probability pool question

[davidh]

There are 30 students in a room. Ten of them are boys and the rest are girls. (20 girls)
a) What is the probability that a randomly made group of 12 students will have 5 boys?

assuming that we have 5 boys that means 7 girls so (working on this i was wrong)

b) What is the probability that a randomly made group of 12 students will have at least 1 boy?

so for B i think im doing it properly by doing
10c1X20c11 + 10c2X20c10 + 10c3X20c9 + 10c4X20c8 + 10c5X20c7 +10c6X20c6 +10c7X20c5 +10c8X20c4 +10c9X20c3 +10c10X20c2 +10c11X20c1 +10c12X20c0
i think this would be correct as at least means you can have 1 2 3 4 5 6 7 8 9 10 11 and 12 boys

c) If you make a group of 12 students, what is the expected number of girls?

this question confuses me as im not sure how to start looking at it as there is no value for boys unless im just missing something

 

[JeffM]

WHOA.

First of all, probabilities never exceed 1. So your answers for a and b are way off. What is the reason for doing the combinatorics? The number you calculated in part a does tell you how many ways you can pick exactly 5 boys and exactly 7 girls, but what must we do with that number to find the probability?

Secondly, you are doing b the very hard way. If I told you that the probability of no boys is p, what of at least 1 boy.

Third, what is the meaning of expected value?

 

[davidh]

im still working on a and b but to answer your question isnt expected value mean?

 

[davidh]

WHOA.

First of all, probabilities never exceed 1. So your answers for a and b are way off. What is the reason for doing the combinatorics? The number you calculated in part a does tell you how many ways you can pick exactly 5 boys and exactly 7 girls, but what must we do with that number to find the probability?

Secondly, you are doing b the very hard way. If I told you that the probability of no boys is p, what of at least 1 boy.

Third, what is the meaning of expected value?

and for probability you need to find the total outcomes and possible outcomes
unless i need to use the binomial formula for these questions?

 

[Romsek]

Hypergeometric multinomial distribution
2 groups, boys and girls, 10 boys, 20 girls

a)
5 boys out of 12 means 5 boys, 7 girls

[MATH]P[\text{5 boys out of 12}] = \dfrac{\dbinom{10}{5}\dbinom{20}{7}}{\dbinom{30}{12}}[/MATH]
b)
[MATH]P[\text{at least 1 boy in random 12}] = 1 - P[\text{0 boys in random 12}] = 1 - \dfrac{\dbinom{10}{0}\dbinom{20}{12}}{\dbinom{30}{12}}[/MATH]

 

[davidh]

Hypergeometric multinomial distribution
2 groups, boys and girls, 10 boys, 20 girls

a)
5 boys out of 12 means 5 boys, 7 girls

[MATH]P[\text{5 boys out of 12}] = \dfrac{\dbinom{10}{5}\dbinom{20}{7}}{\dbinom{30}{12}}[/MATH]
b)
[MATH]P[\text{at least 1 boy in random 12}] = 1 - P[\text{0 boys in random 12}] = 1 - \dfrac{\dbinom{10}{0}\dbinom{20}{12}}{\dbinom{30}{12}}[/MATH]

Thank you it makes so much sense now im still uncertain how to do 2c, any tips on how to approach it?

 

[JeffM]

im still working on a and b but to answer your question isn't expected value mean?

Yes, the expected value is a weighted mean, weighted by the probabilities.

So if we are talking about the number of girls selected from a total of y people, z of whom are girls, and u(x) is the probability that exactly x girls are selected, the expected number of girls is.

[MATH]E(x) = \sum_{x=0}^y u(x) * x.[/MATH]
If there are three girls and seven people, and we choose 2 people

[MATH]u(0) = \dbinom{3}{0} * \dbinom{4}{2} \div \dbinom{7}{2} = \dfrac{3!}{0! * 3!} * \dfrac{4!}{2! * 2!} * \dfrac{2! * 5!}{7!} \approx 28.6\%.[/MATH]
[MATH]u(1) = \dbinom{3}{1} * \dbinom{4}{1} \div \dbinom{7}{2} = \dfrac{3!}{1! * 2!} * \dfrac{4!}{1! * 3!} * \dfrac{2! * 5!}{7!} \approx 57.1\%.[/MATH]
[MATH]u(2) = \dbinom{3}{2} * \dbinom{4}{0} \div \dbinom{7}{2} = \dfrac{3!}{1! * 2!} * \dfrac{4!}{1! * 3!} * \dfrac{2! * 5!}{7!} \approx 14.3\%.[/MATH]
Let's check our work.

