Question

When the probabilty of only a certain combination rials may be employed, as indicated in the following formula: in a given-sized group ts retquilieu, I (n-x) where P is the probability to be calculated, ǐs the product of the number of integers in the group of items [n!-n x (n-1) x (n-2) × ( n-3) ×·.. × 1; that is, if a group consists of five items (such as the five babies in Section 111-1), m! = 5! 5 × 4 × 3 × 2 × 1-120], xt is the product of the integers for one class (p) and (n -x) is the product of the integers in the other class (a), and p is the probability for one occurrence (i.e., .boys) and g is the probability for the other (i.e., girls). Note: Factorial 0 (0!) 1 and any number raised to the 0 power- 1. 2. If six babies are born in a given hospital on the same day, what is the probability that two will be boys and four will be girls? Substitute in the following formula: . al 241 Now, using the same procedure, calculate the probabilities that, among six children bom in a given hospital on a particular day a, one will be a boy and five will be girls. -P リ2) iAs, tome, she b. three will be boys and three will be girls. 813 c. all six will be girls, wr-1%)IAP-V129 291 30 3!3 3.2l し!
Checking to see if I did A & B correctly? I am not sure how to solve C.

Please show all work, thank you.

Answer 1

yes, first two parts are correct :)

c)

since, all six are girls,so, x will be zero,

P(all six will be girls) = 6!/(0!*6!) * (1/2)⁰ * (1/2)⁶ = 1*(1/2)^6 =1/64= 0.015625

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