A Fair Coin Is Flipped 9 Times What Is The Probabilit 360922

1 A Fair Coin Is Flipped 9 Times What Is The Probability Of Getting

1. A fair coin is flipped 9 times. What is the probability of getting exactly 6 heads? 2. You flip a coin three times. (a) What is the probability of getting heads on only one of your flips? (b) What is the probability of getting heads on at least one flip? 3. A jar contains 10 blue marbles, 5 red marbles, 4 green marbles, and 1 yellow marble. Two marbles are chosen (without replacement). (a) What is the probability that one will be green and the other red? (b) What is the probability that one will be blue and the other yellow? 4. A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2 mangos. a) Imagine you stick your hand in this refrigerator and pull out a piece of fruit at random. What is the probability that you will pull out a pear? b) Imagine now that you put your hand in the refrigerator and pull out a piece of fruit. You decide you do not want to eat that fruit so you put it back into the refrigerator and pull out another piece of fruit. What is the probability that the first piece of fruit you pull out is a banana and the second piece you pull out is an apple? c) What is the probability that you stick your hand in the refrigerator one time and pull out a mango or an orange? 5. Roll two fair dice. Each die has six faces. a) List the sample space. b) Let A be the event that either a three or four is rolled first, followed by an even number. Find P(A). c) Let B be the event that the sum of the two rolls is at most seven. Find P(B). d) In words, explain what “P(AB)” represents. Find P(AB). e) Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification. f) Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification 6. At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper. a) Find the probability that a course has a final exam or a research project. b) Find the probability that a course has NEITHER of these two requirements. 7. You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $500 prize, two $100 prizes, and four $25 prizes. Find your expected gain or loss. 8. Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for100 students. a) What is the average class size assuming each class is filled to capacity? b) Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable X equal the size of the student’s class. Define the PDF for X. c) Find the mean of X. d) Find the standard deviation of X. 9. A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities. a) In words, define the random variable X b) List the values that X may take on. c) Give the distribution of X. X ~ _____(_____,_____) e) How many of the 12 students do we expect to attend the festivities? f) Find the probability that at most four students will attend. g) Find the probability that more than two students will attend.

Paper For Above instruction

The collection of probability problems presented spans a diverse range of scenarios, each requiring different methods of solving, from basic probability calculations to more advanced concepts like expected value, variance, and probability distributions. This paper will analyze and solve each problem systematically, illustrating the core principles of probability theory and statistics.

Problem 1: Probability of Exactly 6 Heads in 9 Coin Flips

When flipping a fair coin nine times, the probability of obtaining exactly six heads follows the binomial distribution, which is appropriate considering each flip is independent and has two outcomes—heads or tails—with equal probability (0.5). The formula for the binomial probability is:

P(X = k) = C(n, k) p^k (1-p)^{n-k}

Here, n=9, k=6, p=0.5, and C(n, k) is the combination of n items taken k at a time.

Calculating the binomial coefficient:

C(9,6) = 84

Thus, the probability:

P(6 heads) = 84 (0.5)^6 (0.5)^3 = 84 * (0.5)^9 = 84/512 ≈ 0.164

Hence, there is approximately a 16.4% chance of flipping exactly six heads in nine flips.

Problem 2: Flipping a Coin Three Times

2a) The probability of getting heads on only one flip involves three possible arrangements: HTT, THT, TTH. Each has a probability of (0.5)^3 = 0.125. Since there are three such arrangements, the probability is:

P(only one head) = 3 * 0.125 = 0.375

2b) The probability of getting at least one head is the complement of getting no heads (i.e., all tails):

P(at least one head) = 1 - P(no heads) = 1 - (0.5)^3 = 1 - 0.125 = 0.875

Problem 3: Marble Selection Without Replacement

Total marbles: 10 blue + 5 red + 4 green + 1 yellow = 20 marbles.

