Question

question 4
the smith 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.5, find the probability that the smith family had:
at least 5 girls?
at most 3 girls?
question help: message instructor

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single - trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Here, \(n = 6\), \(p=0.5\), \(1 - p = 0.5\)

Step2: Calculate probability of at least 5 girls

\(P(X\geq5)=P(X = 5)+P(X = 6)\)

For \(k = 5\):
\(C(6,5)=\frac{6!}{5!(6 - 5)!}=\frac{6!}{5!1!}=6\)
\(P(X = 5)=C(6,5)\times(0.5)^{5}\times(0.5)^{6 - 5}=6\times(0.5)^{6}\)

For \(k = 6\):
\(C(6,6)=\frac{6!}{6!(6 - 6)!}=1\)
\(P(X = 6)=C(6,6)\times(0.5)^{6}\times(0.5)^{6 - 6}=(0.5)^{6}\)

\(P(X\geq5)=6\times(0.5)^{6}+(0.5)^{6}=(6 + 1)\times(0.5)^{6}=7\times\frac{1}{64}=\frac{7}{64}\)

Step3: Calculate probability of at most 3 girls

\(P(X\leq3)=P(X=0)+P(X = 1)+P(X = 2)+P(X = 3)\)

For \(k = 0\):
\(C(6,0)=\frac{6!}{0!(6-0)!}=1\)
\(P(X = 0)=C(6,0)\times(0.5)^{0}\times(0.5)^{6}=(0.5)^{6}\)

For \(k = 1\):
\(C(6,1)=\frac{6!}{1!(6 - 1)!}=6\)
\(P(X = 1)=C(6,1)\times(0.5)^{1}\times(0.5)^{5}=6\times(0.5)^{6}\)

For \(k = 2\):
\(C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15\)
\(P(X = 2)=C(6,2)\times(0.5)^{2}\times(0.5)^{4}=15\times(0.5)^{6}\)

For \(k = 3\):
\(C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4}{3\times2\times1}=20\)
\(P(X = 3)=C(6,3)\times(0.5)^{3}\times(0.5)^{3}=20\times(0.5)^{6}\)

\(P(X\leq3)=(1 + 6+15 + 20)\times(0.5)^{6}=42\times\frac{1}{64}=\frac{21}{32}\)

Answer:

The probability of at least 5 girls is \(\frac{7}{64}\)
The probability of at most 3 girls is \(\frac{21}{32}\)