One day, thirteen babies are born at a hospital. Assuming each baby has an equal chance of being a boy or girl, what is the probability that at most eleven of the thirteen babies are girls?A. 11/64B. 4089/4096C. 29/128D. 743/1024

Answer :

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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