Probability A Battery Lasts 10 Hours Or More Is 7

The Probability That A Battery Will Last 10 Hr Or More Is 7

QUESTION 1 The probability that a battery will last 10 hr or more is .75, and the probability that it will last 29 hr or more is .18. Given that a battery has lasted 10 hr, find the probability that it will last 29 hr or more. a.0.205 b.0.22 c.0.24 d.0. points

QUESTION 2 A pair of fair dice is cast. Let E denote the event that the number landing uppermost on the first die is a 6 and let F denote the event that the sum of the numbers falling uppermost is 8. Determine whether E and F are independent events. a.dependent b.independent 1 points

QUESTION 3 The proprietor of Cunningham's Hardware Store has decided to install floodlights on the premises as a measure against vandalism and theft. If the probability is 0.1 that a certain brand of floodlight will burn out within a year, find the minimum number of floodlights that must be installed to ensure that the probability that at least one of them will remain functional within the year is at least 0.999. (Assume that the floodlights operate independently.) a.100 b.4 c.2 d.11 e.3 1 points

QUESTION 4 An automobile manufacturer obtains the microprocessor used to regulate fuel consumption in its automobiles from three microelectronic firms: A, B, and C. The quality-control department of the company has determined that 1.5% of the microprocessors produced by firm A are defective, 1% of those produced by firm B are defective, and 2% of those produced by the firm C are defective. Firms A, B, and C supply 40%, 30%, and 30%, respectively, of the microprocessors used by the company. What is the probability that a randomly selected automobile manufactured by the company will have a defective microprocessor? a.0.01000 b.0.01500 c.0.00700 d.0.02000 e.0. points

QUESTION 5 Let A and B be events in a sample space S such that and . Find: . ​a. b. c. d. 1 points

QUESTION 6 A tax specialist has estimated the probability that a tax return selected at random will be audited is .02. Furthermore, he estimates the probability that an audited return will result in additional assessments being levied on the taxpayer is .60. What is the probability that a tax return selected at random will result in additional assessments being levied on the taxpayer? a.0.037 b.0.038 c.0.012 d.0. points

QUESTION 7 A pair of fair dice is cast. Let E denote the event that the number falling uppermost in the first die is 1 and let F denote the event that the sum of the numbers falling uppermost is 5. Compute . Are E and F dependent events? a. , yes b. , yes c. , no d. , no 1 points

QUESTION 8 Determine whether the given events A and B are independent. , , and . a.independent b.not independent 1 points

QUESTION 9 A card is drawn from a well-shuffled deck of 52 playing cards. Let E denote the event that the card drawn is red and let F denote the event that the card drawn is a hearts. Determine whether E and F are independent events. a.independent b.not independent 1 points

QUESTION 10 Determine whether the given events A and B are independent. , , and .​ a.dependent b.independent

Paper For Above instruction

The provided questions involve fundamental concepts in probability theory, including conditional probability, independence of events, and combined probability calculations involving multiple random variables. This analysis aims to systematically address these questions by exploring the mathematical principles behind each scenario and applying them to arrive at accurate solutions.

Question 1: Conditional Probability of a Battery Lasting 29 Hours Given it Lasted 10 Hours

The problem states that the probability a battery lasts at least 10 hours is 0.75, and the probability it lasts at least 29 hours is 0.18. The goal is to find the probability that it lasts 29 hours or more, given that it has already lasted 10 hours. This is a classic case for conditional probability, expressed as:

P(Lasts ≥ 29 hours | Lasts ≥ 10 hours) = P(Lasts ≥ 29 hours ∩ Lasts ≥ 10 hours) / P(Lasts ≥ 10 hours)

Since lasting 29 hours or more implies lasting 10 hours or more, the intersection is simply P(Lasts ≥ 29 hours). Therefore:

P(Lasts ≥ 29 hours | Lasts ≥ 10 hours) = P(Lasts ≥ 29 hours) / P(Lasts ≥ 10 hours) = 0.18 / 0.75 ≈ 0.24

Thus, the probability that a battery, having lasted at least 10 hours, will last at least 29 hours is approximately 0.24, matching option c.

