To solve this problem, we can model the situation using a binomial distribution. We're looking at the probability of fewer than 2 couples having both partners in the labor force out of 5 couples.
Parameters:
- n (number of trials) = 5 (the number of couples)
- p (probability of success) = 0.521 (the probability that both parties in the couple are in the labor force)
We are looking for \( P(X < 2) \), where \( X \) is the number of couples with both parties working. This is the same as:
\[ P(X < 2) = P(X = 0) + P(X = 1) \]
Step 1: Calculate \( P(X = k) \)
The probability mass function for a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
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Calculate \( P(X = 0) \): \[ P(X = 0) = \binom{5}{0} (0.521)^0 (0.479)^5 \]
The binomial coefficient \( \binom{5}{0} = 1 \), thus: \[ P(X = 0) = 1 \cdot 1 \cdot (0.479)^5 \approx (0.479)^5 \approx 0.04741 \]
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Calculate \( P(X = 1) \): \[ P(X = 1) = \binom{5}{1} (0.521)^1 (0.479)^4 \]
The binomial coefficient \( \binom{5}{1} = 5 \), thus: \[ P(X = 1) = 5 \cdot (0.521) \cdot (0.479)^4 \approx 5 \cdot 0.521 \cdot 0.05555 \approx 0.14445 \]
Step 2: Combine the probabilities
Now we can sum these two probabilities up: \[ P(X < 2) = P(X = 0) + P(X = 1) \approx 0.04741 + 0.14445 \approx 0.19186 \]
Final Step: Round the answer
Thus, the probability that fewer than 2 of the couples have both parties working is: \[ \boxed{0.192} \]