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
