213k views
5 votes
Below are the names of ten students:

Dorothy
Anthony
Harold
Margaret
Tiffany
Nancy
Angela
Paul
For the following, assume the probability of a student being chosen is the same for each student. Also, assume that
simple random sampling is being used. (If necessary, use an exact decimal value for all probabilities.)
a.) What is the probability of choosing a student whose name ends with the letter "k"?
b.) What is the probability of choosing a student whose name begins with the letter "C"?
Patrick
Jeremy
c.) What is the probability of choosing a student whose name contains the letters "m" or "M"?
d.) A Math Club is formed using these ten students, and the club must consist of seven students. How many ways
can students be assigned to the Math Club?
e.) The ten students enter an art competition where first place wins $100, second place wins $50, third place wins
$25, and fourth place wins $10. How many different ways can prizes be distributed?


It’s not multiple choice you have to work the math problem!!

1 Answer

5 votes

Explanation:

a probability is always the ratio

desired cases / totally possible cases

let's relist the 10 names, as there was some "mixing" going on in your text :

Dorothy

Anthony

Harold

Margaret

Tiffany

Nancy

Angela

Paul

Patrick

Jeremy

so, whatever happens, the totally possible cases are 10 (as we are only taking about these 10 names, no other name can suddenly appear in all the scenarios).

in each scenario one student is randomly chosen.

a) the probabilty to pick a student with name ending "k".

we look through the list and find only 1 student, whose name ends with "k" : Patrick.

that means the number of desired cases is 1.

and the probabilty is therefore

1/10 = 0.1

b) the probability to pick a student with name beginning "C".

we look and look and look. none of the 10 names start with a "C".

that means the number of desired cases is 0.

and the probabilty is

0/10 = 0

c) the probability that the name of the picked student contains "m" or "M".

we find 2 names : Margaret and Jeremy.

that means that the number of desired cases is 2.

and the probabilty is therefore

2/10 = 1/5 = 0.2

d) in how many ways can we pick 7 elements out of 10, when the sequence of the pulled 7 elements does not matter (e.g. ... Nancy, Paul ... is the same as ... Paul, Nancy, ...). and no student can be pulled more than once (no repetitions).

that means we have to calculate the combinations of 7 items out of 10 without repetition :

C(10, 7) = 10! / (7! × (10-7)!) = 10! / (7! × 3!) =

= 10×9×8 / (3×2) = 5×3×8 = 120

there are 120 possibilities to build the math club.

e) now we pick 4 students out of 10. but as we give them prices based on ranking, the sequence matters (e.g. Nacy first, Paul second is different to Paul first, Nancy second).

but we still have no repetition, as nobody can win more than one price.

that means we need to calculate the permutations of 4 items out of 10 without repetition :

P(10, 4) = 10! / (10-4)! = 10! / 6! = 10×9×8×7 = 5,040

there are 5040 different ways to distribute the 4 prices among the 10 students.

User Allan Ojala
by
8.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories