Question

In a family of 5 children what is the probability of have 1 boy and then 4 girls in that order. (Exclude multiple births and assume all outcomes)

Answer 1

Assuming that the probability of having a boy or a girl is equal (i.e., 0.5), the probability of having 1 boy and 4 girls in that order in a family of 5 children can be calculated as follows:

P(1 boy, 4 girls) = P(boy) x P(girl) x P(girl) x P(girl) x P(girl)

where P(boy) is the probability of having a boy and P(girl) is the probability of having a girl.

Therefore, we have:

P(1 boy, 4 girls) = 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.5^5

= 0.03125 or 3.125%

So, the probability of having 1 boy and 4 girls in that order in a family of 5 children is 3.125%.

Answer 2

The probability of having 1 boy followed by 4 girls in a family of 5 children, assuming equal chance for both genders, is 3.125%, or 1/32, computed by multiplying the probability of each event occurring independently.

Step-by-step explanation:

The question asks for the probability of having 1 boy followed by 4 girls in a family of 5 children. Assuming the probability of having a boy or a girl is equal (since no specific information is provided), which is commonly approximated as 1/2 or 50%, we use basic principles of probability to calculate this scenario. The probability of a specific sequence of gender birth can be found by multiplying the probabilities of each individual event. In this case, having a boy (B) first and then four girls (GGGG) would be calculated as:

P(BGGGG) = P(B) × P(G) × P(G) × P(G) × P(G)

P(BGGGG) = (1/2) × (1/2) × (1/2) × (1/2) × (1/2) = (1/32)

So, the probability of having 1 boy followed by 4 girls, in that order, is 1/32 or approximately 3.125%.