Probability of independent and dependent events

Mutually Exclusive Events
Two events are mutually exclusive if they cannot occur at the same time. Another word that means mutually exclusive is disjoint.

If two events are disjoint, then the probability of them both occurring at the same time is 0.

Disjoint: P(A and B) = 0

If two events are mutually exclusive, then the probability of either occurring is the sum of the probabilities of each occurring.

Specific Addition Rule
Only valid when the events are mutually exclusive.


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

Non-Mutually Exclusive Events
In events which aren't mutually exclusive, there is some overlap. When P(A) and P(B) are added, the probability of the intersection (and) is added twice. To compensate for that double addition, the intersection needs to be subtracted.

General Addition Rule
Always valid.

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



Independent Events
Two events are independent if the occurrence of one does not change the probability of the other occurring.

An example would be rolling a 2 on a die and flipping a head on a coin. Rolling the 2 does not affect the probability of flipping the head.

If events are independent, then the probability of them both occurring is the product of the probabilities of each occurring.

Specific Multiplication Rule
Only valid for independent events

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



Dependent Events
If the occurrence of one event does affect the probability of the other occurring, then the events are dependent.

Conditional Probability
The probability of event B occurring that event A has already occurred is read "the probability of B given A" and is written: P(BA)

General Multiplication Rule
Always works.

P(A and B) = P(A) * P(BA)

EXAMPLES
A fair die is tossed twice. Find the probability of getting a 4 or 5 on the first toss and a 1, 2, or 3 in the second toss.

Answer

P(E1) = P(4 or 5) = 2/6 = 1/3

P(E2) = P(1, 2 or 3) = 3/6 = 1/2

They are independent events, so

P(E1 and E2) = P(E1) × P(E2) = 1/3 × 1/2 = 1/6

EXAMPLE -
Two balls are drawn successively without replacement from a box which contains 4 white balls and 3 red balls. Find the probability that

(a) the first ball drawn is white and the second is red;

(b) both balls are red.



(a) The second event is dependent on the first.

P(E1) = P(white) = 4/7

There are 6 balls left and out of those 6, three of them are red. So the probability that the second one is red is given by:

P(E2 E1) = P(red) = 3/6 = 1/2

Dependent events, so

P(E1 and E2) = P(E1) × P(E2 E1) = 4/7 × 1/2 = 2/7


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(b) Also dependent events. Using similar reasoning, but realising there will be 2 red balls on the second draw, we have:

P(RR)=3/7*2/6=1/7


EXAMPLE-

A bag contains 5 white marbles, 3 black marbles and 2 green marbles. In each draw, a marble is drawn from the bag and not replaced. In three draws, find the probability of obtaining white, black and green in that order.

Answer

We have 3 dependent events.
P(W1)*P(B2W1)*P(G3B2 AND W1)
=5/10*3/9*2/8
=1/24