Question

A student committee consists of 20 boys and 15 girls. A team of 8 students is selected randomly from the committee to participate in an activity and the team must consists of at least 3 boys and at least 3 girls. Find the probability that the teams can be formed if the team consists of 4 boys already.

Answer 1

This will take me a few mins sorry im just thinking it through my head

Answer 2

THe probability that they can be or will be?

Answer 3

can be

Answer 4

Well with 15 Girls left 4 slots on the team open and 16 boys left i would think 100%

Answer 5

There seems to be no rules whether or not they can join a team

Answer 6

" the team must consists of at least 3 boys and at least 3 girls."

Answer 7

ATLEAST not ONLY so it would be 100% wouldnt it?

Answer 8

If there are 7 boys, only 1 girl, then the team cannot be formed. So, it will NOT be 100%

Answer 9

There arent multiple teams so there wouldnt be a number limit on how many of each we can add

Answer 10

I suggest you to read the question carefully again.

Answer 11

IT DOES NOT SAY IT HAS TO HAVE 3 MORE BOYS AND GIRLS ATLEAST MEANING THAT 4 BOYS AND 4 GIRLS WOULD WORK

Answer 12

4 boys 4 girls is just one way
It can be 3 boys 1 girl, then it fails.

Answer 13

Oh i am sorry it is written in such a way that it was slightly dumbfounding me how my way wouldnt work

Answer 14

But since there are 4 boys already there could be 4/4 5/3 6/2 7/1 8/0 so the probability would be 40% or 2/5

Answer 15

Im sorry

Answer 16

No...

Answer 17

?

Answer 18

Your assumption in your calculation is that boys and girls will be picked with equal probability but they are not. There are more boys than girls.

Answer 19

@kewlgeek555 Can you help me with this kewl?

Answer 20

And for now i will use 40% as my base and will edit that to get the correct percent im sorry for not doing too well i stopped taking statistics last year but im pretty good compared to most people when it comes to problems that have an equation

Answer 21

Oh, maybe you can help me with my other problem then. Let me solve the problem first though. :)

Answer 22

Ok so selecting a girl has a 75% chance compared to selecting a boy not the full chance.

Answer 23

GTG Be back in a bit

Answer 24

LOL I think I got it.

Answer 25

Let B = boy(s), G = girl(s)
\[P(4B) = \frac{16}{31}\times\frac{15}{30}\times\frac{14}{29}\times\frac{13}{28}\times\frac{4!}{4!}=\frac{52}{899}\]\[P(3B1G) = \frac{16}{31}\times\frac{15}{30}\times\frac{14}{29}\times\frac{15}{28}\times\frac{4!}{3!}=\frac{240}{899}\]\[P(2B2G) = \frac{16}{31}\times\frac{15}{30}\times\frac{15}{29}\times\frac{14}{28}\times\frac{4!}{2!2!}=\frac{360}{899}\]\[P(1B3G) = \frac{16}{31}\times\frac{15}{30}\times\frac{14}{29}\times\frac{13}{28}\times\frac{4!}{3!}=\frac{208}{899}\]\[P(4G) = \frac{15}{31}\times\frac{14}{30}\times\frac{13}{29}\times\frac{12}{28}\times\frac{4!}{4!}=\frac{39}{899}\]
So, \[\text{P(can form a team) = }\frac{P(1B3G) + P(4G)}{P(4B)+P(3B1G)+P(2B2G)+P(1B3G) + P(4G)} = \frac{247}{899}\]

Does this make sense?

Answer 26

The last fraction after the equal sign in the second last line is \(\large\frac{247}{899}\).

Answer 27

Ok good lol

Answer 28

The other one?

Answer 29

Haha! You want some challenges?

Answer 30

Sure just not ones that cant be solved with a percentage or a fraction i prefer not to do large equations in my head (I do much worse when i try to work on paper) please

Answer 31

Hmm... I do have something I cannot solve at the moment... Hold on.

Answer 32

Please show me

Answer 33

Yeah hes making a new one kewl