5 important Probability Fact

 5 Probability FACT 

1. A probability of 0 means that an event is impossible.
So if you find that P(E) = 0, that means that E will not occur.

Example:- when rolling a six-sided die, 

the event that we roll a 7 is possible ?

NO 7 event is impossible.

it does not occur in any of our outcomes. Thus, 

P(of 1 roll is )= 1/6

P(of 2 roll is )= 1/6

P(of 3 roll is )= 1/6

P(of 4 roll is )= 1/6

P(of 5 roll is )= 1/6

P(of 6 roll is )= 1/6

P(of 7 roll is ) = 0.

2. A probability of 1 means that an event is certain.
EXAMPLE :-So, when rolling a six-sided die, 

the event that we roll some number is a certainty --

 it occurs in all of our outcomes. 

Thus, P(Roll a number) = 1.

3. An event with a higher probability is more likely to occur.
SO , if the probability of choices is 20% and 80% than the most higher probability of is more likely to occur that is  80%. 

EXAMPLE:- if the probability of snows is 20% while the probability of rains is 80%, then it is more likely to rain than it is to snow. So, events with a lower probability are less likely to occur.

4. Probabilities are always between 0 and 1. 

5. The probabilities of our different outcomes must sum to 1.

EXAMPLE:- 

if we have 4 different outcomes, then

P(Outcome 1) + P(Outcome 2) + P(Outcome 3) + P(Outcome 4) = 1.

EXAMPLE:- there are six possible outcomes in die. Only 1 outcome is coming at one time by flip the die.   and total outcome is 6..

Roll a "1": Probability is $\frac{1}{6}$
Roll a "2": Probability is $\frac{1}{6}$
Roll a "3": Probability is $\frac{1}{6}$
Roll a "4": Probability is $\frac{1}{6}$
Roll a "5": Probability is $\frac{1}{6}$
Roll a "6": Probability is $\frac{1}{6}$

And the sum of all the probabilities:
$\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=1$,

Now, for the GRE, there are three main types of probability problems:

1. The probability of a single event occurring: P(A)
2. The probability that two events both occur: P(A and B)

EXAMPLE:-

P(A and B)$P(A\text{ and }B)$.

Using the definition of conditional probability, we find,

P(A and B)=P(A)P(B|A)=P(first card is an ace)P(second card is an ace given the first card is an ace)$$P(A\text{ and }B)=P(A)P(B|A)=P(\text{first card is an ace})P(\text{second card is an ace given the first card is an ace)}$$

=452⋅351=1221.

3. The probability that one or another event occurs: P(A or B)

p(A or B) = p(A) + p(B).

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