Solved Problems in Probability and Statistics

Posted on Jan 26, 2025 in Mathematics

Probability and Statistics Problems

Problem 1: Selecting a Group

Florida State University (FSU) wants to select 3 men and 2 women from a group of 8 men and 5 women. How many ways can the group of five persons be selected?

Solution:

• Ways to select 3 men from 8: 8! / (3! * (8-3)!) = 56
• Ways to select 2 women from 5: 5! / (2! * (5-2)!) = 10
• Total ways to select the group: 56 * 10 = 560

Therefore, there are 560 ways to select the group.

Problem 2: DEBORA

Problem 3: New Toy Success Rate

A toy company is planning to introduce 3 new toys next Christmas. The toy sales can be a success or a failure. What is the probability that exactly 2 of the toys are a success?

Solution:

Possible outcomes (S = Success, F = Failure):

• SSS
• SSF
• SFS
• SFF
• FSS
• FFS
• FSF
• FFF

Outcomes with exactly 2 successes: SSF, SFS, FSS (3 outcomes)

Total possible outcomes: 8

Probability of exactly 2 successes: 3/8 = 0.375

Therefore, the probability is 0.375.

Problem 4: TOYA

Problem 5: TOYA

Problem 6: Combination Lock Possibilities

A combination lock system has four ten-digit number wheels side by side. How many four-digit opening numbers are possible if a digit cannot be used more than once and the first digit cannot be 9 or 0?

Solution:

For the first digit, there are 8 possibilities (1-8). For the second digit, there are 9 possibilities (0-9 excluding the first digit). For the third digit, there are 8 possibilities, and for the fourth digit, there are 7 possibilities.

Total possible combinations: 8 * 9 * 8 * 7 = 4032

Therefore, there are 4032 possible opening numbers.

Problem 7: TOYA

Problem 8: DEBORA

Problem 9: DEBORA

Problems 10-13: Real Estate Data Analysis

A listing of 100 houses for sale by a realty firm produced the following breakdown:

If a house is selected out of this group, what is the probability that:

Problem 10: House has 4 or More Bedrooms

Solution:

P(4 or more bedrooms) = 34/100 = 0.34

The probability is 0.34.

Problem 11: House has 3 Bedrooms Given 1 Bathroom

Solution:

P(3 bedrooms | 1 bathroom) = 14/22 = 0.64

The probability is approximately 0.64.

Problem 12: House has 1 Bathroom or 2 Bedrooms

Solution:

P(1 bathroom or 2 bedrooms) = P(1 bathroom) + P(2 bedrooms) – P(1 bathroom and 2 bedrooms)

P = 22/100 + 18/100 – 8/100 = 32/100 = 0.32

The probability is 0.32.

Problem 13: House has 2 or Less Bathrooms and 3 or More Bedrooms

Solution:

P(2 or less bathrooms and 3 or more bedrooms) = (14 + 26 + 22) / 100 = 62/100 = 0.62

The probability is 0.62.

Problem 14: Weighted Average Textbook Cost

FSU bookstore has purchased Fundamental Business Statistics textbooks as follows: 40 units at $65, 30 units at $75, and 25 units at $85. What is the weighted average cost paid for a textbook?

Solution:

Sum of weighted values (Σwx) = (40 * $65) + (30 * $75) + (25 * $85) = $2600 + $2250 + $2125 = $6975

Sum of weights (Σw) = 40 + 30 + 25 = 95

Weighted average (Xw) = Σwx / Σw = $6975 / 95 = $73.42

The weighted average cost is $73.42.

Problem 15: Probability of Product Rejection

Apple Computers is planning to introduce two new products: a wireless MP3 player and a cell phone. Previous studies indicate that the wireless MP3 player will have a probability of acceptance of 0.7 and the cell phone a probability of acceptance of 0.3. If the selection of the products is independent, what is the probability that both products will be rejected?

Solution:

P(MP3 rejection) = 1 – 0.7 = 0.3

P(Cell phone rejection) = 1 – 0.3 = 0.7

P(Both rejected) = P(MP3 rejection) * P(Cell phone rejection) = 0.3 * 0.7 = 0.21

The probability that both products will be rejected is 0.21.

Problem 16: Probability of Drawing Marbles

Two marbles are drawn at random and without replacement from a box containing seven blue marbles and three red marbles. What is the probability that a red and a blue marble are drawn/selected?

Solution:

P(Red then Blue) = (3/10) * (7/9) = 21/90

P(Blue then Red) = (7/10) * (3/9) = 21/90

P(Red and Blue) = P(Red then Blue) + P(Blue then Red) = 21/90 + 21/90 = 42/90 = 0.467

The probability is approximately 0.467.

Problem 17: Defective Cars Distribution

Of six cars produced at a particular factory, three are known to be defective. Three of the six cars are shipped to dealer A and the others to dealer B. If the cars are selected at random, what is the probability that dealer A will receive the three defective cars?

Solution:

P(Dealer A gets all 3 defective cars) = (3/6) * (2/5) * (1/4) = 6/120 = 0.05

The probability is 0.05.

Problems 18-20: Automobile Speed Analysis

The top speeds for a sample of eight automobile brands are the following: 160, 146, 144, 161, 156, 188, 170, 186 km/h

Problem 18: Mean Speed

Solution:

Mean = (160 + 146 + 144 + 161 + 156 + 188 + 170 + 186) / 8 = 1311 / 8 = 163.875

The mean speed is 163.875 km/h.

Problem 19: Median Speed

Solution:

First, order the speeds: 144, 146, 156, 160, 161, 170, 186, 188

Median = (160 + 161) / 2 = 160.5

The median speed is 160.5 km/h.

Problem 20: Standard Deviation