Respuesta :
We need to use Binomial Probability.
Of 6 births, we want to find the probability of at least 2 of them being girls.
To solve this, we need to find:
Probability of exactly 2 girls
Probability of exactly 3 girls
Probability of exactly 4 girls
Probability of exactly 5 girls
Probability of exactly 6 girls
If we add all these probabilities, we get the probability of at least 2 girls.
To find the probabilities, we can use the formula:
[tex]_nC_r\cdot p^r(1-p)^{n-r}[/tex]Where:
n is the number of trials (in this case, the number of total births)
r is the number of girls we want to find the probability
p is the probability of the event occurring
[tex]_nC_r\text{ }is\text{ }the\text{ }combinatoric\text{ }"n\text{ }choose\text{ }r"[/tex]The formula for "n choose r" is:
[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex]Then, let's find the probability of exactly 2 girls:
The probability of the event occurring is:
[tex]P(girl)=\frac{1}{2}[/tex]Because there is a 50% probability of being a girl or a boy.
let's find "6 choose 2":
[tex]_6C_2=\frac{6!}{2!(6-2)!}=\frac{720}{2\cdot24}=15[/tex]Now we can find the probability of exactly 2 girls:
[tex]Exactly\text{ }2\text{ }girls=15\cdot(\frac{1}{2})^2(1-\frac{1}{2})^{6-2}=15\cdot\frac{1}{4}\cdot(\frac{1}{2})^4=\frac{15}{4}\cdot\frac{1}{16}=\frac{15}{64}[/tex]We need to repeat these calculations for exactly 3, 4, 5, and 6 girls:
Exactly 3 girls:
let's find "6 choose 3":
[tex]_6C_3=\frac{6!}{3!(6-3)!}=\frac{720}{6\cdot6}=20[/tex]Thus:
[tex]Exactly\text{ }3\text{ }girls=20\cdot(\frac{1}{2})^3(1-\frac{1}{2})^{6-3}=20\cdot\frac{1}{8}\cdot\frac{1}{8}=\frac{5}{16}[/tex]Exactly 4 girls:
"6 choose 4":
[tex]_6C_4=\frac{6!}{4!(6-4)!}=\frac{720}{24\cdot2}=15[/tex]Thus:
[tex]Exactly\text{ }4\text{ }girls=15\cdot(\frac{1}{2})^4(1-\frac{1}{2})^{6-4}=15\cdot\frac{1}{16}\cdot\frac{1}{4}=\frac{15}{64}[/tex]Exactly 5 girls:
"6 choose 5"
[tex]_6C_5=\frac{6!}{5!(6-5)!}=\frac{720}{120}=6[/tex]Thus:
[tex]Exactly\text{ }5\text{ }girls=6\cdot(\frac{1}{2})^5(1-\frac{1}{2})^{6-5}=6\cdot\frac{1}{32}\cdot\frac{1}{2}=\frac{3}{32}[/tex]Exactly 6 girls:
"6 choose 6"
[tex]_6C_6=\frac{6!}{6!(6-6)!}=\frac{720}{720\cdot0!}=\frac{720}{720}=1[/tex]Thus:
[tex]Exactly\text{ }6\text{ }girls=1\cdot(\frac{1}{2})^6(1-\frac{1}{2})^{6-6}=\frac{1}{64}\cdot(\frac{1}{2})^0=\frac{1}{64}[/tex]now, to find the answer we need to add these 5 values:
[tex]\frac{15}{64}+\frac{5}{16}+\frac{15}{64}+\frac{3}{32}+\frac{1}{64}=\frac{57}{64}=0.890625[/tex]To the nearest tenth, the probability of at least 3 girls is 0.891, thus, the last option is the correct one.
