A Bag Contains 5 Purple Jelly Beans, 3 Pink Jelly Beans, And

A Bag Contains 5 Purple Jelly Beans 3 Pink Jelly Beans And 2 Yello

A bag contains 5 purple jelly beans, 3 pink jelly beans, and 2 yellow jelly beans. Two consecutive draws are made from the bag with replacement. Find the probability of drawing two pink jelly beans. Additionally, analyze the probability of drawing two purple jelly beans without replacement. Other problems involve permutations, combinations, probability of student demographics, and geometric probability. These problems collectively test understanding of basic probability, combinatorics, and geometric concepts in probability.

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The problem set presented involves various probability and combinatorics scenarios that are fundamental to understanding how to analyze random events and sample spaces. This comprehensive analysis discusses each problem, applying appropriate probability rules, counting principles, and geometric probability concepts, with detailed calculations and explanations.

Problem 1: Probability of Drawing Two Pink Jelly Beans with Replacement

The first problem involves selecting two pink jelly beans from a bag containing 5 purple, 3 pink, and 2 yellow jelly beans. Since the draws are with replacement, the composition of the bag remains the same after each draw. Therefore, the probability of drawing a pink jelly bean on each individual draw is the ratio of pink jelly beans to the total number of jelly beans.

The total number of jelly beans in the bag is 5 + 3 + 2 = 10. The probability of drawing a pink jelly bean in one draw is therefore 3/10. Since the draws are independent due to replacement, the probability of drawing pink jelly beans twice consecutively is the product of the individual probabilities:

P(pink twice) = P(pink on first draw) × P(pink on second draw) = (3/10) × (3/10) = 9/100 = 0.09.

Problem 2: Probability of Drawing Two Purple Jelly Beans Without Replacement

In contrast to the first problem, here the draws are without replacement, affecting the probabilities on subsequent draws. The initial probability of drawing a purple jelly bean is 5/10. After removing one purple jelly bean, there are 4 purple jelly beans left in a total of 9 remaining jelly beans.

Therefore, the probability of drawing two purple jelly beans without replacement is the product of probabilities on each step:

P(purple first) = 5/10 = 1/2.

P(purple second | first purple) = 4/9.

Thus, combined probability:

P(two purple without replacement) = (5/10) × (4/9) = (1/2) × (4/9) = 4/18 = 2/9 ≈ 0.2222.

Problem 3: Number of Ways to Select 3 Colors from 8

In choosing 3 colors out of 8 available, the order of selection does not matter, so we employ combinations. The number of ways is given by the binomial coefficient:

C(8,3) = 8! / (3! × (8-3)!) = 8! / (3! × 5!) = (8×7×6) / (3×2×1) = 56.

Hence, Katie can select her colors in 56 different ways.

Problem 4: Probability that a Randomly Chosen Graduate Was in Band or Took Art

Given data: Total students = 200, students took art = 45, students in band = 89, students who took art but not band = 20.

To find the number of students in both art and band, use the principle of inclusion-exclusion:

Students in both art and band = (Students in art) + (Students in band) – (Students in either art or band).

But since we need the students in band or art, the formula becomes:

Number in band or art = (# in art) + (# in band) – (# in both).

To find # in both, note that students in art but not band = 20, so students in both art and band = 45 – 20 = 25.

Now, total students in band or art:

45 + 89 – 25 = 109.

Probability: 109/200 = 0.545.

Thus, the probability that a randomly selected student was in the band or took art is approximately 54.5%.

Problem 5: Number of Ways to Award 1st, 2nd, 3rd, and 4th Places Among 9 Contestants

This is a permutation problem since the order of winners matters. Total arrangements are calculated by permutations of 9 contestants taken 4 at a time:

P(9,4) = 9×8×7×6 = 3024.

Therefore, there are 3024 different ways to award the prizes.

Problem 6: Probability that a Respondent is Male or Does Not Oppose the Change

Given: Total respondents = 100, males = 50, oppose change = 37, females oppose change = 25.

Number of females opposing = 25, hence females not opposing = 37 – 25 = 12 (incorrect assumption), but considering the data more carefully:

• Number of oppose = 37
• Number of female oppose = 25
• Number of male oppose = 37 – 25 = 12

Number of males = 50, and males oppose = 12, thus males not oppose = 38. The total number who do not oppose the change (either favor or no opinion):

Total respondents not oppose = 100 – 37 = 63.

