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Consider the experiment of flipping a balanced coin three times independently. The questions involve basic probability calculations, including determining sample space size, computing probabilities of specific outcomes, analyzing independence of events, and understanding related probability concepts. The tasks include identifying the size of the sample space, calculating the probability of getting exactly two heads, assessing if certain events are independent or disjoint, and exploring probabilities of other related events.

Paper For Above instruction

Probability theory is a fundamental area of mathematics applicable in numerous fields such as statistics, finance, engineering, and science. It deals with quantifying uncertainty and predicting the likelihood of various outcomes within a defined sample space. This paper explores concepts related to conditional probability, independence, and specific probability calculations through the context of a simple coin-flipping experiment and other probability scenarios involving dice and events.

Analysis of Coin Toss Experiment

The first problem involves flipping a fair coin three times independently. The sample space, outcomes, and probability of specific events are assessed. For a fair coin, each flip has two possible outcomes: heads (H) or tails (T). Since the flips are independent, the total number of possible outcomes for three flips is calculated using the multiplication principle: 2 x 2 x 2 = 8. Consequently, the sample space contains 8 outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

The probability of obtaining exactly two heads in three flips is calculated by counting the outcomes with exactly two Hs: {HHT, HTH, THH}. Each outcome has a probability of (1/2)^3 = 1/8, since each flip is independent and has a probability of 1/2. There are 3 such outcomes, so the probability is 3/8 = 0.375. This matches the given option in the multiple-choice question.

The question of whether the events “exactly two heads” and “exactly three heads” are independent involves understanding that the occurrence of one affects the probability of the other. Since these two events are mutually exclusive (they cannot happen simultaneously), they are disjoint. Disjoint events are dependent because the occurrence of one precludes the occurrence of the other, meaning their probabilities are not independent.

In the case of the events “the first coin is heads” and “the second and third coins are tails,” the calculation involves examining if knowing the outcome of the first flip influences the probability that the second and third are tails. Since each flip is independent, the outcome of the first flip does not affect the others. Therefore, these events are independent. The probability that the first coin is heads is 1/2, and the probability that the second and third are tails is (1/2) x (1/2) = 1/4. The joint probability under independence is (1/2) x (1/4) = 1/8, aligning with the principles of independence.

Dice Throwing Scenarios

The second problem involves rolling two fair six-sided dice independently. Several probabilities are computed, including the chance that the sum is less than or equal to 4, at least one die shows 4, and exactly one die shows 2 while the total sum is 4. The total number of outcomes when rolling two dice is 6 x 6 = 36.

Calculating the probability that the sum of the two dice is at most 4 involves enumerating the favorable outcomes: (1,1), (1,2), (2,1), and (2,2). The sum-≤4 outcomes are (1,1), (1,2), (2,1). There are 3 such outcomes, so the probability is 3/36 = 1/12 ≈ 0.0833, which indicates a close approximation to 0.0833. However, the multiple-choice options suggest considering all outcomes with sum ≤4, indicating the inclusion of (2,2), making a total of four outcomes and a probability of 4/36 = 1/9 ≈ 0.1111, which better fits option A.

The probability that at least one die shows 4 requires calculating the complementary probability that neither die shows 4, then subtracting from 1: total outcomes with no 4 are 5 options for each die, leading to 5 x 5 = 25 outcomes. Thus, probability that at least one die shows 4 is 1 - 25/36 = 11/36 ≈ 0.3056, matching option B.

When calculating the probability that one die shows 1 and the sum is 4, valid outcomes are (1,3) and (3,1). These are 2 outcomes, so the probability is 2/36 = 1/18 ≈ 0.0556, aligning with option A.

Assessing whether events A: sum of dice is 4, and B: exactly one die shows 2, are independent involves examining if P(A ∩ B) equals P(A) x P(B). The event A includes outcomes: (1,3), (3,1), (2,2), so P(A) = 3/36 = 1/12.

Event B, “exactly one die shows 2,” includes outcomes: (2,1), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2). There are 10 outcomes, so P(B) = 10/36 ≈ 0.2778. The intersection A ∩ B involves outcomes where the sum is 4 and exactly one die shows 2, i.e., (2,2) (sum=4, but both dice show 2), which is not included in "exactly one die shows 2." Outcome (2,2) has both dice showing 2, so it’s not in B. Therefore, A ∩ B is empty, probability zero. Since P(A ∩ B) = 0, but P(A) x P(B) ≈ 0.077, the two are dependent, not independent.

Event Dependence and Probabilities

The third problem involves a probability event where the nature of dependence between events A and B is analyzed, with calculations indicating the likelihood of combined occurrences. If the events are independent, then the probability of their intersection equals the product of their individual probabilities. Calculations often involve using conditional probabilities, joint probabilities, and marginal probabilities, emphasizing the importance of understanding event relationships in probabilistic models.

Rain Tomorrow Scenario

Another scenario considers the probability of rain tomorrow being 0.23. The probability that it does not rain is simply 1 - 0.23 = 0.77, utilizing complement rule. This straightforward example emphasizes how complements simplify probability calculations when an event’s probability is known.

Probability of Opening Branches in Cities

The final problem involves calculating the probability of a factory opening branches in Riyadh and Jeddah. Given probabilities for each city individually and the probability for either, we can employ the inclusion-exclusion principle: P(Riyadh ∪ Jeddah) = P(Riyadh) + P(Jeddah) - P(Riyadh ∩ Jeddah). Using the provided probabilities, P(Riyadh ∩ Jeddah) = P(Riyadh) + P(Jeddah) - P(Riyadh ∪ Jeddah) = 0.7 + 0.4 - 0.8 = 0.3.

The probability that the factory opens branches in both cities is 0.3, and the probability that it opens in neither city can be found by considering the complement of opening in either city: 1 - P(Riyadh ∪ Jeddah) = 1 - 0.8 = 0.2. Alternatively, the probability of neither city being chosen is (1 - 0.7) x (1 - 0.4) = 0.3 x 0.6 = 0.18, suggesting a slight discrepancy indicating the importance of precise calculation with known joint probabilities.

Conclusion

This analysis demonstrates that probability theory provides essential tools for understanding random experiments, whether involving coin flips, dice throws, or real-world decision-making scenarios such as marketing or environmental prediction. Independence, conditional probability, and event relationships are crucial concepts for interpreting probabilistic data accurately and making informed decisions based on statistical evidence.

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