- The probability that it is Friday and that a student is
absent is 0.03. Since there are 5 school days in a week, the
probability that it is Friday is 0.2. What is the probability
that a student is absent given that today is Friday?
- At Kennedy Middle School, the probability that a student
takes Technology and Spanish is 0.087. The probability
that a student takes Technology is 0.68. What is the
probability that a student takes Spanish given that the
student is taking Technology?
- A car dealership is giving away a trip to Rome to one of
their 120 best customers. In this group, 65 are women,
80 are married and 45 married women. If the winner is
married, what is the probability that it is a woman?
- A card is chosen at random from a deck of 52 cards. It is
then replaced and a second card is chosen. What is the
probability of choosing a jack and then an eight?
- A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles.
A marble is chosen at random from the jar. After replacing
it, a second marble is chosen. What is the probability of
choosing a green and then a yellow marble?
- A school survey found that 9 out of 10 students like
pizza. If three students are chosen at random with
replacement, what is the probability that all the three
students like pizza?
- A nationwide survey found that 72% of people in the
United States like pizza. If 3 people are selected at
random, what is the probability that all the three like
pizza?
Below are the solutions to the probability problems stated in your request:
Problem 1
Question: What is the probability that a student is absent given that today is Friday?
Given Data:
- Probability of being absent on Friday, ( P(A \cap F) = 0.03 )
- Probability that it is Friday, ( P(F) = 0.2 )
Using Conditional Probability:
To find the probability that a student is absent given that today is Friday, we use the formula for conditional probability:
[
P(A | F) = \frac{P(A \cap F)}{P(F)}
]
Calculation:
[
P(A | F) = \frac{0.03}{0.2} = 0.15
]
Answer: The probability that a student is absent given that today is Friday is 0.15.
Problem 2
Question: What is the probability that a student takes Spanish given that the student is taking Technology?
Given Data:
- Probability of taking Technology and Spanish, ( P(T \cap S) = 0.087 )
- Probability of taking Technology, ( P(T) = 0.68 )
Using Conditional Probability:
[
P(S | T) = \frac{P(T \cap S)}{P(T)}
]
Calculation:
[
P(S | T) = \frac{0.087}{0.68} \approx 0.128
]
Answer: The probability that a student takes Spanish given that the student is taking Technology is approximately 0.128.
Problem 3
Question: If the winner is married, what is the probability that it is a woman?
Given Data:
- Total customers: 120
- Women: 65
- Married: 80
- Married women: 45
Using Conditional Probability:
Let ( W ) be the event of being a woman and ( M ) be the event of being married.
[
P(W | M) = \frac{P(W \cap M)}{P(M)}
]
Where:
- ( P(W \cap M) = \frac{45}{120} )
- ( P(M) = \frac{80}{120} )
Calculation:
[
P(W | M) = \frac{\frac{45}{120}}{\frac{80}{120}} = \frac{45}{80} = 0.5625
]
Answer: The probability that the winner is a woman given that they are married is 0.5625.
Problem 4
Question: What is the probability of choosing a jack and then an eight?
Given Data:
A standard deck of cards has 52 cards.
Calculating Individual Probabilities:
1. Probability of choosing a Jack first:
- There are 4 Jacks in a deck.
- ( P(J) = \frac{4}{52} = \frac{1}{13} )
2. Since the card is replaced, the probability of choosing an Eight:
- There are also 4 Eights in a deck.
- ( P(8) = \frac{4}{52} = \frac{1}{13} )
Combined Probability:
[
P(J \text{ and } 8) = P(J) \times P(8) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}
]
Answer: The probability of choosing a Jack and then an Eight is (\frac{1}{169}).
Problem 5
Question: What is the probability of choosing a green and then a yellow marble?
Given Data:
- Red: 3
- Green: 5
- Blue: 2
- Yellow: 6
Total marbles: ( 3 + 5 + 2 + 6 = 16 )
Calculating Individual Probabilities:
1. Probability of choosing a Green marble first:
- ( P(G) = \frac{5}{16} )
2. Since the marble is replaced, the probability of choosing a Yellow marble:
- ( P(Y) = \frac{6}{16} = \frac{3}{8} )
Combined Probability:
[
P(G \text{ and } Y) = P(G) \times P(Y) = \frac{5}{16} \times \frac{3}{8} = \frac{15}{128}
]
Answer: The probability of choosing a green and then a yellow marble is (\frac{15}{128}).
Problem 6
Question: What is the probability that all three students like pizza if 9 out of 10 students like pizza?
Given Data:
- Probability of liking pizza, ( P(L) = 0.9 )
Using Independence:
For independent events, the probability of all three liking pizza:
[
P(L)^3 = (0.9)^3
]
Calculation:
[
P(L)^3 = 0.729
]
Answer: The probability that all three students like pizza is 0.729.
Problem 7
Question: What is the probability that all three people like pizza if 72% of people like pizza?
Given Data:
- Probability of liking pizza, ( P(L) = 0.72 )
Using Independence:
For independent events, the probability of all three liking pizza:
[
P(L)^3 = (0.72)^3
]
Calculation:
[
P(L)^3 = 0.373248
]
Answer: The probability that all three people like pizza is approximately 0.3732.
These solutions provide a comprehensive view of each scenario, employing appropriate mathematical techniques and principles of probability to arrive at clear answers.