Question 2: Independence of Events E and F in Rolling Dice

Event E: The first die shows a 6. Event F: The sum of both dice is 8. The probability of E is 1/6, and the probability of F is determined by counting the outcomes summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2). There are 5 such outcomes out of 36 total outcomes, so P(F)=5/36.

The probability that both E and F occur (first die is 6, sum is 8) is when first die is 6, and the total sum is 8: the only such outcome is (6,2). Thus, P(E ∩ F) = 1/36.

Check independence: P(E) P(F) = (1/6) (5/36) ≈ 0.0231, while P(E ∩ F) = 1/36 ≈ 0.0278. Since these are not equal, E and F are dependent events.

Answer: dependent (a).

Question 3: Floodlights to Achieve a High Reliability

The probability that a single floodlight burns out within a year is 0.1. We want at least one to remain functional with probability ≥ 0.999. The complement is all floodlights failing: (0.1)^n, so the probability at least one survives is 1 - (0.9)^n. Set:

1 - (0.9)^n ≥ 0.999

(0.9)^n ≤ 0.001. Taking natural logs:

n * ln(0.9) ≤ ln(0.001)

Since ln(0.9) ≈ -0.1054, and ln(0.001) ≈ -6.9078:

n ≥ -6.9078 / -0.1054 ≈ 65.5

Thus, at minimum, 66 floodlights are needed. Since 66 is not among options, approximate options are checked. Given options, choice d (11) is too small; none match the calculation precisely. The closest practical choice that ensures reliability is a higher number, but since options are limited, the correct mathematical calculation indicates that the minimum number is around 66, not listed. Suggested: this question anticipates approximations, hence likely answer is either the closest large number; but in real tests, choice a (100) may be suggestive.

Question 4: Probability of a Defective Microprocessor

Given the defective rates and manufacturer proportions, the overall probability that a randomly selected microprocessor is defective is computed using the Law of Total Probability:

P(defective) = P(A) P(defective | A) + P(B) P(defective | B) + P(C) * P(defective | C)

= 0.4 0.015 + 0.3 0.01 + 0.3 * 0.02 = 0.006 + 0.003 + 0.006 = 0.015

The probability is 0.015, aligning with option b.

Question 5: Calculating Probability for Events A and B

Details are incomplete in the prompt; assuming typical probability calculations involving intersection and union for events A and B, standard formulas apply:

P(A ∩ B) = P(A) * P(B | A), depending on independence or dependence.

Without specific values, the question cannot be fully answered. Assuming independence and known probabilities, the calculation would follow standard probability rules.

Question 6: Probability of Additional Assessments Based on Audit Probability

The probability that a randomly selected tax return results in additional assessments is:

P(Occurred) = P(Audited) P(Additional | Audited) = 0.02 0.60 = 0.012

Answer: 0.012 (option c).

Question 7: Dependence Between Dice Events

The events: E (first die shows 1), F (sum is 5). The probability of E is 1/6. The probability of F (sum to 5) includes outcomes: (1,4), (2,3), (3,2), (4,1). P(F) = 4/36 = 1/9.

The joint event E ∩ F occurs only if first die is 1 and sum is 5: only (1,4), which has probability 1/36.

Check independence: P(E) P(F) = (1/6)(1/9)= 1/54 ≈ 0.0185, P(E ∩ F)=1/36≈0.0278. Not equal, thus dependent events.

Answer: yes (a).

Question 8: Independence of Events A and B

Without specific data, standard approach: verify P(A ∩ B) = P(A) * P(B). If true, independent; if not, dependent. Given the incomplete data, the general assumption suggests they are not independent.

Answer: not independent.

Question 9: Independence of Drawing Red and Hearts from a Deck

Probability card is red: 26/52= 1/2. Probability card is a heart: 13/52= 1/4. The probability a card is both red and a heart: 13/52= 1/4. Since P(E ∩ F) = P(F), and P(E)=1/2, P(F)=1/4, P(E) * P(F)=1/8=0.125, but actual joint probability is 0.25, thus events are dependent.

Answer: not independent (b).

Question 10: Dependence of Events A and B

Insufficient details given. Based on typical probability principles, if P(A ∩ B) ≠ P(A) * P(B), they are dependent. Without specific data, the default assumption is that the events are dependent.

References

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