Ahigh school holds a charity raffle every year that awards a randomly chosen student a small gift certificate to a local mall. the first year’s winner of the raffle is a student named jessica. some students in the school say jessica will likely win a second year in a row because of “winner’s luck,” or the idea that being chosen randomly the first year makes jessica more likely to be chosen the second year. other students believe jessica is less likely than any other student to win because it is unlikely that one student would win two years in a row. the concept of lucky streaks, or winner’s luck, is a common one. state whether you agree or disagree with each group of students, and give reasons for your opinion. be sure to use some of the terms you’ve learned in this unit, such as independent event, dependent event, and conditional probability, in your answer.

Neither of the two groups of students is right.

Step-by-step explanation:

If the probability that Jessica will win the raffle this year was  , the probability that she will win the raffle in subsequent years is given that the number of students do not change. Since this is a random win, she as likely to win next year as she was this year.

Another way to explain this is to say that the probability of Jessica winning the raffle this year is independent of the probability that she wins the raffle next year.  Suppose event A, is the event that Jessica wins the raffle this year, let event B be the event that she wins the raffle next year, event A and B are independent because the occurrence of event A does not affect the occurrence of event B in any way

sorry, i can't understand your question. will later if you describe more clearly!

: )

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.