Ourrert learning objective: Binomial Distribution

Questions

Some:
Qof 4 points

The dones family was one of the first to come to the US. They had 6 chilidren. Assurting that the probsblity of a child heing a gin is Quy, find the protitlity that the dotes farnily that
a. at least 5 ginls?

The probability that the sones family had at least 5 ginls is
Exteryour answer
(Repont answer acourate to four cecimal ylaces.)
b. at most 3 ginls?

ORequinements
Emer an imeper or cecimal number
The probability that the sones familiy had at most 3 girls is
Enteryour answer - (ineport answer accurate to four decimal places.)

Question

Ourrert learning objective: Binomial Distribution

Questions

Some:
Qof 4 points

The dones family was one of the first to come to the US. They had 6 chilidren. Assurting that the probsblity of a child heing a gin is Quy, find the protitlity that the dotes farnily that
a. at least 5 ginls?

The probability that the sones family had at least 5 ginls is
Exteryour answer
(Repont answer acourate to four cecimal ylaces.)
b. at most 3 ginls?

ORequinements
Emer an imeper or cecimal number
The probability that the sones familiy had at most 3 girls is
Enteryour answer

- (ineport answer accurate to four decimal places.)

UpStudy AI · Accepted Answer

Let $$ X $$ be the number of girls among the 6 children. Assuming that each child is equally likely to be a girl with probability $$ 0.5 $$, $$ X $$ follows a binomial distribution:
$$
X \sim \text{Binomial}(n=6,\, p=0.5)
$$

### a. At least 5 girls

We need to determine
$$
P(X \ge 5) = P(X=5) + P(X=6).
$$

The probability mass function for a binomial random variable is:
$$
P(X=k) = \binom{6}{k} (0.5)^k (0.5)^{6-k} = \binom{6}{k} (0.5)^6.
$$

Calculate each term:

- For $$ k = 5 $$:
  $$
  P(X=5) = \binom{6}{5} (0.5)^6 = 6 \cdot \frac{1}{64} = \frac{6}{64}.
  $$

- For $$ k = 6 $$:
  $$
  P(X=6) = \binom{6}{6} (0.5)^6 = 1 \cdot \frac{1}{64} = \frac{1}{64}.
  $$

Then,
$$
P(X \ge 5) = \frac{6}{64} + \frac{1}{64} = \frac{7}{64} \approx 0.1094.
$$

### b. At most 3 girls

We need to determine
$$
P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3).
$$

Calculate each term:

- For $$ k = 0 $$:
  $$
  P(X=0) = \binom{6}{0} (0.5)^6 = 1 \cdot \frac{1}{64} = \frac{1}{64}.
  $$

- For $$ k = 1 $$:
  $$
  P(X=1) = \binom{6}{1} (0.5)^6 = 6 \cdot \frac{1}{64} = \frac{6}{64}.
  $$

- For $$ k = 2 $$:
  $$
  P(X=2) = \binom{6}{2} (0.5)^6 = 15 \cdot \frac{1}{64} = \frac{15}{64}.
  $$

- For $$ k = 3 $$:
  $$
  P(X=3) = \binom{6}{3} (0.5)^6 = 20 \cdot \frac{1}{64} = \frac{20}{64}.
  $$

Then, sum the probabilities:
$$
P(X \le 3) = \frac{1+6+15+20}{64} = \frac{42}{64} \approx 0.6563.
$$

### Final Answers

a. The probability that the family has at least 5 girls is $$ 0.1094 $$.

b. The probability that the family has at most 3 girls is $$ 0.6563 $$.
a. The probability of having at least 5 girls is 0.1094. b. The probability of having at most 3 girls is 0.6563.

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