I need help with these counting/probability questions. ° You may use ONE page of
ID: 3363813 • Letter: I
Question
I need help with these counting/probability questions.
° You may use ONE page of prepared notes (both sides), but otherwise the test is closed book. All work must be your own. ° Show ALL your work. You will get little or no credit for an unexplained answer. ° Do all your work in the blue exam books. Please WRITE your answers IN THE GIVEN 126 ORDER, though you may SOLVE problems in ANY order. The value of each question appears in parentheses. Use t your time. this as a guide in allocating There are 75 points, so plan on averaging about a point a minute NO CELLPHONES - NO CALCULATORS. There is no need to reduce answers to simplest terms. 1. (20 pts) We have three families, each with a mother, a father, and a child. -> (e) Five of the people will be chosen for a basketball team. How many different teams are possible (b) They are randomly placed in a line. In how many different ways can this be done? (give the specific integer)? Then they line up around a large table, each holding the hand of both neighbors. Again, how many different outcomes (orderings) are there? (c) In first case above (standing in line), find the probability that none of the families is separated (i.e., the mother, father, and child are adjacent), and explain your answer. child stands between his/her parents. elses child. Explain both answers. (d) As above, the families are separated, but now we want the probability that in addition, each In addition, find the the probability that between each husband and wife there is ONLY someaond In addition, find the the probabiility that between each husband and wife there isOExplanation / Answer
There are total 15 people in the sample space
a)
5 people from a set of 15 can be selected in 15C5 ways
i.e. 3003
b)
15 people can be arranged in a line randomly, in 15! ways.
If these 15 people are arrange around a table, then this would be circular arrangement. number of possible arrangements are (15 - 1)!
This is because, the one person has to be pivoted and others can be arranged considering him/her as reference.
c)
If we want all families to be together, there are 5! ways to arrange them.
However, we can have different arrangements among the families which can be done in 3! ways.
Hence total possible combinations are 5!*3!
Required probability = 5!*3! / 15!
d)
If each child stands between his parents, this can be done in 5!*2!
Here we can arrange 2 parents among themseleves, child has his poition fixed
Required probability = 5!*2!/15!
Now, consider 5 children will have their fixed positions(with 5! arrangements), and there aer 10 empty positons to be filled by parents. Hence there are 10!*5! arrangements where we have a child between two adults.
There are 5!*2! arrangements where we have child between her own parents.
Arrangements where parents have someone elses child between them are 10!*5! - 5!*2!
Required probability = (10!*5! - 5!*2!)/15!