i have no clue how to solve the following questiom, help is appreciated
Phil is running a stall at the local Primary School Fair involving lucky dips. It costs $2 to have a go, and contained in a large box are 80 lucky dips from which to choose. Phil claims that one in 5 lucky dips contains a prize. By the end of the day, all 80 have been sold. Calculate the probability, correct to 4 decimal places that:
a) the first four people to select a lucky dip don't win a prize, but the next two do
b) there are at least 10 winners
c) there are no more than 18 prize winners, given that at least 10 people won a prize.
Always state given conditions:
The $2 is irrevelant in this question.
"One in 5 lucky dips contains a prize" -> p = 1/5 = 0.2
"80 lucky dips from which to choose" -> n = 80
a) You can draw a tree diagram or you can just work it out.
0.8^4 * 0.2^2 = 0.0164
b) "At least 10 winners" i.e. Pr (X greater than or equal to 10)
Use BinomialCDf on CAS:
Lower: 10
Upper: 80
Numtrial: 80
pos: 0.2
= 0.9713
c) "there are no more than 18 prize winners, given that at least 10 people won a prize."
i.e. Pr (X less than or equal to 18 | X greater than or equal to 10)
Pull out your formula sheet:
Intersection between the two conditions above divided by your answer in part b.
Intersection between less than or equal to 18 and greater than or equal to 10 is Pr([10,18]).
i.e. BinomialCDf once again:
Lower: 10
Upper: 18
Numtrial: 80
pos: 0.2
= 0.7333623149
Note: When using previous answer, get as many decimal places as you can so there are no rounding errors.
--> 0.7333623149/0.9712824327 = 0.7550