SOLUTION: Four unmarked packages have lost their shipping labels, and you must reapply them. What is the probability that you apply the labels and get all four of them correct? Exactly thre

Algebra ->  Probability-and-statistics -> SOLUTION: Four unmarked packages have lost their shipping labels, and you must reapply them. What is the probability that you apply the labels and get all four of them correct? Exactly thre      Log On


   

Question 373867: Four unmarked packages have lost their shipping labels, and you must reapply them. What is the probability that you apply the labels and get all four of them correct? Exactly three correct? Exactly two correct? At least one correct?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Look at all of the possible outcomes for incorrect(I) and correct(C) placement of labels on each of the 4 boxes.
CCCC
CCCI
CCIC
CCII
CICC
CICI
CIIC
CIII
ICCC
ICCI
ICIC
ICII
IICC
IICI
IIIC
IIII
There are 4%5E2=16 possible outcomes.
1 has exactly 4 correct.
4 have exactly 3 correct.
6 have exactly 2 correct.
4 have exactly 1 correct.
1 has exactly 0 correct.
.
.
The probability that the label is correct is P%28C%29=1%2F4.
The probability that the label is incorrect is P%28I%29=3%2F4.
.
.
P%283C%29=4%2A%281%2F4%29%281%2F4%29%281%2F4%29%283%2F4%29=3%2F64
P%282C%29=6%2A%281%2F4%29%281%2F4%29%283%2F4%29%283%2F4%29=27%2F128
P%281C%29=4%2A%281%2F4%29%283%2F4%29%283%2F4%29%283%2F4%29=27%2F64
At least one correct means that its not the case that all four are incorrect.
P%284I%29=1%2A%283%2F4%29%283%2F4%29%283%2F4%29%283%2F4%29=81%2F256
So then either all 4 are incorrect or at least one is correct.
P(at least 1 C)+P(4I)=1
P(at least 1 C)=1-P%284I%29
P(at least 1 C)=1-81%2F256
P(at least 1 C)=175%2F256