P. B. Sreenivas was born on September 22, 1930.

P. B. Sreenivas was born on September 22, 1930.

P. B. Sreenivas is 80 years old (birthdate: September 22, 1930).

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

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.

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.

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)

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)

SCALE

- S. Balaji Sreenivas

Hermann Minkowski

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).

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)

P=B×R

B=P÷R

R=P÷B

If A and B are two events then

P(A or B) = P(A) + P(B) - P(A and B)

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.

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)

P(A'/B)=P(A'nB)/P(B)

is it illegal to put a squirrel in a t-shirt cannon and shoot it at a pedestrian

Well for one you cant buy Alex Rider the Gemini Project because it doesn't exist what your after is Alex Rider Point Blanc and as for that you can buy it in many places like amazon or got to the libary and get it

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Consider events A and B. P(A or B)= P(A) + P(B) - P(A and B)

The rule refers to the probability that A can happen, or B can happen, or both can happen together. That is what is stated in the addition rule.

Often P(A and B ) is zero, if they are mutually exclusive. In this case the rule just becomes P(A or B)= P(A) + P(B).

If B is between P and Q, then:

P<B<Q

There are symbols missing from your question which I cam struggling to guess and re-insert.

p(a) = 2/3

p(b ??? a) = 1/2

p(a ∪ b) = 4/5

p(b) = ?

Why use the set notation of Union on the third given probability whereas the second probability has something missing but the "sets" are in the other order, and the order wouldn't matter in sets.

There are two possibilities:

1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2

→ p(a) + p(b) = p(a ∪ b) + p(a ∩ b)

→ p(b) = p(a ∪ b) + p(a ∩ b) - p(a)

= 4/5 + 1/2 - 2/3

= 24/30 + 15/30 - 20/30

= 19/30

2) The second and third probabilities are probabilities of "given that", ie:

p(b|a) = 1/2

p(a|b) = 4/5

→ Use Bayes theorem:

p(b)p(a|b) = p(a)p(b|a)

→ p(b) = (p(a)p(b|a))/p(a|b)

= (2/3 × 1/2) / (4/5)

= 2/3 × 1/2 × 5/4 = 5/12

P(A given B')=[P(A)-P(AnB)]/[1-P(B)].

The short answer is that if you don't do so, the intersection is double counted. The longer answer follows:

Suppose you have two events A and B which overlap. Then,

Event A = (only A happens) and (both A and B happen).

These two are mutually exclusive so that P(A) = P(A only) + P(both).

Similarly, P(B) = P(B only) + P(both).

Now, the event that A or B happen is the event that only A happens or only B happens or both happen.

That is, P(A or B) = P(A only) + P(both) + P(B only)

Adding and subtracting P(both) gives

P(A or B) = P(A only) + P(both) + P(B only) + P(both) - P(both)

Now, the first two are P(A) and the next two are P(B)

So P(A or B) = P(A) + P(B) - P(both).

first prove *: if A intersect B is independent, then A intersect B' is independent. (this is on wiki answers)

P(A' intersect B') = P(B')P(A'|B') by definition

= P(B')[1-P(A|B')] since 1 = P(A) + P(A')

= P(B')[1 - P(A)] from the first proof *

= P(B')P(A') since 1 = P(A) + P(A')

conclude with P(A' intersect B') = P(B')P(A') and is therefore independent by definition.

***note*** i am a student in my first semester of probability so this may be incorrect, but i used the first proof* so i figured i would proof this one to kinda "give back".

p=b+3a+c

p-3a-c=b+3a-3a+c-c

p-3a-c=b

b=p-3a-c

Range

-S.Balaji Sreenivas

This is a tricky question, but if you study probability, re-writing it in mathematical notations and using simple definitions and algebra would show you that the answer is yes.

We could rewrite the question as:

if B occur, A is more likely to occur,

Probability of A happening given B happened > probability of A happening

P( A given B) > P(A)

Is it true that in this case, P(B given A) > P(B)?

by definition,

P( B given A) = P(A and B) / P( A) = P( A given B) * P(B) / P(A)

Since P( A given B) > P(A),

then P( A given B)/ P(A) > 1

thus

P( B given A) > P(B)

and the answer is true, occurence of A makes B more likely.

Say A = lung cancer, B = smoking. If someone is smoking, they are more likely to have lung cancer. If someone has lung cancer, they are more likely to be a smoker or have smoked (compared to being a non-smoker).

A compound event is any event combining two or more simple events.

The notation for addition rule is: P(A or B) = P(event A occurs or event B occurs or they both occur).

When finding the probability that event A occurs or event B occurs, find the total numbers of ways A can occurs and the number of ways B can occurs, but find the total in such a way that no outcome is counted more than once.

General addition rule is :

P(A or B) = P(A) + P(B) - P(A and B), where P(A and B) denotes that A and B both occur at the same time as an outcome in a trial procedure.

It is a special addition rule that shows that A and B cannot both occur together, so P(A and B) becomes 0:

If A and B are mutually exclusive, then P(A) or P(B)= P(A or B) = P(A) + P(B)

a and b are both multiplied by p, so take the p out. This changes the equation to p (a - b). The two factors are p and (a - b), so you have factored (a - b) out of the ap - bp.

p=a+b+c for a

Is there a cure for Depression B/p?

P(A)=.35 P(B given A)=0.6 P(A and B)= ?

P. B. Sawant was born in 1930.

B. P. Paquette was born in 1975.

P. B. Chatwin was born in 1873.

P. B. Chatwin died in 1964.

B. P. Govinda was born in 1951.

P. B. Desai died in 1974.

P. B. Desai was born in 1910.

B. P. Schulberg died in 1957.

B. P. Schulberg was born in 1892.

B. P. Mandal died in 1982.

B. P. Mandal was born in 1918.

B. P. Wadia died in 1958.

B. P. Wadia was born in 1881.

P(A given B')=[P(A)-P(AnB)]/[1-P(B)].

In words: Probability of A given B compliment is equal to the Probability of A minus the Probability of A intersect B, divided by 1 minus the probability of B.

dependent because your changing

P = a + b + c + d, where a, b, c, d are sides.

P = 2 × (a + b), where a and b are adjacent sides.

P=2*(15.1+16)

P=2*31.1

P=62.2

Answer: P= 62.2

Consider the three events:

A = rolling 5, 6, 8 or 9.

B = rolling 7

C = rolling any other number.

Let P be the probability of these events in one roll of a pair of dice.

Then P(A) = P(5) + P(6) + P(8) + P(9) = 18/36 = 1/2

P(B) = P(7) = 6/36 = 1/6

and P(C) = 1 - [P(A) + P(B)] = 1/3

Now P(A before B) = P(A or C followed by A before B)

= P(A) + P(C)*P(A before B)

= 1/2 + 1/3*P(A before B)

That is, P(A before B) = 1/2 + 1/3*P(A before B)

or 2/3*P(A before B) = 1/2

so that P(A before B) = 1/2*3/2 = 3/4