The probability that a respondent is male or does not oppose the change is:

P(male or not oppose) = P(male) + P(not oppose) – P(male and oppose).

P(male) = 50/100 = 0.5.

P(not oppose) = 63/100 = 0.63.

P(male and oppose) = 12/100 = 0.12.

Therefore,

P(male or not oppose) = 0.5 + 0.63 – 0.12 = 1.01 – 0.12 = 0.89.

Problem 7: Probability at Least Two Friends Choose the Same Costume

There are 15 costumes, and four friends choosing independently. To find the probability that at least two friends choose the same costume, it's easier to compute the complement—the probability that all four friends select different costumes—and subtract this from 1.

The total number of outcomes where all friends pick different costumes:

For the first friend: 15 choices.

For the second: 14 choices (excluding the first's choice).

For the third: 13 choices.

For the fourth: 12 choices.

Number of outcomes with all different choices: 15×14×13×12.

Total number of possible outcomes, with repetition allowed: 15^4.

Thus, probability all choose different costumes:

P(all different) = (15×14×13×12) / 15^4 = (15×14×13×12) / (15×15×15×15) = (14×13×12) / (15×15×15).

Calculating numerator: 14×13=182, 182×12=2184.

Denominator: 15×15=225, 225×15=3375.

Therefore, P(all different) = 2184 / 3375 ≈ 0.646.

Finally, probability that at least two friends select the same costume:

1 – 0.646 ≈ 0.354.

Problem 8: Probability License Plate Begins with a Vowel and Ends with a Multiple of 3

License plate structure: 2 letters followed by 4 digits. Letters and digits can be repeated.

Number of possible vowels: A, E, I, O, U = 5.

Number of total letters: 26.

Number of total digits: 10 (0-9).

Probability that the first letter is a vowel:

P(first letter vowel) = 5/26.

Probability that the last digit is a multiple of 3 (digits 0-9): 0, 3, 6, 9, totaling 4 options, so:

P(last digit multiple of 3) = 4/10 = 2/5.

Since choices are independent, the combined probability:

P = (5/26) × (26/26) × (26/26) × (26/26) × (2/5). -- correction: for digits, only the last digit's probability matters, and the other digits can be any digit: so total possibilities for the four digits are 10^4, but the probability for the last digit being a multiple of 3 is 4/10. The initial 2 letters can be any letters, with no restriction.

Number of total license plates: 26×26×10×10×10×10 = 26^2 × 10^4.

Number of plates beginning with a vowel and ending with a multiple of 3: 5 (vowels) × 26 (second letter) × 10^3 (middle digits) × 4 (last digit multiple of 3).

Probability:

P = (5/26) × (26/26) × (10/10)^3 × (4/10) = (5/26) × 1 × 1 × 4/10 = (5/26) × 4/10 = (5×4)/(26×10) = 20/260 ≈ 0.077.

Problem 9: Probability Number Cube > 4 and Both Coins Heads

Number cube has outcomes 1-6. The favorable outcomes for the cube are numbers > 4, i.e., 5 and 6. Probability: 2/6 = 1/3.

Two coins are tossed, and both must be heads: probability (H, H) = (1/2) × (1/2) = 1/4.

Since the dice roll and coin tosses are independent, combined probability:

P = (1/3) × (1/4) = 1/12 ≈ 0.0833.

Problem 10: Probability a Point is in the Larger Circle but Not in the Smaller

The two circles share the same center; the larger circle has radius 5 cm, and the smaller has radius 3 cm. The probability of randomly selecting a point in a specific area within a circle, assuming uniform distribution over the area, is proportional to that area.

Area of larger circle: π × 5^2 = 25π.

Area of smaller circle: π × 3^2 = 9π.

Probability that a point is in the larger circle but not in the smaller circle:

P = (Area of larger segment not overlapping smaller) / (Area of larger):

= (25π – 9π) / 25π = 16π / 25π = 16 / 25 = 0.64.

Thus, the probability is 0.64, indicating a 64% chance that a point randomly chosen within the larger circle falls outside the smaller circle.

Conclusion

These problems encompass a broad spectrum of probability and combinatorial topics, including basic probability calculations, permutations, combinations, and geometric probability. Successfully solving such problems requires a clear understanding of foundational concepts, such as independence, conditional probability, the principle of counting, and area ratios. These skills are vital for analyzing real-world scenarios—from understanding probabilities in chance events to understanding sampling distributions and geometric measures, essential in statistics, operations research, and decision-making processes.

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