Twenty years ago Juan Medina and Bertha Zenith got married. The Medina-Zenith fa

Question

Twenty years ago Juan Medina and Bertha Zenith got married. The Medina-Zenith family's biggest life aspiration was to have a baby girl. Luck was not on their side. They had eight boys in a row. They were determined to follow their aspiration and the ninth child was a boy. According to the family's doctor, it is very unlikely for a family to haw eight ben's in a row followed by a girl. However, according to Professor X, if a couple has nine children, it is guarantee that at least one of the children will be a girl. Who is correct? Is having a girl the same event as having at least one girl? Explain. Is it possible to calculate these precise probabilities? If yes, how? If no, . why? Can Professor X guarantee that a family with nine children will have at least one girl? The Johnson family just got married and they are planning to have four children. Are they more or less likely than the Medina-Zenith to have one girl? Explain.

Explanation / Answer

Part (a)

NO. These 2 events are different. Event A, say, One girl child out of 9 children => just 1 out of 9 is a girl.

But, event B, say, At least one girl would imply 1 or more girls => it could be 1or 2 or ,,,,,,, or 8 or 9.

In terms of probability, if probability of a girl is p, P(A) = 9p(1- p)8 and P(B) = 1- (1- p)9.

If p = ½ which what is normally assumed, P(A) = 0.1758 whereas P(B) = 0.998.   

Part (b)

It is possible to calculate these probabilities using Binomial Distribution. If X = number of girls out of n children, then X ~ B(n, p), where p = probability of a girl. This probability is:

p(x) = P(X = x) = P(x girls out of n children) = nCxpx(1 - p)(n - x).

If p = ½ which what is normally assumed, p(x) = nCx(1/2)n.

Part (c)

As shown in Part (a), P(B) = 0.998 which is as good as 1 which in turn implies that is sure (rather almost sure) that at least one out of 9 children would be a girl. So, Professor’s claim is valid.

Part (d)

Probability of at least one girl out of 4 children

= P(B) = 1- (1- p)4[refer explanation in Part (a)]

= 1 – ½)4 = 0.9375 which is less than 0.998 [refer Part (c)]

So, Johnsons family is less likely. ANSWER

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