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No. If both parents are B+ then their children will be B+ (75%) or O+ (25%).

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Q: Can parents be B pos and have a kid A B p?
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15 b b in p?

sry dudes and dudets i am a little kid i dnt even know what that means just playing


If A and B are independent events then are A and B' independent?

if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof


What is the product rule and the sum rule of probability?

Sum Rule: P(A) = \sum_{B} P(A,B) Product Rule: P(A , B) = P(A) P(B|A) or P(A, B)=P(B) P(A|B) [P(A|B) means probability of A given that B has occurred] P(A, B) = P(A) P(B) , if A and B are independent events.


What is the formula for inclusive events?

The probability of inclusive events A or B occurring is given by P(A or B) = P(A) + P(B) - P(A and B), where P(A) and P(B) represent the probabilities of events A and B occurring, respectively.


How do you find P A given B?

P(A|B)= P(A n B) / P(B) P(A n B) = probability of both A and B happening to check for independence you see if P(A|B) = P(B)


Addition rule for probability of events A and B?

If they're disjoint events: P(A and B) = P(A) + P(B) Generally: P(A and B) = P(A) + P(B) - P(A|B)


Give the example of why probabilities of A given B and B given A are not same?

Let's try this example (best conceived of as a squared 2x2 table with sums to the side). The comma here is an AND logical operator. P(A, B) = 0.1 P(A, non-B) = 0.4 P(non-A, B) = 0.3 P(non-A, non-B) = 0.2 then P(A) and P(B) are obtained by summing on the different sides of the table: P(A) = P(A, B) + P(A, non-B) = 0.1 + 0.4 = 0.5 P(B) = P(A,B) + P(non-A, B) = 0.1 + 0.3 = 0.4 so P(A given B) = P (A, B) / P (B) = 0.1 / 0.4 = 0.25 also written P(A|B) P(B given A) = P (A,B) / P (A) = 0.1 / 0.5 = 0.2 The difference comes from the different negated events added to form the whole P(A) and P(B). If P(A, non-B) = P (B, non-A) then P(A) = P(B) and also P(A|B) = P(B|A).


Definition of additive law in probability?

This has to do with the union of events. If events A and B are in the set S, then the union of A and B is the set of outcomes in A or B. This means that either event A or event B, or both, can occur. P(A or B) = P(A) + P(B) - P(A and B) **P(A and B) is subtracted, since by taking P(A) + P(B), their intersection, P(A and B), has already been included. In other words, if you did not subtract it, you would be including their intersection twice. Draw a Venn Diagram to visualize. If A and B can only happen separately, i.e., they are independent events and thus P(A and B) = 0, then, P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - 0 = P(A) + P(B)


Formulas on Percentage Base and Rate?

P=B×RB=P÷RR=P÷B


How might two probabilities be added?

If A and B are two events then P(A or B) = P(A) + P(B) - P(A and B)


What is the Formula for odds for either of two events?

If A and B are mutually exclusive event then Probability of A or B is P(A)+P(B). If they are not mutually exclusive then it is that minus the probability of the P(A)+P(B) That is to say P( A or B)= P(A)+P(B)- P(A and B). Of course it is clear that if they are mutually exclusive, P(A and B)=0 and we have the first formula.


What is multiplication rule in probability?

Given two events, A and B, the conditional probability rule states that P(A and B) = P(A given that B has occurred)*P(B) If A and B are independent, then the occurrence (or not) of B makes no difference to the probability of A happening. So that P(A given that B has occurred) = P(A) and therefore, you get P(A and B) = P(A)*P(B)