Janet will wrap a birthday gift. She can choose from 8 different patterns of gif
ID: 3204845 • Letter: J
Question
Janet will wrap a birthday gift. She can choose from 8 different patterns of gift wrap, 7 different color ribbons, and 11 different gift tags. Only one of each category will be used to prepare the gift. Her choice of each item is independent of the others. Each combination of gift wrap, ribbon and gift tags is equally likely.
a) How many ways can Janet prepare the gift?
b) Four of the ribbon choices are shiny, what is the probability that the gift will have a shiny ribbon?
c) If 3 of the gift wraps are shiny and 4 of the ribbons are shiny, what is the probability that the birthday present will have both shiny wrapping paper and shiny ribbon?
d) Janet will arrange the ribbons in a line. How many ways can she arrange the ribbons?
e) Four of the ribbon choices are shiny. What is the probability that the first and last ribbons in the line (as described in part d) will be shiny?
f) Janet has 6 presents to wrap and will use 6 of the 11 gift tags. How many different sets of gift tags can she use? g) If 5 of the 11 original gift tags are shades of blue, what is the probability that 2 of those Janet uses on the 6 presents are blue?
Explanation / Answer
Solution:-
a) The number of ways can Janet prepare the gift is 616.
Different patterns of gift wrap = 8 ,
Different color ribbons = 7
Different gift tags = 11
The number of ways can Janet prepare the gift = 8 × 7 × 11 = 616
b) Four of the ribbon choices are shiny, the probability that the gift will have a shiny ribbon is 0.5714
Different patterns of gift wrap = 8
Number of shinny ribbons = 4
Different gift tags = 11
Total ways of selceting from 4 shinny ribbons = 8 × 4 × 11 = 352
The total number of ways can Janet prepare the gift = 8 × 7 × 11 = 616
The probability that the gift will have a shiny ribbon = 352/616 = 0.571428
c) If 3 of the gift wraps are shiny and 4 of the ribbons are shiny,the probability that the birthday present will have both shiny wrapping paper and shiny ribbon is 0.2143
Number of shinny gift wrap = 3
Number of shinny ribbons = 4
Different gift tags = 11
Total ways of selecting from 4 shinny ribbons and 3 shinny gift wraps = 3 × 4 × 11 = 132
The total number of ways can Janet prepare the gift = 8 × 7 × 11 = 616
The probability that the gift will have a shiny ribbon and shinny gift wrap = 132/616 = 0.21428
d) The number of ways can she arrange the ribbons is 5040.
Total number of ribons = 7
Number of ways of arranging ribbons = 7! = 5040
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