Assertion: probability | /r/HasProperty | equal
Facets:
value: to 5/16 or 31.25%
type: OTHER
value: to p(ab)/p(b
type: OTHER
value: to the probability of b.
type: OTHER
value: to this number
type: OTHER
value: to a
type: OTHER
value: to the number of heads
type: OTHER
value: to the area of that subregion
type: LOCATION
value: to 1 ∞
type: OTHER
value: to 1/2*1/2 = 1/4
type: OTHER
value: to 1/2
type: OTHER
value: to 2/5
type: OTHER
value: to 3/5
type: OTHER
value: to 2/5 = 0.4
type: OTHER
value: to three
type: OTHER
value: to the product of the probabilities
type: OTHER
value: to its relative frequency
type: OTHER
Source sentences:
text: Since all six probabilities on the left are equal, that common probability must be 1/6.
text: We learned that the probability of an event is equal to its relative frequency for a large (infinite) number of trials.
text: First, recall that the probability of an outcome is equal to its relative frequency; also recall that the sum of all the relative frequencies is unity
text: Fourth and finally, the probability of an event E is equal to unity minus the probability of the event's complement, EC.
text: x {\displaystyle x} : in fact, by definition, a continuous random variable can have an infinite set of possible values and thus the probability it has a single particular value x is equal to 1 ∞ = 0
text: that is, the probability of both \(\phi\) and \(\psi\) is equal to the product of the probabilities of \(\phi\) and \(\psi\).
text: Since one of these must be selected, the probability of choosing any marble is equal to the probability of the sample space S = 1.
text: If there are three red marbles and two blue marbles, then the probability of drawing any red marble is the number of outcomes in the event " red ," which is equal to three, divided by the total number of outcomes, 5, or 3/5 = 0.6.
text: The sample space is this case is { red , blue }, which must have total probability equal to 1, so the probability of drawing a blue marble is equal to 2/5 = 0.4.
text: If two events A and B are not disjoint, then the probability of their union (the event that A or B occurs) is equal to the sum of their probabilities minus the sum of their intersection.
text: In the example corresponding to the second Venn diagram above, we know that the probability of drawing a red marble on the first draw (event A) is equal to 3/5, and the probability of drawing a blue marble on the second draw (event B) is equal to 2/5.
text: In the example corresponding to the second Venn diagram above, we know that the probability of drawing a red marble on the first draw (event A) is equal to 3/5, and the probability of drawing a blue marble on the second draw (event B) is equal to 2/5.
text: The probability of a head on any toss is equal to 1/2.
text: Since the tosses are independent, the probability of a head on both tosses (the intersection) is equal to 1/2*1/2 = 1/4.
text: Thus we find that an event having occurred successively any number of times, the probability that it will happen again the next time is equal to this number increased by unity divided by the same number, increased by two units.
text: The probability that the series of events leading to a particular node will occur is equal to the product of that node and its parents' probabilities.
text: Since the square has area 1, the probability that the dart will hit any particular subregion of the square is equal to the area of that subregion.
text: The probability is then equal to the number of heads actually found divided by the total number of flips.
text: Show that the probability of C being equal to A is 1/6.
text: The probability of A conditioned on B, denoted P(A|B), is equal to P(AB)/P(B).
text: If A is a subset of, or equal to B, then the probability of A is less than, or equal to the probability of B.
text: The Probability is equal to the number of yellow balls in the bowl divided by the total number of balls, i.e. 2/6 = 1/3.
text: For example, if performing activity X has a probability of 0.01 of suffering an accident of A, with a loss of 1000, and a probability of 0.000001 of suffering an accident of type B, with a loss of 2,000,000, then total loss expectancy is 12, which is equal to a loss of 10 from an accident of type A and 2 from an accident of type B.
text: So the probability of getting at least 3 heads in four tosses is equal to 5/16 or 31.25%.