[MATH]28.6\% + 57.1\% + 14.3\% = 100.0%. \checkmark[/MATH]
The probability of picking one girl is the sum of the probabilities of picking a girl on the first pick and a boy on the second plus the probability of picking a boy on the first pick and a girl on the second, which is

[MATH]\dfrac{3}{7} * \dfrac{4}{6} + \dfrac{4}{7} * \dfrac{3}{6} = \dfrac{24}{42} \approx 57.1\%. \checkmark[/MATH]
So the expected number of girls is

[MATH]0 * 28.6 + 1 * 0.571 + 2 * 0.143 = 0.857[/MATH]
Does that make sense? Sure it does, if we had an equal number of boys and girls and picked two people at random, it is intuitive that, on average, we would select one girl. Here, where there are more boys than girls, it is intuitive that on average we shall get fewer than one girl.

 

[HallsofIvy]

Imagine choosing 12 students in order. You have 10 boys and 20 girls so the probability the first child is a boy is 10/30. If that is true we have left 29 students, 9 of whom are boys. The probability the second child is a boy is 9/29. The probability the third student is a boy is 8/28, the probability the fourth student is a boy is 7/27, and the probability the fifth student is a boy is 6/26. After that, there are 25 students left, five boys and 20 girls. The probability the sixth student chosen is a girl is 20/25, the probability the seventh student chosen is a girl is 19/24, the probability the eighth chosen is a girl is 18/23, ninth. 17/22, tenth, 16/21, eleventh, 15/20, and twelfth, 14/19. Several of those fractions can be reduced but I left them that way so that we can say that the probability of "five boys and 7 girls, in that order" is (10/30)(9/29)(8/28)(7/27)(6/26)(20/25)(19/24)(18/23)(17/22)(16/21)(15/20)(14/19). The denominator, 30(29)(28)...(19)= (30)(29)(28)...(19)(18)(17)...(2)(1)/(18)(17)...(2)(1) can be written as 30!/18!. The first part of the numerator, (10)(9)(8)(7)(6)= (10)(9)(8)...(2)(1)/(5)(4)...(2)(1)= 10!/5! and the other part is (20)(19)(18)...(15)(14)= (20)(19)(18)... (3)(2)(1)/(13)(12)...(2)(1)= 20!/13!. So the probability of "five boys and 7 girls, in that order" is $\displaystyle \frac{\frac{10!}{5!}\frac{20!}{13!}}{\frac{30!}{18!}}= \frac{10!20!18!}{5!13!30!}$.

Now, the same calculation shows that the seven girls and five boys chosen in any specific order the same. Further there are $\displaystyle \frac{7!10!}{12!}$ different orders so the probability of choosing 7 girls and 5 boys in any order is $\displaystyle \frac{7!10!10!20!18!}{12!5!13!30!}$.

b) is much easier. The opposite of "at least one boy" is "all 12 girls" which is $\displaystyle \frac{20}{30}\frac{19}{20}\frac{18}{19}...\frac{9}{19}= \frac{20!}{8!}\frac{30!}{18!}$ so "at least one boy" is one minus that.

 

[pka]

There are 30 students in a room. Ten of them are boys and the rest are girls. (20 girls)
a) What is the probability that a randomly made group of 12 students will have 5 boys?
assuming that we have 5 boys that means 7 girls so (working on this i was wrong)

b) What is the probability that a randomly made group of 12 students will have at least 1 boy?

I am answering this to encourage davidh to use links to $Wolfram|Apha$.
There are $\dbinom{30}{12}$ ways to select twelve from thirty. See here.
There are $\dbinom{20}{12}$ ways to select all girls to form the group. See here.
To see the probability at least one boy in the group is See here Can you explain the reasoning in that result.

 

[Romsek]

Thank you it makes so much sense now im still uncertain how to do 2c, any tips on how to approach it?

[MATH]E[\text{#girls in random group of 12}] = \sum \limits_{k=1}^{12} k P[\text{$k$ girls in random group of 12}][/MATH]
You should have the tools to compute that.

However a little thought will reveal a much simpler solution. There are twice as many girls as boys.
So on average what do you think the composition of 12 random picks would be genderwise.