Please answer using proper notation and 4 decimal places. thank you. Janet will
ID: 3205411 • Letter: P
Question
Please answer using proper notation and 4 decimal places. thank you.
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
Given,
Gift wrap=8
Colour ribbons=7
Gift tags=11
each of the events are independent
A.a) total combinations=8C1*7C1*11C1
total combinations= 8*7*11
total combinations =616
A.b) shiny ribbons=4
total ribbons=7
probability of having shiny ribbon=4/7
probability of having shiny ribbon =0.57
A.c) shiny gift wraps=3
shiny ribbons=4
total gift wraps=8
total ribbons=7
probability of shiny gift wraps and shiny ribbon=3/8*4/7
probability of shiny gift wraps and shiny ribbon =3/14=0.21
A.d) let the line positions be P1,P2,………P7
total ribbons=7
P1 can be filled in 7 ways
P2 can be filled in 6 ways
.
.
.
P7 can be filled in 1 way
therefore, total no of ways=7*6*5*4*3*2*1
total no of ways of arranging ribbons is 5040 ways
A.e)
A.f) Given,
6 presents
11 gift tags
different set of gift tag combinations=11C6
different set of gift tag combinations =462 combinations
A.g) Total gift tags=11
blue gift tags=5
Probability of blue gift tags is 2 out of 6 presents
probability=5C2*6C4/11C6=0.3247
Probability of blue gift tags is 2 out of 6 presents is 0.3247