I was recently trying to solve a rather simple probability question. Notwithstanding, when I compared both approaches, I noticed that the outcomes were different, and was wandering if someone could explain me where it went wrong.
The problem. A cooking course is given, attended by four experienced participants and eight new ones. It is known that new participants attend the course with a $90\%$ probability, while those who have attended in previous years do so with a $75\%$ probability.
Determine the probability, that at least $11$ participants will attend today's lesson.
Approach 1. Notice that at least $11$ participants will attend the course today iff
- $8$ new and $4$ experienced participants do so.
- $8$ new and $3$ experienced participants do so.
- $7$ new and $4$ experienced participants do so.
Define the variables $A="\text{number of new participants attending todays course}"$ and $B$ similarly for experienced participants. Then we are asked to evaluate $$P(A=8)\cdot P(B=4)+P(A=8)\cdot P(B=3)+P(A=7)\cdot P(B=4)=\ldots\approx 43,8\%$$
Approach 2. We consider the probability that a random selected participant will attend the course today. This is $$p=\frac{0.9\cdot8+0.75\cdot 4}{12}=0.85$$ Thus, if $X$ denotes the amount of participants attending today's lesson we should have $$P(X\geqslant 11)=P(X=11)+P(X=12)=\binom{12}{11}\cdot (0.85)^{11}\cdot 0.15+0.85^{12}\approx 44.3\%$$
I am confident that the first approach is correct, and, hence, the answer should be $43,8\%$. But, what's wrong with the second approach? I am looking for both rigorous and intuitive ways of showing that the latter approach does not work. Thanks :)