# Probabiliy distributions

1. Nov 21, 2004

### kikar

hey guys..

Im having trouble answering the following questions I tried everything I can. I JUSS dont get it

1)In a common carnival game the player tosses a penny from a distance of about 5 feet onto a table ruled in 1-inch squares. If the penny (3/4 inch in diameter) falls entirely inside a square, the player receives 5 cents but does not get his penny back; otherwise he loses his penny. If the penny lands on the table, what is his chance to win?

2) 2If a chord is selected at random on a fixed circle, what is the probability that its length exceeds the radius of the circle?

3a) You roll a single die 3 times. Let X be the random variable which measures how many of each roll turns up (e.g. 3 one’s, 2 two’s, 1 three, 1 four, 1 six). Determine the probability distribution of the situation.
b) The above random variable is said to exhibit a multinomial distribution. Use the above question to derive an expression (equation) for a particular outcome of X.

4) From a shuffled deck, cards are laid out on a table one at a time, face up from left to right, and then another deck is laid out so that each of its cards are beneath a card from the first deck. What is the probability that exactly ‘r’ matches occur (a match being the same value and suit matching)

please help me out, im stuck, i got the other 16 questions right, and i am not satisfied with just that, cuz im determined to find out what i did wrong.
thanks

2. Nov 21, 2004

### mathman

1. The center of the penny (in a square) has to be 3/4 in. or more away from any edge. Therefore the center has to be inside a 1/4 in. by 1/4 in. square in the middle of the 1 in. square. Assuming uniform, the answer is 1/16.

2. There may be an ambiguity. When you say random, what is random and what is its distribution?
A possible distribution is the center of the chord is uniform within the circle and the chord is perpendicular to the radius at the center.
Another would be one end of the chord is fixed and the other end is uniform on the edge of the circle.
I haven't worked it out, but I suspect the results would be different.

3. and 4. Too much work. Should be straightforward.

3. Nov 21, 2004

### Galileo

Hin for 4: There are 52! possible configurations of the second card deck.
If we select r cards from the above deck, there are (52-r)! configurations of the lower deck which have those r cards matching. (do you see why?)

4. Nov 21, 2004

### kikar

oh man
Im confused even more!

these questions are driving me insane, i want to know how to do them :(

5. Nov 21, 2004

### kikar

ne oen? :tongue:

6. Nov 21, 2004

1) is definitly zero, since it's a carnie game.

7. Nov 22, 2004

### TenaliRaman

Kikar,
read mathmans post carefully again and post exactly what u did not understand in his post.

As regards to question 2, mathman is correct ...
Its an old problem which has become a classic on its own and is known as the Bertrands Paradox. To read more on this go here,
http://mathworld.wolfram.com/BertrandsProblem.html

-- AI