These questions are conditional probabilities.
There is useful information in this regard on Wikipedia, and on Wolfram, but basically, conditional probabilities go like this:
$P(A|B)$ is the probability that event $A$ occurs given that event $B$ has already occurred. If you take the entire sample space, $S$, (the set of all possible events, or outcomes), then $A$ and $B$ represent sub-sets of that space.
Visually, you can think of these as Venn diagrams where $S$ is the entire area, and $A$ and $B$ are two smaller areas within the space. The intersecting, or shared space, between $A$ and $B$ is notated $AB$ (or $A\cap B$), while the combination of $A$ and $B$ is notated $A\cup B$.
Well, if we have that $B$ has occurred, and we want the probability that $A$ will occur, this must relate to the $AB$ sub-set, the set of events in which both $A$ and $B$ occur. We want to measure this relative to the initial probability that $B$ occurred, so:
$P(A|B) = \frac{P(AB)}{P(B)}$
Based on this, we can look at part (a) of the question.
a. What is the probability that the dwarf named Bashful gets kissed first on Monday?
Let's assign $A$ as the event that Bashful gets kissed first, and $B$ as the event that it is Monday. We are then looking for $P(A|B)$, the probability of Bashful being kissed first, given that it is Monday.
What is $P(AB)$? It is the probability that it is both Monday ($\frac{1}{5}$) and that Bashful is first to be kissed ($\frac{1}{7}$).
What is $P(B)$? It is the probability that it is Monday ($\frac{1}{5}$).
Therefore, $P(A|B)$ = $\frac{\frac{1}{5}\times \frac{1}{7}}{\frac{1}{5}}$ = $\frac{1}{7}$.
The remaining parts of the question follow the same process.
So far you have properly identified $P(D|M), \dots, P(D|F).$ But you need to look at Bayes' Theorem and figure out
how to get the 'inverse' or 'posterior' probability $P(F|D).$ Here is an
outline; you should match the steps with Bayes' Theorem.
Notice that $P(D \cap F),\, P(D|F)$ and $P(F|D)$ are all probabilities
involving events $D$ and $F.$ But they refer to different populations:
$P(D \cap F)$ refers to the entire weekly production of ball bearings;
$P(D|F)$ refers to the population of ball bearings made on Friday; and $P(F|D)$ refers to the population of all defective ball bearings, asking what
proportion of them were made on Friday.
$P(D \cap M) = P(M)P(D|M) = .2(.08) = .16.$ This uses what is sometimes
called the 'general multiplication rule'.
$P(D) = P(D \cap M) + P(D \cap Tu) + \cdots + P(D \cap F) = 0.064.$ This uses
what is sometimes called the 'law of total probability'.
Using R statisical software as a calculator:
p.d = sum(.2*c(.08, .04, .04, .04, .12)); p.d
## 0.064
$P(F|D) = P(D \cap F)/P(D) = .024/.064 = 0.375.$ This uses Bayes' Theorem.
While only 20% of all ball bearings are made on Friday, 37.5% of defective ball bearings were made on Friday.
Best Answer
The probability changes depending on what was eaten in the prior days. Say $v$ is a meal with a vegetable, $n$ is one without.
The possible combinations for Monday and Tuesday are:
In each case, what is the chance of $v$ on Wednesday?
To account for each case, weight the probability of $v$ on Wednesday according to the likelihood of the given Monday-Tuesday scenario, then combine:
$P = \frac 1{10}\cdot 0 + \frac 3{10}\cdot\frac 1 3+\frac 3{10}\cdot\frac 1 3 + \frac 3{10}\cdot\frac 2 3$, $= \boxed{\frac {2}{5}}$
Why does this work? Well, the probability of any event (such as eating vegetables Wednesday) equals the probability of the event occurring in any possible scenario, combined. For example: You go to work on rainy days and on sunny days. So the number of days you go to work is the number of days you go to work when it rains, plus the number of days you go to work when it is sunny.
So for eating vegetables Wednesday, the probability is the same as the probability of eating vegetables Wednesday combined for each possible Monday-Tuesday scenario.
$\mathrm P(v) = \mathrm P(vv\mapsto v) + P(vn\mapsto v) + P(nv\mapsto v) + P(nn\mapsto v)$
I know it can be confusing learning how to work with probabilities. In my experience, introductory class questions were designed to trick the student rather than focusing on developing strong intuition or modeling ability. Please do feel free to follow up in the comments if having further questions or concerns.