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Sagot :
Step 1: Using the theorem of binomial distribution
[tex]^nC_r(p)^{n-r}q^r[/tex]n = 13
p = probability that the baby is a girl
q = probability that the baby is not a girl
Step 2: To find the probability that most eleven of the thirteen babies are girls
= 1 - [(probability that exactly twelves girls) + (probability that exactly thirteen girls)]
Step 3: Get the value of p and q
[tex]\begin{gathered} \text{Probability that it is a girl p = }\frac{1}{2} \\ \text{Probability that it is not a girl q = }\frac{1}{2} \end{gathered}[/tex]Step 4: Find the probability that at most eleven of the thirteen babies are girls.
[tex]\begin{gathered} =^{}1-(^{13}C_{12}(\frac{1}{2})^{13-12}(\frac{1}{2})^{12}^{}+^{13}C_{13}(\frac{1}{2})^{13-13}(\frac{1}{2})^{13}) \\ ^nC_r\text{ = }\frac{n!}{(n-r)!r!} \end{gathered}[/tex]Step 5: Simplify the expression
[tex]\begin{gathered} =\text{ 1 - }\lbrack13\text{ }\times\text{ (}\frac{1}{2})^1(\frac{1}{2})^{12}\text{ + 1 x (}\frac{1}{2})^0(\frac{1}{2})^{13}\rbrack \\ =\text{ 1 - \lbrack 13 }\times\frac{1}{2}\text{ }\times\text{ }\frac{1}{4096}\text{ + 1 }\times\text{ 1 }\times\text{ }\frac{1}{8192}\rbrack \\ =\text{ 1 - }\frac{13}{8192}\text{ - }\frac{1}{8192} \\ =\text{ 1 - }\frac{14}{8192} \\ =\text{ }\frac{8192\text{ - 14}}{8192} \\ =\text{ }\frac{8178}{8192} \\ To\text{ the lowest term} \\ =\text{ }\frac{4089}{4098} \end{gathered}[/tex]Option B is the correct answer
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