NAME:

1. Give an example of a discrete random variable. Explain why it is both discrete and random. Give its values and its probability density function.
 

Let X = 1 if fair coin lands heads when flipped, 0 if tails.  X is discrete because it takes on only two values, and is random because the outcome is unknowable beforehand.  The pdf is f(x) = .5  for x = 0,1, and f(x) = 0 for all other x.

There are many other choices, of course, but as long as you're invited to choose your own, why make it hard on yourself?  Also, please note that discrete variables do NOT have to have finitely many values.  For example, let X = nmber of flips of a coin before the first Head appears.  X could take on all values from 0 to infinity, but is still discrete because the values are discrete.
 
 

2. Select two cards WITHOUT REPLACEMENT from a standard deck of playing cards. Let A be the event that the first card is a heart. Let B be the event that the second card is a club. Are A and B independent? Prove your answer. (A good first step to this proof is to state rather clearly what the definition or test of independence is.)

There are 3 "tests" for independence:
1) P(A and B) = P(A) * P(B)
2) P(A|B) = P(A)
3) P(B|A) = P(B)

Method 2 doesn't make too much sense, because it's hard to understand what is meant by "the probability that the first card is a heart given that the second card is a club."  But you can actually still do it this way, because the order of the cards isn't important, really.  Just imagine that the first card is dealt but hidden, and the second card is turned face up.

Mumber 1 and Mumber 3, however, are quite do-able.

Method 3:  P(B|A) = 13/51  (because there are 51 cards left and 31 of them are clubs.)
P(B) = 1/4  (Two ways to think of this:  if the first card is dealt face down, and you are asked, what's the probability the second card is a club, then it is just as if the first card had never been dealt at all.  So the probability is 1/4.  Or, you can define a new event: B1 = event that the first card is a club.  Then P(B) = P(B1)P(B| B1) + P(B1^c) P(B | B1^c)
= 1/4 * (12/51) + (3/4) * (13/51) = 51/204 = 1/4)
Because P(B) does not equal P(B|A), A and B are not independent.

Method 1: P(A and B) = P(A) * P(B |A)  = (1/4) *( 13/51)
P(A) = P(B) = 1/4
So P(A) * P(B) = (1/16)
But (1/16) does not equal (1/4)*(13/51) So A and B are NOT independent.