Question

Your friend texted you a question. You did some research and found that a Minifig is a package containing one figurine. You also found out that, for every 60 packs produced, 4 of those are Chip and 4 are Dale.

A) What is the probability of opening one box and getting Chip?

B) What is the probability of opening one box and getting Chip OR Dale?

C) What is the probability of opening two boxes and specifically getting Chip in the first box and Dale in the second box?

D) What is the probability of opening two boxes and getting one Chip and one Dale (in either order)?

ANSWER ALL PARTS A-D

Answer 1

Total number of boxes = 60

number of boxes having Chip = 4

number of boxes having Dale = 4

A) What is the probability of opening one box and getting Chip?

P[ opening one box and getting Chip ] = number of boxes having Chip / Total number of boxes

P[ opening one box and getting Chip ] = 4/60

P[ opening one box and getting Chip ] = 1/15

P[ opening one box and getting Chip ] = 0.0667

B) What is the probability of opening one box and getting Chip OR Dale?

P[ opening one box and getting Chip or Dale ] = number of boxes having Chip or Dale / Total number of boxes

number of boxes having Chip or Dale = 4 + 4 = 8

P[ opening one box and getting Chip or Dale ] = 8/60

P[ opening one box and getting Chip or Dale ] = 2/15

P[ opening one box and getting Chip or Dale ] = 0.1333

C) What is the probability of opening two boxes and specifically getting Chip in the first box and Dale in the second box?

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = P[ getting Chip in the first box ]*P[ getting Dale in the second box ]

P[ getting Chip in the first box ] = number of boxes having Chip / Total number of boxes

P[ getting Chip in the first box ] = 4/60

P[ getting Chip in the first box ] = 1/15

P[ getting Dale in the second box ] = number of boxes having Dale / Total number of boxes left

number of boxes having Dale = 4

Total number of boxes left = 59

P[ getting Dale in the second box ] = 4/59

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = (1/15)*(4/59)

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = 4/885

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = 0.0045

D) What is the probability of opening two boxes and getting one Chip and one Dale (in either order)?

P[ opening two boxes and getting one Chip and one Dale (in either order) ] = P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] + P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ]

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = 0.0045 ( previous part )

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = P[ getting Dale in the first box ]*P[ getting Chip in the second box ]

P[ getting Dale in the first box ] = number of boxes having Dale / Total number of boxes

P[ getting Dale in the first box ] = 4/60

P[ getting Dale in the first box ] = 1/15

P[ getting Chip in the second box ] = number of boxes having Chip / Total number of boxes left

number of boxes having Chip = 4

Total number of boxes left = 59

P[ getting Chip in the second box ] = 4/59

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = (1/15)*(4/59)

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = 4/885

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = 0.0045

P[ opening two boxes and getting one Chip and one Dale (in either order) ] = 0.0045 + 0.0045

P[ opening two boxes and getting one Chip and one Dale (in either order) ] = 0.009

I am considering with out replacement case.

In case of replacement in part C and D. Answer will change

With replacement

C) What is the probability of opening two boxes and specifically getting Chip in the first box and Dale in the second box?

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = P[ getting Chip in the first box ]*P[ getting Dale in the second box ]

P[ getting Chip in the first box ] = number of boxes having Chip / Total number of boxes

P[ getting Chip in the first box ] = 4/60

P[ getting Chip in the first box ] = 1/15

P[ getting Dale in the second box ] = number of boxes having Dale / Total number of boxes

number of boxes having Dale = 4

Total number of boxes left = 60

P[ getting Dale in the second box ] = 4/60

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = (1/15)*(1/15)

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = 1/225

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = 0.0044

D) What is the probability of opening two boxes and getting one Chip and one Dale (in either order)?

P[ opening two boxes and getting one Chip and one Dale (in either order) ] = P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] + P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ]

P[ opening two boxes and specifically getting Chip in the first box and Dale in the second box ] = 0.0044 ( previous part )

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = P[ getting Dale in the first box ]*P[ getting Chip in the second box ]

P[ getting Dale in the first box ] = number of boxes having Dale / Total number of boxes

P[ getting Dale in the first box ] = 4/60

P[ getting Dale in the first box ] = 1/15

P[ getting Chip in the second box ] = number of boxes having Chip / Total number of boxes

number of boxes having Chip = 4

Total number of boxes left = 60

P[ getting Chip in the second box ] = 4/60

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = (1/15)*(1/15)

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = 1/225

P[ opening two boxes and specifically getting Dale in the first box and Chip in the second box ] = 0.0044

P[ opening two boxes and getting one Chip and one Dale (in either order) ] = 0.0044 + 0.0044

P[ opening two boxes and getting one Chip and one Dale (in either order) ] = 0.0088