4. Janet will wrap a birthday gift. She can choose from 8 different patterns of
ID: 3205486 • Letter: 4
Question
4. 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 (a described in part d) will be shiny? f) Janet has 6 presents to wrap and will use 6 the 11 gift tags. How many different sets of gift tags can of 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? 5. In a poker game, 5 cards are dealt from a standard 52 card deck that has been well shuffled. You are the only player in this scenario. (Note: if you are not familiar with poker hands, you may want to look up what some of these are also check out Chapter 23 in the textbook.) a) How many possible 5 card hands are possible? b) What is the probability that you are dealt two pairs? c) What is the probability that you are dealt a three of a kind or 4 of a kind? (Note: three of a kind means that the other 2 cards are NOT a pair.) d) What is the probability that you are dealt a full house? (Note: a full house means 3 of a kind and a pair.) e) what is the probability that you are dealt a flush, three of a kind, or 4 of a kind? (any type of flush is acceptable)Explanation / Answer
Solution:-
4)
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
5) a) Total number of combinations of different hands = 52C5 = 2,598,960
b) Probability of two pairs = 0.047539
This hand has the pattern AABBC where A, B, and C are from distinct kinds.
The number of such hands = 13C2 × 4C2 × 4C2 ×11C1 × 4C1 = 123,552
Total number of combinations of different hands = 52C5 = 2,598,960
Probability of two pairs = 123,552/2,598,960 = 0.047539
c) Probability of three of a kind or a 4 of a kind = 55,536 /2,598,960 = 0.02137
3 of a kind has the pattern AAABC where A, B, and C are from distinct kinds.
The number of such hands = 13C1 × 4C3 ×12C2 × (4C1)2 = 54,912
4 of a kind has the pattern AAAAB where A and B are from distinct kinds.
The number of such hands= 13C1 × 4C4 × 12C1 × 4C1 = 624
Total hands of three of a kind or a 4 of a kind = 624 + 54,912 = 55,536
Total number of combinations of different hands = 52C5 = 2,598,960
Probability of three of a kind or a 4 of a kind = 55,536 /2,598,960 = 0.02137
d) Probability of getting a full house = 0.001441
This hand has the pattern AAABB where A and B are from distinct kinds.
The number of such hands = 13C1 × 4C3 × 12C1 × 4C2. = 3774
Total number of combinations of different hands = 52C5 = 2,598,960
Probability of getting a full house = 3774/2,598,960 = 0.001441.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.