probability?

Question

it is known that 40% of the students in a certain college are girls and 50% of the students are above the median height.if 2/3 of the boys are above median height,what is the probability that a randomly selected student who is below median height is a girl?

Answer 1

Let
p(b) = probability student is a boy
p(g) = probability student is a girl
p(m+) = probability student is above median height
p(m-) = probability student is below median height

Given
p(g) = 4/10 = 2/5
p(m+) = 1/2
p(m+ | b) = 2/3

Find
p(g | m-)
_________

p(b) = 1 - p(g) = 1 - 2/5 = 3/5
p(b ∩ m+) = p(m+ | b) p(b) = (2/3)(3/5) = 2/5

p(g ∩ m+) = 1 - p(b ∩ m+) = 1 - 2/5 = 1/10
p(g ∩ m-) = p(g) - p(g ∩ m+) = 2/5 - 1/10 = 3/10

p(g | m-) = p(g ∩ m-) / p(m-) = (3/10) / (1/2) = 3/5

Answer 2

I find that drawing a Venn Diagram always helps, though it's a bit difficult in this question.
Nevertheless, assume there are 100 students in the school.

You know that 40 of the students are girls, meaning that 60 are boys.

You're also told that 2/3 of the boys are above median height. In other words, 2 x 60/3 = 40 of the boys are above median heigh.

Your final hint is that 50 of the students are above median height, and since 40 of the are boys, 10 of them must be girls.

So now, the number of girls below median height must be total # of girls minus girls above median height, or 40 - 10 = 30.

Finally, 30 out of the 50 students who are below median height are girls, so if I were to select a random student from these 50, I have a 3/5 x 100% = 60% chance of having selected a girl below median height.

Of course there is a chance that a student could be exactly median height, but since we're not given that piece of information, it's safe to assume none exist. If they did, we'd simply subtract that percentage from total number of girls, along with the girls of above median height, to find those below median.

Hope this helped.

Answer 3

let
P(b) = pr. of boy
P(g) = pr. of girl
P(>) = pr. of above median height
P(<) = pr. of below median height
assume none are hermaphrodite nor equal the median height.

given to you
P(g) = 40% = 2/5
P(>) = 50% = 1/2
P(> l b) = 2/3

P(g l <) is requested from the question.

P(b)
= 1 - P(g)
= 1 - 2/5
= 3/5

P(<)
= 1 - P(>)
= 1 - 1/2
= 1/2

we already know that
P(g l <) + P(b l <) = 1
and
P(> l b) + P(< l b) = 1
and also
P(< l b)*P(b) = P(b l <)*P(<)

so from those 3 equation,
P(> l b) + P(< l b) = 1
P(< l b) = 1 - 2/3 = 1/3

P(< l b)*P(b) = P(b l <)*P(<)
1/3 * 3/5 = P(b l <) * 1/2
P(b l <) = 2/1 * 1/3 * 3/5 = 2/5

P(g l <) + P(b l <) = 1
P(g l <) = 1 - 2/5 = 3/5

so the probability that a randomly selected student who is below median height is a girl, or in math notation
P(g l <) = 3/5 = 60%

so in that college there's
60% girls in "short" category
20% girls in "tall" category
66.7% "tall" in boys
33.3% "short" in boys

in general, the composition of students in the college
30% are short girls
10% are tall girls
40% are tall boys
20% are short boys.

Answer 4

60% of students are boys.
Therefore, boys above median height are 40% of the population. Thus, 10% of the population are girls above median height.

So, 30% of population are girls BELOW median height, and 20% of population is boys below median height.

So, the probability that a randomly selected person below median height would be a girl is 3/5, or 60%.

Answer 5

Probability of selecting a girl P(G) = 0.4
Probability of selecting a boy P(B) = 1-P(G) = 0.6
Probability of selecting a student above medium height P(Samh) =0.5
2/3 of boys above medium height (Bamh) implies that 40% of the 50% of all Samh are boys.
therefore 10% of all students are Gamh.
implying that 60%- 40% of all students are boys below medium height (Bbmh) = 20% and
40% - 10% of all students are Gbmh = 30%.
Hence:
P(selecting a girl below medium height)= P( G and Gbmh)
= P(G) x P(Gbmh) = 0.4 x 0.3 = 0.12.