Question

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 1

Step 1: Using the theorem of binomial distribution

`^nC_r(p)^(n-r)q^r`

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

`\begin{gathered} \text{Probability that it is a girl p = }(1)/(2) \\ \text{Probability that it is not a girl q = }(1)/(2) \end{gathered}`

Step 4: Find the probability that at most eleven of the thirteen babies are girls.

`\begin{gathered} =^{}1-(^(13)C_(12)((1)/(2))^(13-12)((1)/(2))^(12)^{}+^(13)C_(13)((1)/(2))^(13-13)((1)/(2))^(13)) \\ ^nC_r\text{ = }(n!)/((n-r)!r!) \end{gathered}`

Step 5: Simplify the expression

`\begin{gathered} =\text{ 1 - }\lbrack13\text{ }*\text{ (}(1)/(2))^1((1)/(2))^(12)\text{ + 1 x (}(1)/(2))^0((1)/(2))^(13)\rbrack \\ =\text{ 1 - \lbrack 13 }*(1)/(2)\text{ }*\text{ }(1)/(4096)\text{ + 1 }*\text{ 1 }*\text{ }(1)/(8192)\rbrack \\ =\text{ 1 - }(13)/(8192)\text{ - }(1)/(8192) \\ =\text{ 1 - }(14)/(8192) \\ =\text{ }\frac{8192\text{ - 14}}{8192} \\ =\text{ }(8178)/(8192) \\ To\text{ the lowest term} \\ =\text{ }(4089)/(4098) \end{gathered}`

Option B is the correct answer