A Martian couple has children until they have 2 males (sexes of children are independent). Compute the expected number of children the couple will have if, on Mars, males are: (a) Half as likely as females. (b) Just as likely as females. (c) Twice as likely as females.

Answer:a) 6b) 4c) 3Step-by-step explanation:Let p be the probability of having a female Martian, and of course, 1-p the probability of having a male Martian. To compute the expected total number of trials before 2 males are born, imagine an experiment simulating the fact that 2 males are born is performed n times. Let ak be the number of trials performed until 2 males are born in experiment k. That is, a1= number of trials performed until 2 males are born in experiment 1 a2= number of trials performed until 2 males are born in experiment 2 and so on. If a1 + a2 + … + an = N we would expect Np females.  Since the experiment was performed n times, there 2n males (recall that the experiment stops when 2 males are born). So we would expect 2n = N(1-p), or N/n = 2/(1-p) But N/n is the average number of trials per experiment, that is, the expectation. We have then that the expected number of trials before 2 males are born is 2/(1-p) where p is the probability of having a female. a) Here we have the probability of having a male is half as likely as females. So 1-p = p/2 hence p=2/3 The expected number of trials would be 2/(1-2/3) = 2/(1/3) =6 This means the couple would have 6 children: 4 females (the first 4 trials) and 2 males (the last 2 trials). b) Here the probability of having a female = probability of having a male = 1/2 The expected number of trials would be 2/(1/2) = 4 This means the couple would have 4 children: 2 females (the first 2 trials) and 2 males (the last 2 trials). c) Here, 1-p = 2p so p=1/3 The expected number of trials would be 2/(1-1/3) = 2/(2/3) = 6/2 =3 This means the couple would have 3 children: 1 female (the first trial) and 2 males (the last 2 trials).