binomial distribution?

Question

The sex ratio of newborn human infants is about 105 males to 100 females. If four infants are chosen at a time, what is the probability that

a. two are male and two are female
b. all four are male
c all four are the same sex

Someone please help it would be greatly appreciated!

Answer 1

The last two are not difficult if you assume that the probabilities are independent (i.e., the probability of choosing a female on the second draw is unaffected by whether you drew a female in the first draw). In that case you convert the birth rates into probabilities 105/205 = 0.5122 for males and 0.4878 for females. The probabilty of choosing four successive individuals of the same sex is equal to the probability of a single draw of that sex raised to the power four (or to the power of whatever number of draws you make.) E.g., for females the probability of choosing four females in four draws is 100/205^4 = 0.0566.

The first question relies on the binomial expansion. To answer this assume that p = the probability of choosing a female = 0.4878. The number of 'trials' is the number of newborns selected ('n') and the number of successes (times two females is chosen) is 'k'. The probability of choosing two females is [n! / k! (n-k)!] (p^k) (1-p^n-k). In this case 0.3746. This is also the probabilty of choosing two males, since any grouping of four containing two females will also contain four males.

Answer 2

You will have to check these answers for yourself.
a. 66.90%
b. 6.88%
Chance of all four female - 5.66%
C. 6.88% + 5.66%