The probability of having homework over the weekend in math is 0.9,history is 0.6, and Spanish is 0.75.

1. What is the probability of having exactly one class that gives homework?
Is this a nCx problem?
I got an illogical answer: 10.54=1054% THIS IS NOT RIGHT!!!!!!!
2. What is the probability of having homework in at least 1 class?
My answer : 0.492

1.

could be 1 in math or 1 in history or 1 in Spanish
that is (math)(no history)(no Spanish) + ....

prob = .9(.4)(.25) + .6(.1)(.25) + .25(.1)(.4)
= .115

Exclude the case of no homework in any
which is (.1)(.4)(.25) = .01

prob(homework in at least 1 class) 1 - .01 = .99

To find the probability of having exactly one class that gives homework, you need to use the concept of combinations (nCx), where n is the total number of classes (in this case, 3) and x is the number of classes that give homework (in this case, 1).

1. Probability of having exactly one class that gives homework:
To calculate this probability, you can use the formula for combinations:

P(X = 1) = nCx * (p^x) * ((1 - p)^(n - x))

Let's calculate it step by step:

nCx = n! / ((n - x)! * x!)
= 3! / ((3-1)! * 1!)
= 3! / (2! * 1!)
= 3

p = probability of having homework in a class
For math: p(math) = 0.9
For history: p(history) = 0.6
For Spanish: p(Spanish) = 0.75

Substituting the values into the formula:
P(X = 1) = 3 * (0.9^1) * ((1 - 0.9)^(3 - 1))
= 3 * 0.9 * 0.1^2
= 3 * 0.9 * 0.01
= 0.027

Therefore, the probability of having exactly one class that gives homework is 0.027, or 2.7%.

Regarding your second question, the probability of having homework in at least 1 class can be found by adding the probabilities of having homework in each individual class and subtracting the probability of having no homework in any class.

2. Probability of having homework in at least 1 class (math, history, or Spanish):
P(at least 1 class with homework) = 1 - P(no class with homework)

To calculate P(no class with homework), you need to find the complement of the probability of having homework in each class:

P(no class with homework) = (1 - p(math)) * (1 - p(history)) * (1 - p(Spanish))

= (1 - 0.9) * (1 - 0.6) * (1 - 0.75)
= 0.1 * 0.4 * 0.25
= 0.01

Now, calculating P(at least 1 class with homework):
P(at least 1 class with homework) = 1 - 0.01
= 0.99

So, the probability of having homework in at least 1 class is 0.99, or 99%.