What is the probability that Ramona catches her bus on Monday, Tuesday and Thursday, but misses it on Wednesday and Friday? Problem statement -
Every morning Ramona misses her bus with probability $\frac {1}{10}$, independently of other mornings. What is the probability that next week she catches her bus on Monday, Tuesday and Thursday, but misses her bus on Wednesday and Friday?
My attempt -
M = {catches bus on Monday}
T = {catches bus on Tuesday}
TR = {catches bus on Thursday}
$W^C$ = {misses bus on Wednesday}
$F^C$ = {misses bus on Friday}
We want to find P(M $\bigcap$ T $\bigcap$ TR $\bigcap$ $W^C$ $\bigcap$ $F^C$). We know all of these events are independent with respect to one another. Thus,
P(M $\bigcap$ T $\bigcap$ TR $\bigcap$ $W^C$ $\bigcap$ $F^C$) = P(M)P(T)P(TR)P($W^C$)P($F^C$)
Which equals $\frac{9^3}{10^3}$$\frac{1}{10^2}$. This simplifies to $\frac{729}{100000}$. I looked at the back of the book to check my answer, and it answer it had was $\frac{729}{10000}$.
Is the book's answer incorrect? If not, then what did I do wrong?
 A: The other poster's answer is incorrect as well as the textbook. It is a given that the events are pairwise independent. Thus, we can readily calculate the answer by computing $P(M) * P(Tu) * P(W^c) * P(Th) * P(F^c) = 9/10 * 9/10 * 1/10 * 9/10 * 1/10 = \dfrac{729}{10^5}$, where $P(day)$ is the probability that Ramona DOES catch her bus on that day.
EDIT: This follows from the fact that if a collection of events are independent, then the probability of their intersection is equal to the product of the individual probabilities, which is included in that chapter of the textbook.
A: The answer is almost correct.
The book answer is also wrong (probably a typo).
Your answer is a solution another problem: What is the probability of riding 3 days of the week and not riding 2 days of the week?
But the problem asks for a specific choice of days, mainly ride M, T, and Th and not ride W and F.
This means that we need to consider which days and therefore use combinations as well! *Note: we do not use permutations because the order of the days of the week is fixed. In other words, we do not want to count these as separate choices:

*

*Ride M T Th, not ride W F


*Ride M Th T, not ride W F


*Ride T Th M, not ride W F


*Ride T M Th, not ride W F


*Ride Th T M, not ride W F


*Ride Th M T, not ride W F
Rather, we want to count all of these choices permutations as 1 choice! Therefore, we use combinations.
So to find the permutations, we need to choose 3 out of 5 days to ride the bus: $$\frac{5!}{\left(5-3\right)!3!} = \textbf{10 ways}$$
We can choose 3 days out of 5 to ride the bus. Note: this implies that we are also choosing 2 days where we do not ride the bus.
Therefore, we want only 1 out of 10 ways we can pick 3 days to ride the bus in a week. So, we must divide by the number of ways, or multiply by 1/10.
Therefore, the answer is $$\frac{1}{10} * \left(\frac{9}{10}\right)^3 * \left(\frac{1}{10}\right)^2 = \frac{729}{1000000}$$
Intuition: We need to consider the number of ways we can ride 3 days a week, otherwise, our probability includes possibilities out of the scope of the problem. For example, only considering the probabilities and not the number of combinations. An answer of $$\left(\frac{9}{10}\right)^3 * \left(\frac{1}{10}\right)^2$$ would include valid sequences:
Ride M T W, not ride Th, F
Ride W Th F, not ride M, T
etc...
We can clearly see why ignoring the choices we have — when we want to pick 3 days to ride and 2 days to not ride — is problematic! Therefore, we must also consider the number of possibilities (i.e. combinations in this case)!