3a) Probability that one marble is green and the other red can be approached as:

P(green then red) + P(red then green):

P(green then red) = (4/20) (5/19) = (4/20)(5/19) = (1/5)*(5/19) = 1/19

P(red then green) = (5/20) * (4/19) = same as above = 1/19

Total probability: 2/19 ≈ 0.105

3b) Probability that one marble is blue and the other yellow:

P(blue then yellow) + P(yellow then blue):

P(blue then yellow) = (10/20)(1/19) = (1/2)(1/19) = 1/38

P(yellow then blue) = (1/20)*(10/19) = same as above = 1/38

Total = 2/38 = 1/19 ≈ 0.053

Problem 4: Fruits in Refrigerator

Total fruits: 6 + 5 + 10 + 3 + 7 + 11 + 2 = 44

4a) Probability of pulling out a pear:

P(pear) = 3/44 ≈ 0.068

4b) Probability that first draw is a banana (10/44), then an apple (6/43):

P = (10/44) * (6/43) ≈ 0.0318

4c) Probability that the fruit drawn is a mango or an orange:

P(mango or orange) = P(mango) + P(orange) since mutually exclusive:

P(mango) = 2/44 ≈ 0.045; P(orange) = 5/44 ≈ 0.114; Total ≈ 0.159

Problem 5: Dice Roll Events

Sample Space

Since each die has 6 faces, total outcomes are 36, represented as pairs (d1, d2), with d1 and d2 from 1 to 6.

Event A: First die is 3 or 4, followed by an even number

Number of outcomes where first roll is 3 or 4: 2 * 6 = 12, and second roll is even (2,4,6): 3 outcomes per first roll, total 6.

P(A) = 6/36 = 1/6 ≈ 0.167

Event B: Sum at most 7

Number of outcomes where sum ≤ 7: count systematically or realize it's symmetrical; total are 15 outcomes.

P(B) = 15/36 = 5/12 ≈ 0.417

Event AB: both A and B happen

Calculate intersection by identifying outcomes satisfying both conditions, then find P(AB).

Mutually Exclusive & Independence

Events A and B are not mutually exclusive as some outcomes may satisfy both. To test independence, compare P(AB) with P(A)*P(B), and see if they are equal.

Problem 6: Course Requirements Probability

Given:

  • P(Final exam) = 0.72
  • P(research paper) = 0.46
  • P(both) = 0.32

a) Probability that a course has a final or research: P(F ∪ R) = P(F) + P(R) - P(F ∩ R) = 0.72 + 0.46 - 0.32 = 0.86

b) Probability that a course has neither: P(neither) = 1 - P(F ∪ R) = 1 - 0.86 = 0.14

Problem 7: Expected Gain or Loss in Lottery

Cost per ticket: $10. Prizes:

  • $500: 1 ticket
  • $100: 2 tickets
  • $25: 4 tickets

Total tickets: 100.

Expected value E = sum of (prize value * probability of winning that prize) - cost

Calculations:

E = (1/100)($500) + (2/100)($100) + (4/100)*($25) - $10 = ($5) + ($2) + ($1) - $10 = -$2

Expected loss of $2 per ticket.

Problem 8: Class Sizes

Total classes: 1 with 30, 8 with 60, 1 with 70, 4 with 100 students.

a) Average class size:

= (130 + 860 + 170 + 4100) / 14 = (30 + 480 + 70 + 400)/14 = 980/14 ≈ 70 students

b) Total capacity: 980 students across 14 classes. Randomly selecting a student, PDF for X (class size):

  • X=30 with probability 1/14
  • X=60 with probability 8/14
  • X=70 with probability 1/14
  • X=100 with probability 4/14

c) Mean of X:

= (30(1/14) + 60(8/14) +70(1/14) +100(4/14)) ≈ 70

d) Standard deviation involves calculating variance using probabilities and then taking the square root.

Problem 9: Attendance at Tet Festivities

Define X as the number of students out of 12 attending Tet.

X can take values 0 through 12.

X follows a Binomial distribution with n=12, p=0.18: X ~ Binomial(12, 0.18).

Expected number attending: E[X] = 12 * 0.18 ≈ 2.16 students.

Calculations for probabilities of at most four and more than two students involve binomial probability formulas and cumulative sums.

The probability that at most four students attend is sum of P(X=k) for k=0 to 4.

The probability that more than two attend is 1 minus P(X ≤ 2).

References

  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences (9th ed.). Brooks/Cole.
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