Question 8 Fer Scrore: O of 4 points The Jones family was one of the fist to come to the US. They had 6 childran. Assuming that the probability of a child being a ginl is Q5, find the protability that the dones family fact a. at least 5 girls? The probability that the Jones farmily had at lieast 5 girls is Enter your answer - Peport answer accurate to four decimal places) th. at most 3 girls? The probability that the Jones family had at most 3 girls is Enter your answer (5) - (Feport answer accurate to four decimal places))

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UpStudy AI · Accepted Answer

**Step 1. Define the problem parameters**

The Jones family has $$ n = 6 $$ children. The probability that any given child is a girl is $$ p = 0.5 $$. The number of girls in the family follows a binomial distribution

$$
X \sim \text{Binomial}(n=6, p=0.5).
$$

The probability mass function is given by

$$
P(X = k) = \binom{6}{k}(0.5)^k (0.5)^{6-k} = \binom{6}{k}(0.5)^6.
$$

---

**Step 2. (a) Probability of at least 5 girls**

"At least 5 girls" means $$ X \ge 5 $$. So we need to calculate:

$$
P(X \ge 5) = P(X = 5) + P(X = 6).
$$

Using the binomial probability:

- For $$ k = 5 $$:

$$
P(X = 5) = \binom{6}{5}(0.5)^6 = 6 \times (0.5)^6.
$$

- For $$ k = 6 $$:

$$
P(X = 6) = \binom{6}{6}(0.5)^6 = 1 \times (0.5)^6.
$$

Adding these up:

$$
P(X \ge 5) = 6(0.5)^6 + 1(0.5)^6 = (6+1)(0.5)^6 = 7(0.5)^6.
$$

Since $$(0.5)^6 = \frac{1}{64}$$, we obtain:

$$
P(X \ge 5) = \frac{7}{64} \approx 0.109375.
$$

Rounding to four decimal places, the answer is

$$
0.1094.
$$

---

**Step 3. (b) Probability of at most 3 girls**

"At most 3 girls" means $$ X \le 3 $$. So we need:

$$
P(X \le 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3).
$$

Calculate each term:

- For $$ k = 0 $$:

$$
P(X = 0) = \binom{6}{0}(0.5)^6 = 1 \times (0.5)^6.
$$

- For $$ k = 1 $$:

$$
P(X = 1) = \binom{6}{1}(0.5)^6 = 6 \times (0.5)^6.
$$

- For $$ k = 2 $$:

$$
P(X = 2) = \binom{6}{2}(0.5)^6 = 15 \times (0.5)^6.
$$

- For $$ k = 3 $$:

$$
P(X = 3) = \binom{6}{3}(0.5)^6 = 20 \times (0.5)^6.
$$

Now sum these probabilities:

$$
P(X \le 3) = \big(1 + 6 + 15 + 20\big)(0.5)^6 = 42 \times \frac{1}{64}.
$$

Thus,

$$
P(X \le 3) = \frac{42}{64} = \frac{21}{32} \approx 0.65625.
$$

Rounded to four decimal places, the answer is

$$
0.6563.
$$

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**Final Answers**

- The probability that the Jones family had at least 5 girls is $$0.1094$$.

- The probability that the Jones family had at most 3 girls is $$0.6563$$.
The probability that the Jones family had at least 5 girls is 0.1094, and the probability that they had at most 3 girls is 0.6563.

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