Question

A young couple falls in love and decides to marry and start a family. Assume that with each birth, they are equally like to have a boy or girl child - and only one at a time (no twins or other multiple births). They decide to keep having children until they have one girl. 5. a. On average, how many children will be born to this family? b. If there are no girls in the first two births, what is the probability that at least four children total are born to this family? On average, how many boys will be born to this family? On average, how many girls will be born to this family? c. d.

Answer 1

Let the random variable X denote the children born before first girl. That is, X follows geometric distribution. Let p denote the probability of birth is girl and a denote the probability of birth is boy. Assume that each birth is equally likely to have a boy or girl. That is, p = (a) If the first-born baby is girl with probability p. That is, 1x p =p If the first-born baby is boy and second born baby is girl with probability qp. That is, 2xqp = 2ạp If the first born is boy, second born is boy and the third born is girl with probability qqp. That is, 3x qqp = 399p And so, on up to they keep having children until they have one girl which is represented in the below table Number of children probability qр qap=qp gagp=qp gagqp=44p + . . The series is, 1.p+2.qp +3.02p+49p+... (j)+{}))+(*)++f

Tet s=+2(E)+= {])++) +.. (1) 25=1+2(1)+3(4) +-()+ (I) Subtract (I) from (II) 23-s={}+2(3)+=(+ (16) +..) -{1+2{4}+{})++ ().) (The series is in geometric progression. The formula for sum of in finite series is, here a is the first term a=1/2 r = 1/2 s=1+1 s=1.5 The average number of children required to born is 1.5

(b) The probability of having three children (no girls in the first two births) is. P(X=3) = sqp=(1) 09/09) The probability that at least four children is, P(X>4) =1-P(XS3)=1= =0.875 The number of boys is,

Number of boys probability am in . The series is, qp+2q2p+3q2p+4q* p.. (194)+{}\(\)-{)*b++*)+ Let 45=(4+2( )+()+*()+) 4s=1+2(4)+3[)+447+ (10)

Subtract (II) from (IV) 35=[(-4) +(4-)-> (4_1_)+*(76 )+..) (The series is in geometric progression. The formula for sum of in finite series is, here a is the first term a=1/4 r= 1/4 3s = 5 36 s=0.3611 Therefore, the average number of boys born to a family is 0.3611. (d) The average number of girls is, 1(p) +2(0) they stop to give birth if they have one girl) Therefore, the average number of girls born to a family is 0.5.