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.
Your thinking for a) is correct ... but your formula is not quite right. If you define $P(A)$ as the probability of a student attending and passing , and $P(B)$ the probability of a student not attending and passing, then the probability of passing is simply $P(A)+P(B)$.
That is, you don't further multiply $P(A)$ by $0.79$, but rather, you figure out $P(A)$ by multiplying the $0.97$ by $0.79$. That is:
$P(A)=0.97\cdot 0.79$
Likewise:
$P(B)=0.19\cdot 0.21$
For b): Note that $P(A)+P(B)$ gives you the percentage of students that passed the exam. So, what part of that is made up by students who attended regularly? That will be the probability of Sam attending regularly, given that Sam was one of those who passed. For example, if $P(A)$ turns out to be $.4$, i.e. $40$%, and $P(B)$ turns out to be $0.2$, i.e. $20$%, then out of all those students who passed the exam (which is $20+40=60$% of all stduents), the proportion that attended regularly is $\frac{0.4}{0.6}$
Best Answer
Please check my attempt for correctness:
choosing 3 out of 50 children can be done in 50C3 ways.
And to satisfy the condition, i can make a selection of 1 from 13, 1 from 25 and 1 from 12 students.
so i have 12*13*25/(50C3).