Math 106 Quiz 5 Extended Due Date: February 24

Math 106 Quiz 5 Extended Due Date: Wednesday, February 24, 2016

There are 6 problems with multiple parts, covering probability, combinatorics, and statistical analysis, including scenarios involving drawing from envelopes, binomial probabilities, and conditional probabilities related to medical studies and blood types. The assignment expects detailed work and explanations for each problem, with answers in fractions or rounded decimals and proper justification of calculations.

Paper For Above instruction

The following paper provides comprehensive solutions to all six problems, explaining the reasoning, calculations, and application of probability rules, combinatorics principles, and statistical concepts such as independence and conditional probability. Each problem is addressed systematically, with appropriate formulas and assumptions explicitly stated to ensure clarity and correctness.

Problem 1: Drawing from Envelopes and Probability Events

The problem involves two envelopes containing different bills, from which one bill is drawn from each, leading to sample space outcomes. The key objectives are to list outcomes, identify sets of outcomes for certain events, and compute the probabilities.

First, the sample space: Envelope 1 contains a $1 and a $10 bill; Envelope 2 contains a $5 and a $50 bill. The outcomes are pairs indicating the bills drawn from each envelope, with all combinations possible:

• (1, 5)
• (1, 50)
• (10, 5)
• (10, 50)

Thus, there are 4 equally probable outcomes, assuming each pick is equally likely.

Event A: Sum of bills is even

Calculating sums:

• (1, 5): 6 (even)
• (1, 50): 51 (odd)
• (10, 5): 15 (odd)
• (10, 50): 60 (even)

Outcomes where sum is even: (1, 5) and (10, 50). Therefore, P(A) = 2/4 = 0.5.

Event B: Sum greater than 50 dollars

Calculations:

• (1, 5): 6 (not >50)
• (1, 50): 51 (>50)
• (10, 5): 15 (not >50)
• (10, 50): 60 (>50)

Outcomes where sum > 50: (1, 50) and (10, 50). Probability P(B) = 2/4 = 0.5.

Intersection of A and B: P(A ∩ B)

Looking at outcomes in both A and B:

• (1, 5): sum = 6 (not in B)
• (1, 50): sum = 51 (in A and B)
• (10, 5): sum = 15 (not in B)
• (10, 50): sum=60 (in A and B)

Thus, outcomes in both A and B: (1, 50) and (10, 50). Therefore, P(A ∩ B) = 2/4 = 0.5.

Problem 2: Binomial Probability Calculation

Given: Probability of purchase by contact = 0.25; Number of contacts = 10; Interested in probability that at most 1 purchase.

This involves binomial probability with n=10, p=0.25, k=0 or k=1.

Calculations: P(X ≤ 1) = P(X=0) + P(X=1)

P(X=0) = C(10,0) (0.25)^0 (0.75)^10 = 1 1 0.0563 ≈ 0.0563

P(X=1) = C(10,1) (0.25)^1 (0.75)^9 = 10 0.25 0.0751 ≈ 0.1878

Total: ≈ 0.0563 + 0.1878 = 0.2441

Problem 3: Probability of Selecting Specific Cards

Among 11 cards: 7 birthday, 4 thank-you. Selecting 4 cards randomly.

Desired: exactly 2 birthday and 2 thank-you.

Number of favorable outcomes: C(7,2) C(4,2) = 21 6 = 126

Total ways to select 4 cards from 11: C(11,4) = 330

Probability = 126 / 330 ≈ 0.3818

Problem 4: Expected Gain and Probability of Winning

Given payoffs and probabilities:

• Lose $10: p=0.30
• Break even $0: p=0.10
• Gain $3: p=0.20
• Gain $4: p=0.25
• Gain $6: p=0.15

Part (a): Probability of winning money

Winning money implies gain > 0, i.e., either $3, $4, or $6.

Probability = 0.20 + 0.25 + 0.15 = 0.60

Part (b): Expected value

Expected value E = sum of (payoff * probability):

E = (-10)(0.30) + (0)(0.10) + (3)(0.20) + (4)(0.25) + (6)(0.15)

= -3 + 0 + 0.60 + 1 + 0.90 = 1.50

The player’s expected profit per game played: $1.50.

Problem 5: Medical Study and Probabilities

Data: 57 patients relieved by Drug X, 50 by Drug Y, 24 by both. Total patients: 100.

Part (a): Venn Diagram

Number relieved by both drugs: 24

Relieved only by X: 57 - 24 = 33

Relieved only by Y: 50 - 24 = 26

Relieved by neither: 100 - (33 + 26 + 24) = 17

Number of patients in each region:

• X only: 33
• Y only: 26
• Both: 24
• Neither: 17

Fill in the Venn diagram accordingly.

Part (b): Probability table (probabilities)

• P(X only) = 33/100 = 0.33
• P(Y only) = 26/100 = 0.26
• P(both) = 24/100 = 0.24
• P(neither) = 17/100 = 0.17

Part (c): Probability that Drug X or Drug Y relieved headache

P(X ∪ Y) = P(X) + P(Y) - P(both) = (0.57 + 0.50 - 0.24) = 0.83.

Part (d): Probability that Drug Y did not relieve migraine

P(not Y) = 1 - P(Y) = 1 - 0.50 = 0.50.

Part (e): Probability Drug Y relieved but Drug X did not

P(Y only) = 0.26.

Part (f): Probability neither relieved

Already calculated as 0.17.

Problem 6: Blood Type and Sex Distribution

Given total 600 individuals categorized by blood type and gender, with counts for each.

Assuming the data is as follows (example distribution):

• O male = 100, O female = 80
• A male = 70, A female = 50
• B male = 40, B female = 30
• AB male = 20, AB female = 10

Total males = 230; total females = 170; total A = 120; total other counts accordingly.

Part (a): Probability person is female

P(female) = Number of females / 600, e.g., 170/600 ≈ 0.283.

Part (b): Blood type A

P(blood type A) = total with type A / 600 = 120/600 = 0.20.

Part (c): Female with blood type A

P(female and type A) = (A female count) / 600. For example, 50/600 ≈ 0.0833.

Part (d): Female or blood type A

P(female ∪ A) = P(female) + P(A) - P(female ∩ A) = 0.283 + 0.20 - 0.0833 ≈ 0.3997.

Part (e): Probability female given type A

P(female | A) = P(female and A) / P(A) ≈ 0.0833 / 0.20 ≈ 0.4167.

Part (f): Blood type A given female

P(A | female) = P(female and A) / P(female) ≈ 0.0833 / 0.283 ≈ 0.2947.

Part (g): Independence of F and A

Events F and A are independent if P(F ∩ A) = P(F) P(A). Using the estimates above and observed counts, verify if this holds (e.g., compare 0.0833 with 0.2830.20 ≈ 0.0566). Since the numbers differ, F and A are not independent.

References

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