Qurrent learning objective: Binomial Distribution Question 8 Score: 0 of 4 points The Jones family was one of the first to come to the U.S. They had 6 children. Assuming that the probability of a child being a girl is 0.5 , find the probability that the Jones family had: The probability that the Jones family had at least 5 girls is Enter your answer b. at most 3 girls? The probability that the Jones family had at most 3 girls is Enter your answer

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UpStudy AI · Accepted Answer

1. Let $$ X $$ be the number of girls. Since the probability of a child being a girl is $$ p = 0.5 $$ and there are 6 children, $$ X $$ follows a binomial distribution with parameters $$ n = 6 $$ and $$ p = 0.5 $$.

2. The probability mass function is given by:
   $$
   P(X = k) = \binom{6}{k} (0.5)^k (0.5)^{6-k} = \binom{6}{k} (0.5)^6
   $$

3. **(a) At least 5 girls:**  
   We need to find:
   $$
   P(X \geq 5) = P(X = 5) + P(X = 6)
   $$
   Compute each term:
   $$
   P(X = 5) = \binom{6}{5} (0.5)^6 = 6 \times (0.5)^6
   $$
   $$
   P(X = 6) = \binom{6}{6} (0.5)^6 = 1 \times (0.5)^6
   $$
   Therefore:
   $$
   P(X \geq 5) = 6 \times (0.5)^6 + 1 \times (0.5)^6 = 7 \times (0.5)^6 = \frac{7}{64}
   $$

4. **(b) At most 3 girls:**  
   We need to find:
   $$
   P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
   $$
   Compute each term:
   $$
   P(X = 0) = \binom{6}{0} (0.5)^6 = 1 \times (0.5)^6
   $$
   $$
   P(X = 1) = \binom{6}{1} (0.5)^6 = 6 \times (0.5)^6
   $$
   $$
   P(X = 2) = \binom{6}{2} (0.5)^6 = 15 \times (0.5)^6
   $$
   $$
   P(X = 3) = \binom{6}{3} (0.5)^6 = 20 \times (0.5)^6
   $$
   Summing up:
   $$
   P(X \leq 3) = (1 + 6 + 15 + 20) \times (0.5)^6 = 42 \times (0.5)^6 = \frac{42}{64} = \frac{21}{32}
   $$

5. **Final Answers:**  
   - The probability that the Jones family had at least 5 girls is $$ \frac{7}{64} $$.  
   - The probability that the Jones family had at most 3 girls is $$ \frac{21}{32} $$.
- The probability of having at least 5 girls is $$ \frac{7}{64} $$. - The probability of having at most 3 girls is $$ \frac{21}{32} $$.

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