Quiz 51: Consider A Family Of Three Children Find The Follow

Quiz 51 Consider A Family Of Three Children Find The Following Proba

Consider a family of three children. Find the following probabilities. a. P (two girls and a boy) b. P (at least one girl) c. P (children of both sexes) d. P (at least two girls)

A committee of four is selected from a total of 4 freshmen, 5 sophomores, and 6 juniors. Find the probabilities for the following events. a. At least three freshmen. b. All four of the same class. c. Not all four from the same class. d. Exactly three of the same class.

At a college, 60% of the students pass Accounting, 70% pass English, and 30% pass both of these courses. If a student is selected at random, find the following conditional probabilities. a. The student passes Accounting given that said student passed English. b. The student passes English given that said student passed Accounting.

John's probability of passing statistics is 40%, and Linda's probability of passing the same course is 70%. If the two events are independent, find the following probabilities. a. P(both of them will pass statistics) b. P(at least one of them will pass statistics)

A $1 lottery ticket offers a grand prize of $10,000; 10 runner-up prizes each paying $1000; 100 third-place prizes each paying $100; and 1,000 fourth-place prizes each paying $10. Find the expected value of entering this contest if 1 million tickets are sold.

Paper For Above instruction

The assignment involves calculating various probabilities in different contexts, including familial gender combinations, committee selections, college course passing probabilities, and a lottery's expected value. These problems showcase fundamental applications of probability theory, collectively illustrating how probabilistic calculations underpin decision-making in everyday scenarios, educational settings, and gambling contexts. This comprehensive analysis aims to demonstrate the methods for calculating probabilities using basic principles such as combinations, conditional probabilities, independence, and expected value, thereby highlighting their significance and practical utility.

Probabilities in a Family of Three Children

Considering a family with three children, each child's gender can be modeled as an independent event with two outcomes: boy (B) or girl (G). Assuming the probability of having a boy or girl is equal (each 0.5), the total number of possible gender combinations is 2^3=8, with each equally likely. These combinations are: BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.

a. The probability of having exactly two girls and one boy involves identifying combinations with exactly two G's and one B. These are BGG, GBG, GGB, totaling 3 out of 8. Therefore, P(two girls and a boy) = 3/8 = 0.375.

b. The probability of having at least one girl is 1 minus the probability of having no girls, which is only BBB. Thus, P(at least one girl) = 1 - P(BBB) = 1 - 1/8 = 7/8 = 0.875.

c. The probability of children being of both sexes involves having at least one boy and one girl among the three children, excluding the cases of all boys (BBB) and all girls (GGG). Hence, the probability is 1 - P(all boys) - P(all girls) = 1 - 1/8 - 1/8 = 6/8 = 3/4 = 0.75.

d. The probability of at least two girls involves the cases with two or three girls: BGG, GBG, GGB, GGG, totaling 4 with two G's and 1 with three G's, totaling 5 out of 8. So, P(at least two girls) = 5/8 = 0.625.

Committee Selection from Students

Analyzing the committee of four selected from 4 freshmen, 5 sophomores, and 6 juniors involves calculating probabilities for specific combinations. The total number of ways to select 4 students from 15 is C(15,4) = 1,365.

a. The probability of having at least three freshmen requires summing the probabilities of selecting exactly three and exactly four freshmen. The number of ways to choose 3 freshmen and 1 from others: C(4,3) C(11,1) = 4 11 = 44; selecting 4 freshmen: C(4,4)*C(11,0)=1. Total favorable ways: 44 + 1=45. So, P(at least three freshmen) = 45/1365 ≈ 0.033.

b. All four members from the same class means either four freshmen, four sophomores, or four juniors. Since only four freshmen are available, only the first case applies (C(4,4)=1). For sophomores: C(5,4)=5, for juniors: C(6,4)=15. Total: 1 + 5 + 15=21. Probability: 21/1365 ≈ 0.015.

c. Not all four from the same class is the complement of the previous, so P(not all same class) = 1 - (probability all four from one class). Hence, 1 - (1/1365 + 5/1365 + 15/1365) = 1 - 21/1365 ≈ 0.984.

d. Exactly three from one class and one from another involve calculating the sum over three possibilities: three freshmen and one from other classes, similarly for other classes, summing all relevant combinations. For exemplification, for freshmen: C(4,3)C(11,1)=44; for sophomores: C(5,3)C(10,1)=1010=100; for juniors: C(6,3)C(9,1)=20*9=180. Summing all: 44 + 100 + 180=324. Probability: 324/1365 ≈ 0.237.

Conditional Probabilities of Course Passing

At a college, 60% of students pass Accounting (A), 70% pass English (E), and 30% pass both. Using these, we can determine individual and joint probabilities to find conditional probabilities.

a. To find P(A|E), i.e., the probability a student passes Accounting given they passed English, we use the formula: P(A|E) = P(A ∩ E) / P(E). Given P(E)=0.70, and P(A ∩ E)=0.30, thus P(A|E)=0.30/0.70 ≈ 0.429.

b. Similarly, P(E|A)=P(A ∩ E) / P(A)=0.30/0.60=0.5.

Independent Events: John's and Linda's Statistics Passing Probabilities

John has a 40% chance of passing, Linda has 70%. Assuming independence, the probability both pass (a) is P(John passes) P(Linda passes) = 0.4 0.7=0.28.

(b) The probability at least one passes is P(John passes or Linda passes) = P(John) + P(Linda) - P(both). So, 0.4 + 0.7 - 0.28= 0.82.

Expected Value of a Lottery Ticket

The lottery offers a grand prize of $10,000 and various smaller prizes with specified quantities: 10 prizes of $1,000, 100 prizes of $100, and 1,000 prizes of $10, totaling 1,110 prizes. Out of 1 million tickets, the probability of winning each prize is respective ratio of prizes to tickets.

Expected value (E) per ticket is calculated as the sum over all prizes of (prize value * probability of winning that prize). The probabilities are:

- Grand prize: 1/1,000,000, expected contribution: $10,000 * 1/1,000,000 = $0.01

- Runner-up: 10 $1,000: 10/1,000,000=0.00001, expected: $1,000 0.00001= $0.01

- Third-place: 100 $100: 100/1,000,000=0.0001, expected: $100 0.0001= $0.01

- Fourth-place: 1,000 $10: 1,000/1,000,000=0.001, expected: $10 0.001= $0.01

Adding these, the total expected value per ticket is approximately $0.04.

Conclusion

In conclusion, probability theory offers powerful tools for analyzing diverse real-world scenarios. The calculated probabilities demonstrate how assumptions about independence and equality influence our predictions. The committee selection problem illustrates combinatorial calculations, while the conditional probability scenarios underpin decision-making in educational contexts. The lottery's expected value showcases how probability informs our understanding of gambling risks and rewards. Together, these insights emphasize the importance of mastering the principles of probability to interpret and navigate uncertainty effectively.

References

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