\bf{P}}[A_i]$ $=$ $\frac{10 \times 7^9}{8^{10}}$, and not equal, as you have claimed.

Answer 2

You over-counted.

Label the balls as B-1, B-2, ..., B-10.

Label the baskets K-1, K-2, ..., K-8.

Consider the very specific case where B-1 and B-2 both go into K-2, and the other $8$ balls all go into K-3.

Any algorithm that you adopt that enumerates each of the satisfying ways (i.e. K-1 empty, K-2 not empty) of distributing the $10$ balls, must count the specific case detailed in the previous paragraph, exactly once.

However, your computation of

$$10 \times 7^9 \tag1 $$

counts the specific situation more than once.

In (1) above, the first factor represents that there are $(10)$ choices for the ball that goes into K-2.

So, the specific case will be counted once when B-1 is the first ball sent to K-2 and then the specific case will be counted a second time, when B-2 is the first ball sent to K-2.

This is no good.

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Although you did not specifically request a remedy, I will provide what I regard as the standard remedy.

The number of ways that basket K-1 can be left empty is $7^{(10)}.$

The number of ways that baskets K-1 and K-2 can both be left empty is $6^{(10)}.$

Therefore, the number of ways that basket K-1 can be left empty, without leaving basket K-2 empty is

$$7^{(10)} - 6^{(10)}.$$

Answer 3

Using conditional probability: Let $N_i=$the number of balls in basket $i$. Let $B$ be the event that $N_1=0$. Let $A$ be the event that $N_2\ge 1$. We're looking for $P(AB)$. We know $P(B)=(7/8)^{10}$. We don't know $A$, but can condition on $B$: $P(AB)=P(A|B)P(B)$. I.e., $P(A|B)=1-P(N_2=0|N_1=0)=1-(6/7)^{10}$. Putting it together to get $(7^{10}-6^{10})/ 8^{10}$.