This question is extremely poorly phrased.
The probability of three boys [sitting] in a row at an all boys school is 1. At an all girls school it is 0 and is otherwise somewhere in between.
If the question is about birth order, do you take account of the fact that nearly half the families have two or fewer children? So that in half the cases the probability is 0.
Finally, 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 giving birth to three boys in a row is 0.523 = 0.1381
(assuming that the probability of having a girl or a boy is 50/50) Looking from beforehand, the probability of having three boys then a girl is the probability of each of these events happening multiplied together. That is 50% x 50% x 50% x 50% or 0.54 This would mean that the chance of having a girl after three boys is 0.0625. If you've already had the three boys though, it is a different story. The point is that previous experiences do not affect future ones; probability has no memory. Thus the probability of having a girl next is 50%, regardless of if you've had boys or girls in the past. To think otherwise is known as the gambler's fallacy, where a gambler says "black has come up 4 times in a row, it must be red next" even though the chance of red is always 50%
Assuming that the probability of having a baby girl is 1/2 and that of having a baby boy is 1/2, the probability of having 3 baby girls in a row is (1/2)(1/2)(1/2)=1/8.
The probability of drawing aces on the first three draws is approx 0.0001810
Assuming that the die is a "normal" one (it has the numbers 1 to 6 and that it is fair), then the probability of rolling six three times in a row is 1/6*1/6*1/6 = 1/216 = 0.00463 The probability of rolling six three times in a row eventually is 1 (ie a certainty).
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you have a 75% chance
The probability of having a boy or a girl is always 50/50 each time, regardless of previous outcomes. So the theoretical probability of having a girl after having three boys in a row is still 50%.
(assuming that the probability of having a girl or a boy is 50/50) Looking from beforehand, the probability of having three boys then a girl is the probability of each of these events happening multiplied together. That is 50% x 50% x 50% x 50% or 0.54 This would mean that the chance of having a girl after three boys is 0.0625. If you've already had the three boys though, it is a different story. The point is that previous experiences do not affect future ones; probability has no memory. Thus the probability of having a girl next is 50%, regardless of if you've had boys or girls in the past. To think otherwise is known as the gambler's fallacy, where a gambler says "black has come up 4 times in a row, it must be red next" even though the chance of red is always 50%
It depends on the context. In a girls' school, it is pretty close to 1 whereas in a boys' school it will be 0.
Assuming that the probability of having a baby girl is 1/2 and that of having a baby boy is 1/2, the probability of having 3 baby girls in a row is (1/2)(1/2)(1/2)=1/8.
The probability of drawing aces on the first three draws is approx 0.0001810
3/18
whats the probability that three times in a row without looking i can pick out an outmeal cookie without replacing them?
The probability of tossing a die and getting three 6's in a row is (1/6)3, or about 0.004630.
Each chance of a having a girl is 50% or 1/2. Three different children are separate events and so multiplying the chance all three together gives 1/2 x 1/2 x 1/2= 1/8. 1 in 8 chances will someone have three girls in a row.
Assuming that the die is a "normal" one (it has the numbers 1 to 6 and that it is fair), then the probability of rolling six three times in a row is 1/6*1/6*1/6 = 1/216 = 0.00463 The probability of rolling six three times in a row eventually is 1 (ie a certainty).
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