Hi Can anyone help please:
Suppose that the births of boys and girls can be modelled by rollling a fair die with 27 faces, 14 of which represent a boy and 13 represent a girl.
1. Using this model, choose the option that is closest to the probability that a couple who continue to have children until they have a boy will have exactly 3 children? I answered this as 0.1394, Is this correct?
2. Using this model, choose the option that is closest to the probability that a couple who continue to have children until they have a boy will have more than 3 children?
3. Using this model, choose the option that is closest to the probability that, in a family with five children, the 1st, 3rd and 5th are girls, while the 2nd, and 4th are boys? I got 0.03456, is this correct?
Options given for questions 1 to 3 are: A=0.0300, B=0.1116 C=0.1202 D=0.1295 E=0.1394 F=0.3221 G= 0.5469 H=06324
any help much appriciated regards Claire
To calculate the probabilities for these questions, we need to use the concept of independent events and the multiplication rule for independent events. Let's go through each question step by step:
1. The probability of having a boy in any given childbirth is 14 out of 27 since there are 14 faces representing a boy out of 27 faces on the die. The probability of having a girl is 13 out of 27. Since each childbirth is an independent event, we can use the multiplication rule for independent events.
To calculate the probability that a couple will have exactly 3 children until they have a boy, we need to calculate the probability of having 2 girls followed by a boy.
P(GGG, B) = P(G) * P(G) * P(B)
= (13/27) * (13/27) * (14/27)
≈ 0.1394
So your answer of 0.1394 for question 1 is correct.
2. To calculate the probability that a couple will have more than 3 children until they have a boy, we need to consider all the possible scenarios of boys and girls until a boy is born. Instead of calculating each scenario individually, it is easier to calculate the probability of having 3 or fewer children and then subtract that from 1 to get the probability of having more than 3 children.
P(more than 3 children) = 1 - P(0 boys and 1 girl) - P(1 boy and 2 girls) - P(2 boys and 1 girl) - P(3 boys and 0 girls)
P(more than 3 children) = 1 - [(13/27)*(13/27)*(13/27)] - [(13/27)*(13/27)*(14/27)] - [(14/27)*(14/27)*(13/27)] - [(14/27)*(14/27)*(14/27)]
≈ 0.5469
So, the option closest to the probability of having more than 3 children is G=0.5469.
3. To calculate the probability that the 1st, 3rd, and 5th children are girls, while the 2nd and 4th children are boys, we again need to use the multiplication rule for independent events.
P(G, B, G, B, G) = P(G) * P(B) * P(G) * P(B) * P(G)
= (13/27) * (14/27) * (13/27) * (14/27) * (13/27)
≈ 0.03456
So your answer of 0.03456 for question 3 is correct.
I hope this helps!