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.