There is no simple answer to the question because children's genders are not independent events. They depend on the parents' ages and their genes.
However, if you assume that they are independent events then, given that the probability of a boy is approx 0.52, the answer is 0.2331.
4
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.
.125
The probability of tossing a coin 9 times and getting at least one tail is: P(9 times, at least 1 tail) = 1 - P(9 heads) = 1 - (0.50)9 = 0.9980... ≈ 99.8%
The probability is 1 and you do not need Matlab to get that answer - only a little bit of thought.
Assuming that boys and girls are equally likely, it is 11/16.
3 out of 7
1 in 2
1/4
http://answerboard.cramster.com/statistics-and-probability-topic-5-292446-0.aspx
There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes. However, if you assume that they are independent events then, given that the probability of a girl is approx 0.48, the probability of 2 or more girls is 0.6617.
There is no simple answer.First of all, the probability of boys is 0.517 not0.5.Second, the probabilities are not independent.If you choose to ignore these important facts, then the answer is 2/3.
There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.However, if you assume that they are independent events then, given that the probability of a boy is approx 0.52, the probability of a daughter and a son in two children is approx 0.4994
The individual probability that a child born will be female is 50% or 0.5.Using this we can calculate the probability that at least one of the children will be female by:calculating the probability that none of the children will be female and then subtracting this from 1.The probability that all the children are male is therefore 0.53 = 0.5 * 0.5 * 0.5 = 0.125.Thus the answer is 1 - 0.125 = 0.875 = 87.5%
Going by the assumption that the probability of a child being a girl is 1/2, and that the events are independent, then the probabilities of different numbers of girls are as follows: P(0 girls) = 1/2 ^ 4 = 1/16 P(1 girl) = 4 * (1/2) * (1/2 ^ 3) = 1/4 P(2 girls) = 6 * (1/2 ^ 2) * (1/2 ^ 2) = 3/8 P(3 girls) = 4 * (1/2 ^ 3) * (1/2) = 1/4 P(4 girls) = 1/2 ^ 4 = 1/16 Therefore the probability of at least 2 girls = 3/8 + 1/4 + 1/16 = 11/16
There are 2 ways to do this problem. 1. Go to a Binomial Distribution Table where n = 4 (4 children) and P=0.5(50% probability of a girl). Probability of at least 1 girl = 1 - probability of no girls. From Binomial Distribution Table n = 0 probability is .0625. So, 1 - 0.0625 = .9375 = probability of at least 1 girl. 2. The other way is to list all the possible ways to have 4 children and count the number of ways at least 1 girl exists divided by the total number of ways to have 4 children. There are 42 ways to have 4 children, all 16 listed below: bbbb bbbg bbgb bgbb gbbb bbgg bggb ggbb gbbg gbgb bgbg bggg gggb ggbg gbgg gggg Since 15 of the 16 have at least 1 girl, the Probability of at least 1 girl = 15/16 = 0.9375, the same answer as above.
There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.However, if you assume that they are independent events then, given that the probability of a girl is approx 0.48, thenProb(at least one girl out of 6 children) = 1 - Prob(no girls out of 6 children)= 1 - Prob(6 boys out of 6 children)= 1 - 0.526 = 0.9809There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.However, if you assume that they are independent events then, given that the probability of a girl is approx 0.48, thenProb(at least one girl out of 6 children) = 1 - Prob(no girls out of 6 children)= 1 - Prob(6 boys out of 6 children)= 1 - 0.526 = 0.9809There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.However, if you assume that they are independent events then, given that the probability of a girl is approx 0.48, thenProb(at least one girl out of 6 children) = 1 - Prob(no girls out of 6 children)= 1 - Prob(6 boys out of 6 children)= 1 - 0.526 = 0.9809There is no simple answer to the question because the children's genders are not independent events. They depend on the parents' ages and their genes.However, if you assume that they are independent events then, given that the probability of a girl is approx 0.48, thenProb(at least one girl out of 6 children) = 1 - Prob(no girls out of 6 children)= 1 - Prob(6 boys out of 6 children)= 1 - 0.526 = 0.9809