One day, eleven 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 nine of the eleven babies are girls?

a.
23/256

b.

191/256

c.
509/512

d.
397/2048

This problem is an example of binomial probability. In this case, we can use the formula:

P (r) = [n! / (n – r)! r!] p^r q^(n-r)

where,

n = total number of babies = 11

r = selected number of babies

p = success of being a girl = 0.5

q = 1 – p = 1 – 0.5 = 0.5

Since we are asked to find for the P (r less than or equal 9), therefore:

P (r less than or equal 9) = 1 – [P(11) + P(10)]

P(11) = [11! / (11 – 11)! 11!] 0.5^11 * 0.5^(11-11) = 4.8828 * 10^-4

P(10) = [11! / (11 – 10)! 10!] 0.5^10 * 0.5^(11-10) = 5.3711 * 10^-3

Therefore:

P (r less than or equal 9) = 1 – [4.8828 * 10^-4 + 5.3711 * 10^-3]

P (r less than or equal 9) = 0.994

The closest answer is letter C:

C = 509/512 = 0.994

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.