Sample Space?

Question

A couple plan to have exactly three children.
This is the sample space: Sample Space: {BBB,BBG,BGG,GGG}

What is Find the probability that the family has at least two girls

Answer 1

Each time they get a child, they have two possibilities, either a boy or a girl, with equal probability of 0.5 for each. So,

BBB = 0.5 X 0.5 X 0.5 = 0.125
BBG = 0.125
BGB = 0.125
BGG = 0.125
GBB = 0.125
GGB = 0.125
GBG = 0.125
GGG = 0.125

Now count all the probabilities which conain 2 or more Gs:

BGG + GGB + GBG + GGG = 0.5

A simpler way is to solve (b + g) ^ 3 where b is the probbility of a boy and g is the probaility of a girl, in this case, each is 0.5:

(b + g)^3 = b^3 + 3 b^2 g + 3 b g^2 + g^3

Probability for atleasat 2 girls is given by 3 b g^2 + g^3 and is 3/8 + 1/8 = 0.5

Answer 2

that's a binomial distribution. you're sampling three times from {B,G} with replace, and without regard to reserve: assume p(B) = p(G) = a million/2. P(a minimum of two G) = P(2 G) + P(3 G)= 3C2*(a million/2)^3 + 3C3*(a million/2)^3 = 3/2^3 + a million/2^3 =4/8 = a million/2 nonetheless, and equivalently, the pattern area has 8 factors with equivalent danger. BBB BBG ...and so on, directly to GGG 4 out of 8 of those have 2 or 3 Gs. That ration 4/8 can grant the risk. notice that this might substitute if p(B) no longer = p(G).

Answer 3

It's a binomial distribution.

You're sampling 3 times from {B,G} with replacement, and without regard to order: Assume p(B) = p(G) = 1/2.

P(at least 2 G) = P(2 G) + P(3 G)= 3C2*(1/2)^3 + 3C3*(1/2)^3
= 3/2^3 + 1/2^3
=4/8 = 1/2

Alternatively, and equivalently, the sample space has 8 elements with equal probability.
BBB
BBG
...etc, on to
GGG

Four out of 8 of these have 2 or 3 Gs. That ration 4/8 gives you the probability. Note that this would change if p(B) not = p(G).

Answer 4

At least 2 girls are in BGG, GGG . since this is half of the possibilities, the probability is 50%

Answer 5

good attempt
sample space also includes BGB, GBB, GGB, GBG