Question 1205980: A school contains 357 boys and 323 girls
If a student is chosen at random, what is the probability of choosing a boy and a girl
Found 4 solutions by josgarithmetic, math_tutor2020, ikleyn, Edwin McCravy:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website! Something is wrong in the question. By the way written, "a student " should mean exactly 1 student. This one student to be both a boy and a girl at the same time seems uncommon.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
P(boy) = probability of a boy
P(girl) = probability of a girl
P(boy) = (number of boys)/(number of students)
P(girl) = (number of girls)/(number of students)
Example:
A very small school has 10 boys and 12 girls
P(boy) = 10/(10+12) = 10/22 = 5/11
P(girl) = 12/(10+12) = 12/22 = 6/11
These results aren't the final answers to your particular problem; however, you can use this example to answer your question.
Answer by ikleyn(52814)  (Show Source):
You can put this solution on YOUR website! .
In mathematical problems, written by professionals, they do not use such form as it is used in this post.
Professional writers take a special care in order for wording be smooth and a reader would not stumble over unsuccessful phrases.
Otherwise, it is seen from the distance of thousand miles, that this composition
is created by an amateur on his knees in a garage, and it's not funny - it's sad, instead.
Math problems should TEACH and INSPIRE - - - it is their major purpose - not to sow disappointment.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
If the problem had been this way:
A school contains 357 boys and 323 girls.
If 2 students are chosen at random, what is the probability of choosing
a boy and a girl.
The numerator of the probability would be the number of successful ways:
Choose a boy 357 ways, then choose a girl 357 ways. 323x357 = 115311
successful ways.
The denominator of the probability would be the number of ways any two
students could be chosen. That's C(357+323,2) = C(680,2) = 230869
So the probability would be 115311/230860 = 969/1940 = 0.4994845361, very
close to 0.5.
Edwin
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