Probability Question

You know for sure that of the 3 children 1 is a boy, so we are concerned only about the remaining 2 children.

You need to get atleast 1 boy of the 2 to make a total of 2 boys (including the one we already know)

Again,

P(Atleast 1 Boy) = 1 - P(both girls)

= 1 - (1/2)(1/2)

3/4

seema19 wrote:A person has 3 children with at least one boy. Find the probablity of having at least 2 boys among the children.

Answer:[spoiler]3/4[/spoiler]

can somebody plz explain how to solve this question.

How I look at the question is this:

Chances of the three kids being:

GGG: 3!/3! = 1
GGB: 3!/2! = 3
GBB: 3!/2! = 3
BBB: 3!/3! = 1

The question states that there is already one boy, so you know GGG is not an option. So then I would just add the GBB and BBB scenarios over total possible and get [spoiler]4/7.[/spoiler] Please advise.

When you know for sure, that a kid is Boy, why would you give it a 1/2 probability again. Though there are 3 kids, you know for sure one is a boy.

What you did would be right if you didnt know if the first kid was a boy or a girl, in that case you could do;

(1+3)/8 = 4/8 (Not sure why the denominator is 7 in your Ans)

dwarzoha001 wrote:How I look at the question is this:

Chances of the three kids being:

GGG: 3!/3! = 1
GGB: 3!/2! = 3
GBB: 3!/2! = 3
BBB: 3!/3! = 1

The question states that there is already one boy, so you know GGG is not an option. So then I would just add the GBB and BBB scenarios over total possible and get [spoiler]4/7.[/spoiler] Please advise.

There are total of 8 possibilities for the gender and order of the children:

GGG BBB
GGB BBG
GBG BGB
GBB BGG

Only 1 of those possibilities is a boy not included. So from there I took the 4 possibilities of 2 boys or more, over the seven remaining scenarios. To me it's important to note that all the children are already born and at least one of the children is a boy as opposed to saying at least the first one is a boy. That's how I read into it at least.

This is not clear. Please help.

Thanks
Mankey