It is not possible to give a proper answer to the question because the children's are not independent. Among other things they depend on the parents' genetics and their age and there is no information on either of those factors. Also, the probability of a male child is not 0.5 as it is often naively assumed to be. It is, in fat, around 0.52.
However, if you suspend reality and assume that
then the answer is (1/2)3 = 1/8.
If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.
It is p3.
The answer to this is 1 minus the probability that they will have 3 or fewer children. This would happen only if they had a boy as the first, second or third child. The probability they have a boy as first child is 0.5 The probability they have a boy as second is 0.25 The probability they have a boy as third is 0.125 Thus the total probability is 0.875 And so the probability they will have more than three children is 1-0.875 or 0.125
In a large enough number of tosses, it is a certainty (probability = 1). In only the first three tosses, it is (0.5)3 = 0.125
In the long run, it is a certainty. In the first three throws it is 1/216.
If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.If a fair die is thrown often enough, the probability is 1.For the first three throws of a fair die, the probability is 1/216.
It is p3.
The probability of drawing aces on the first three draws is approx 0.0001810
The answer to this is 1 minus the probability that they will have 3 or fewer children. This would happen only if they had a boy as the first, second or third child. The probability they have a boy as first child is 0.5 The probability they have a boy as second is 0.25 The probability they have a boy as third is 0.125 Thus the total probability is 0.875 And so the probability they will have more than three children is 1-0.875 or 0.125
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 three out of three being girls is 0.1127.
In a large enough number of tosses, it is a certainty (probability = 1). In only the first three tosses, it is (0.5)3 = 0.125
There are 13 diamonds. Three cards are dealt. The probability of all of them being diamond is (13/52)(12/51)(11/50) = 1716/132600 = 11/850
If you toss the die often enough then the probability of getting the sequence 2-2-1 is 1: a certainty. The probability of getting the result in the first three tosses is 1/216.
The probability is 1/36
two thirds
In the long run, it is a certainty. In the first three throws it is 1/216.
It is 0.00111 approx.