Question

A couple decided to have 5 children.
(a) What is the probability that they will have at least two girls?
(b) What is the probability that all the children will be of the same gender?

Answer 1

what is the probability of a baby being a boy or girl?

Answer 2

1/2 right?

Answer 3

similar to the problem if they decide to have 3 kids with two girls and one boy
{BBB,BBG,BGB,BGG**,GBB,GBG** ,GGB**,GGG} => so probability of two Girls(G) ) and oneBoy(B) P(2G union) 1B) = 3/8 ..do similar way for 5 kids and try to understand the answer.

Answer 4

Use binomial theorem to get the answer. ( take (x+y)^3 x==> represents girls and y==> boys. (x+y)^3 =1 x^3 + 3x^2y + 3xy^2 + 1y^3 the coefficient of the term that represents x^2y(2 girls and one boy) has 3 and total coefficient sum=8 so probability=3/8

The binomial theorem can be used with following characteristics
* There are only two possibilities ...success or failure
* The events are independent
* The probability of each outcome is same for each trial
Such problems are called Bernauli trials . Let p represents the probability of success.
let q represents the probability of failure.
nCrp^(n-r)q^r represents the probability of exactly r successes. Use that here
5C2xy^2 [ 2 girls and one boy] or use tht to expand to get the coefficients
use that for (x+y)^5 fr 5 children = x^4 +

Answer 5

I got the right answer for B but can't get the right answer for A...

Answer 6

(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 5xy^4 + 10x^2y^3 + y^5
let x => represent girl and y=> represent boy. In 5 kids, total same gender (either all boys(y)
or all girls(x) are the term x^5 or y^5 = 1 out of 32 =1/32

Answer 7

1/32 is not the answer for part A

Answer 8

2nd part all boys or all gilrs

Answer 9

GGGGG
GGGGB
GGGBB
GGBBB
GBBBB

if we work out the probability that there is only 1 girl, and take that away from 1, we have the probability that there is at least 2 girls

Answer 10

probabiliy of two girls(x) = 10x^2y^3 , so the coefficient of this term is 10 ==> probability =10/32

Answer 11

10/32 is not the answer for A

Answer 12

I already have the answer for B so you don't need to solve that. But A is wrong

Answer 13

GBBBB - 1/32
BGBBB - 1/32
BBGBB - 1/32
BBBGB - 1/32
BBBBG - 1/32

total = 5/32

answer = 1- 5/32 = 27/32

Answer 14

27/32 is wrong too...

Answer 15

25/32

Answer 16

im pretty sure i did it right... unless p =/= 1/2

Answer 17

I did it the same way as you and got the same answer but it's wrong.

Answer 18

@experimentX maybe you can clear this up

Answer 19

lol ... i am always prone to mistake on first try.

Answer 20

:) im going out in a bit so i thought i should leave him in capable hands, if you're willing to help him

Answer 21

cya guys!

Answer 22

yeah, i will try my best.

Answer 23

Haha thank you :)

Answer 24

5x^4y + 10x^3y^2 + 10 x^2y^3 + x^5(this term all girls)
a). prob tht they will will hve at least 2 girls = 5+10+10+1/32 =26/32

Answer 25

Wow. 26/32 is actually right. Can't believe I was off by only one number.

Answer 26

Thank you all so much! Appreciate it

Answer 27

i kinda think like @rumadutt

Answer 28

z11 you are doing without understanding. try to understand it.

Answer 29

use binomial theorem if it satisfies those conditions

Answer 30

seems like binomial probability or so what it is called i forgot ...

Answer 31

Ok I get it. THANKS

Answer 32

i guess this is it
http://en.wikipedia.org/wiki/Binomial_probability