Ok, lets do the first part to begin.
I don't really want to give you the answer straight out.
How much do you understand probability to begin with?
So EXACTLY 1 white means we want the probability for the event
(W,B,B) OR (B,W,B) OR (B,B,W)
since these are the only possible ways that we could draw exactly one white, agreed?
Can you calculate this probability on your own?
Do you understand how to think about "AND" and "OR" in probability terms? If not I can explain further, but you haven't given many details.
Then for the second part, we want AT LEAST 1 white.
That is, we want either
1 white, OR 2 whites, OR 3 whites
agreed?
So using the method of part a), you should be able to calculate the probability for 2 whites and the probability for 3 whites, and then using the "AND" and/ or "OR" laws for probabilities you should be able to answer this second part of the question.
Does this help?
If you give me some more details of which bit you are getting stuck at I can help you further if you need.
The law of total probability helps here. If the transferred marble was white, which happens $\frac35$ of the time, the probability that the marble drawn from bag B is white is $\frac58$. If the transferred marble was black ($\frac25$ chance), that probability is $\frac12$.
The probability that the drawn marble is white and the transferred marble is white is $\frac58×\frac35=\frac38$. The probability that the drawn marble is white but the transferred marble is black is $\frac12×\frac25=\frac15$. Therefore the probability the transferred marble was white given that the drawn marble is white is
$$\frac{\dfrac38}{\dfrac38+\dfrac15}=\frac{15}{23}$$
Best Answer
Let's start with the colors:
(The other labels aren't relevant to your question)
This means that if the probability of a Y marble being colored black is $P_{black}$ then we are given that $P_{black} = \frac{P_{white}}{2} = \frac{P_{brown}}{2}$ Since these values sum to 1: $P_{black} = .2$
Using this information most of the stuff is unnecessary. All we need to know is that whatever .2Y is it took 20 tries to achieve it. We can use a beta distribution to model this situation since it is the conjugate prior probability distribution to the geometric distribution.
In our case: $ \alpha =2$ and $\beta =20$ since there was 1 event and 19 failures. Then the graph of the distribution looks like this:
The y-value at x represents the probability that x is the probability of the event in our scenario. So assuming the probability distribution of the probability is uniformly randomly distributed aside from the information given, then the point where y is maximized in this distribution between 0 and 1 gives the expected value of $\mathbb P(.2Y)$ which is $\frac{1}{20}$ in this case. (Intuitively it makes sense for the probability of it taking 20 tries for something to happen to be 1 in 20). Finally this gives $\mathbb P(Y) = .25$ or $\frac{1}{4}$.
The advantage of using the beta distribution here is we get to see the probability of $\mathbb P(Y) $ being a different value. For example if we wanted to know the probability of it being .2 it would be about .0576