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Answer:
Assumption 1: All pupils are male and female
With this assumption, we can find the ratio of boys and girls in the class. We can say p= g+b , where p is the number of pupils, g is the number of girls and b is the number of boys. Since there are 40% more girls, we can express this as g=1.4b. Thus, p=1.4b+b=2.4b . Dividing both sides by 2.4, we find that 12.4 or 41.33 of the class are boys, and correspondingly 58.66 of the class are girls.
Assumption 2: Each delegation is done independently
This is a fairly logical assumption, as this is a standard for how “random selections” work. In other words, we are assuming that the first selection does not affect the second and vice versa. Like when drawing names out of a hat, if I pull out Billy’s name I don’t then allow Billy to choose his best friend or something like that. I pull Billy’s name out of the hat, and then I pull another name out of the hat. Having established that each delegation is independent, we can assert that P(A∩B)=P(A)∗P(B) (read as the probability of A and B), which is a basic probability theorem. So the probability of getting a boy and a girl is the probability of choosing a boy times the probability of choosing a girl. You might think we should set this to 0.5 since that was the value given in the problem, but we don’t need to! We already have the ratio of boys to the entire class, which is the same as the probability of choosing a boy from the class, so P(B)=.4133 . The same concept applies to the girls, so P(G)=.5866 . This means that the probability of choosing both a boy and a girl regardless of size of the class is .4133∗.5866=.24244 or about 24.2%.
We can verify that the probability of choosing a boy and a girl cannot be .5 if there are 40% more girls than boys. Here’s how:
Starting from the end of Assumption 1 above (we still are using Assumption 2), we can say that P(B∩G)=P(B)∗P(G) , where
P(B)=number of boys number of students , and
P(G)=number of girls number of students .
Since we know g=1.4b and p=b+g=2.4b , we can substitute these into our equation to leave us with:
P(B∩G)=b2.4b∗1.4b2.4b=1.4b2(2.4b)2
Cancelling out the b2 s leaves us with P(B∩G)=1.42.42=.243 or about 24.3%. Clearly not 50%.