Question

A football team consists of 20 each freshmen and sophomores, 15 juniors and 10 seniors. Four players are selected at random to serve as captains. Find the probability that :
a) all four are seniors
b) There is one each: freshman, sophomore, junior and senior.
c) There are 2 sopomores and 2 fresjhman
d) At least one of the students is a senior

Answer 1

Answer:

a)Probability that all four are seniors is 0.0003102

b)Probability of getting one each is 0.00369

c)Probability of picking 2 sophomores and 2 freshers is 0.00889

d)Probability of at least one senior is .4963

Step-by-step explanation:

total  no persons = summing up the no of people present, that is, 20+20+10+15= 65

a) Probability that all four are seniors is given by Pr( all seniors)

Pr(all seniors) =

?\frac{10}[65}*\frac{9}{64}*\frac{8}{63}*\frac{7}{62} =0.0003102

The probability of an event is defined to be the ratio of the number of cases favorable to the event to the total number of cases

 The event in question is mutually dependent in that each event depends on the former

You would notice that the total no of people were reducing with each selection. That means that taking out one senior reduced the total no of students as well as the events

b)  Probability of getting one each = Pr(1 senior)*Pr(1 sophomore)*Pr(1 freshman)*Pr(1 junior)

Pr(1 senior) = 10/64

Pr(1 junior)= 15/63

Pr(1 sophomore) = 20/62

Pr (1 freshman) = 20/61

Therefore pr(1 each)= frac{10}[65}*\frac{15}{64}*\frac{20}{63}*\frac{20}{62} =0.00369

Same explanation in a applies

C) Probability of picking 2 sophomores and 2 freshers = Pr(1 sophmore)*Pr( 2nd sophmore)*Pr(1 fresher)*Pr ( 2nd fresher)

Pr(1 sophmore)= 20/64

Pr ( 2nd sophmore)=19/63

Pr ( 1 fresher)=20/64

Pr ( 2nd fresher)=19/64

Pr(2 sophomores and freshers) = frac{20}[65}*\frac{19}{64}*\frac{20}{63}*\frac{19}{62} =0.00889

d) Probability of at least one senior = 1- Probability(no senior)

No of players without the seniors = 55

Pr(no senior) = probability of choosing 4 out of 55 people divided by the probability of choosing 5 out of 65= 55C4/65C4 =0.5037

Where C means combination

Therefore probability of atleast one senior = 1- .5037= .4963

Answer 2

Probability of an event is the chance of occurrence of that event.

The needed probabilities are as follows:

• P(All four players are senior) = 0.00033 approx
• P(There is one each: freshman, sophomore, junior and senior) = 0.0944 approx.
• P(There are 2 sophomores and 2 freshman) = 0.0568  approx
• P( At least one of the students is a senior) = 0.4632 approx
  

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}}$

Where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

What is the rule of product in combinatorics?

If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in   $p \times q$  ways.

Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.

Thus, doing A then B is considered same as doing B then A

For the given case, let are given that

Number of freshmen = number of sophomores = 20

Number of juniors = 15

Number of seniors = 10

Thus, total members = 20  + 20 + 15 + 10 = 65

Calculating probabilities for given cases:

a) All four selected members are senior,

Total ways of selecting 4 members out of 65 = $^{65}C_4 =635376$

Total ways of selecting 4 members from 10 seniors = $^{10}C_4 = 210$

Thus, P(All four seniors) = 210/635376 = 0.00033 approx

b) One freshmen, one sophomore, one junior and one senior

Total ways of selecting 4 members out of 65 = $^{65}C_4 =635376$

Taking 1 member out of seniors can be done in 10 ways,

similarly, 1 member out of juniors in 15 ways, and 20 ways each for freshmen and sophomores. Thus, by rule of product from combinatorics, we get total $10 \times 15 \times 20 \times 20 = 60000$

Thus,

P(There is one each: freshman, sophomore, junior and senior) =  

60000/635376 = 0.0944 approx.

c) There are 2 sophomores and 2 freshman

Total ways of selecting 4 members out of 65 = $^{65}C_4 =635376$

Taking 2 members from sophomores can be done in $^{20}C_2 = 190$ ways

Taking 2 members from freshmen can be done in $^{20}C_2 = 190$ ways

By rule of product, total ways of 2 sophomores and 2 freshman is

$(190)^2 =36100$

Thus,

P(There are 2 sophomores and 2 freshman) = 36100/635376 = 0.0568 approx

d)  At least one of the students is a senior

Total ways of selecting 4 members out of 65 = $^{65}C_4 =635376$

P(At least one of the students is a senior) = 1- P(no senior) = $1 - \dfrac{^{20 + 20 + 15}C_4}{635376} = 1 - 341055/635376 \approx 0.4632$

Thus,

The needed probabilities are as follows:

• P(All four players are senior) = 0.00033 approx
• P(There is one each: freshman, sophomore, junior and senior) = 0.0944 approx.
• P(There are 2 sophomores and 2 freshman) = 0.0568  approx
• P( At least one of the students is a senior) = 0.4632 approx

Learn more about probability here:

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