Conditional probability (1 Viewer)

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1. The problem statement, all variables and given/known data
In a neighbourhood, 75% of households speak only English at home, and 5% speak only Spanish at home.

Four households are randomly chosen. Given that at least one of the four households speaks only Spanish at home, what is the probability that only one speaks Spanish at home?


2. Relevant equations
P(B|A) = P(B&A)/P(A)
A = event that at least one of the four households speak only Spanish
B = event that exactly one speaks only Spanish

3. The attempt at a solution
P(A) = 1 - (75/100)(74/99)(73/98)(72/97) (1 - prob. that none speak Spanish)
P(B&A) = ?

Does P(B&A) = P(B)? Which would mean P(B) = 4(75/100)(74/99)(73/98)(5/97)

Attempted solution: P(B|A) = 0.125

Thanks in advance! :)
 
The question is not completely clear, I think. It accounts for 80% of houses, but what about the other 20? Are we to assume that they speak both Spanish and English or do they use another language? I will assume that the 20% speak both Spanish and English. I will also assume that the number of households in the pool is very large so that as each of the 4 houses are taken out of the pool, the probabilities for the next random one chosen stay the same.

With these assumptions, the problem is very simple, I think. At least one of the 4 is Spanish only, so I think the problem is equivalent to this: 3 houses are randomly picked, what is the probability that 0 speak Spanish?

25% of houses speak (at least some) Spanish, and 75% don't. What is the probability that all 3 are from the 75% that don't?
 

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