A basic probability probem I have question and I could not figured out how to proceed solving question:
I think the probability of Success for part (a) is 0.75 and for part (b) it is 0.50. I guess this is a binomial question, but I could not figured out how to proceed with the rest.
Assume that midterm grades of Econ 221 course are summarized in the following stem-and-leaf display.
For a random sample of 5 students,
a.  What is the probability that at most one of them did better than 49?
b.  What is the probability that at most one of them did better then the median grade?
As preparation for (a) and (b), answer the following: 
 What is the Population size (N)?
 What is the sample size (n)?                           
 What is the definition of Success for part (a)?
 What is the probability of success (p) for part (a)?                   

 What is the definition of Success for part (b)?
 What is the median grade?
 What is the probability of success (p) for part (b)?

 (hint: X=number of successes in the population N) 

  Leaf unit =1

  Cummulative               
  Frequency     Stem    Leaf        
         2         1    7 9     
         3         2    9   
         9         3    0 4 4 6 7 8 
        16         4    1 2 4 4 4 6 8   
        31         5    0 0 0 0 1 3 3 3 3 5 6 6 8 8 9   
       (5)         6    2 4 5 8 8       
        28         7    1 1 2 3 3 3 4 6 6 7 8 9 9   
        15         8    0 1 1 2 2 2 2 3 4 4 7       
         4         9    2 2 6 8         

 A: The preliminary questions pretty clearly indicate that you
are to use the binomial distribution for $X$, the number of
Successes out of five selected. But if sampling from
the population of $N = 64$ items is without replacement
(so that there is no possibility of choosing the same
element of the population twice), then the exact distribution
of $X$ is hypergeometric. 
There is a rough rule that says it is OK to approximate
hypergeometric probabilities with binomial ones, if the
sample size is less than 10% of the population size.
Here, $n = 5 < 0.1N = 6.4,$ so the approximation should
be fairly accurate.
Binomial. For (a) you have $X \sim Binom(5, 1/4),$
because you are choosing $n = 5$ and there are 48 out of 64
'Successes' in the population, giving $p = 48/64 = 3/4,$
as you say. By the binomial formula, you want
$$P(X \ge 3) = \sum_{i=3}^5 {5 \choose i}(3/4)^i(1/4)^{5-i}\\
= {5 \choose 3}(3/4)^3(1/4)^2 + {5 \choose 4}(3/4)^4(3/4) + {5 \choose 5}(3/4)^5\\
= 10(27/1024) + 5(81/1024) + 243/1024 =  0.8965$$
Perhaps your textbook has tables from which you can get the
same result. Or you might use a statistical calculator. In R software this result can be computed as:
 i = 3:5;  sum(dbinom(i, 5, 3/4))
 ## 0.8964844

Part (b) is done in a similar way except that $p = 1/2$ and the
computations are, if anything, a little easier, giving the
probability 0.5.
Hypergeometric. For part (a), imagine an urn with 48 balls
marked S and 16 marked F. You are to withdraw five balls without
replacement, and you want to know $P(X \ge 3),$ where $X$ is
the number of S's drawn. The random variable $X$ has a hypergeometric
distribution. Then you want
$$P(X \ge 3) = 
\left[{48 \choose 3}{16 \choose 2} + {48 \choose 4}{16 \choose 1} + {48 \choose 5}{16 \choose 0} \right]/{64 \choose 5} = 0.9051.$$
With an ordinary calculator, the computation using these binomial coefficients is a
little tedious, but manageable; there is a lot of cancellation. To two decimal places, the exact hypergeometric and the
approximate binomial models both give the answer 0.90. In R the hypergeometric computation is
 i = 3:5;  sum(dhyper(i, 48, 16, 5))
 ## 0.9051207

For part (b) there are 32 S's and 32 F's in the 'urn'. The desired
probability is again exactly 0.5.
About the stemplot. Lowest three scores are 27, 29, and 39,
and the top twp scores are 96 and 98. Counts of individuals are accumulated
from the extremes towards the middle; the row marked (5) contains
the median (62 + 64)/2 = 63. The total number of scores is
N = 31 + 5 + 28 = 64.
