Introduction
 Probability is a measure of the likelihood of the occurrence of an event.
 Mathematically, the probability of an event A is calculated using the formula: $P\left(A\right)=\frac{\text{numberoffavourableoutcomes}}{\text{totalnumberofoutcomes}}$
 Probability is of two types  theoretical probability & experimental probability.
 Theoretical probability is what we expect to happen in an experiment (remains the same), whereas experimental probability is what actually happens when we try it out (not always the same).
 The probability of an event ranges from 0 to 1.
 Probability can be expressed in terms of fractions, percentages, or decimals.
 The probability related to the occurrence of two or more events is called Compound probability.
 A combination of such events is called compound events.
Independent and Dependent Events
 Compound events can be of two types  independent & dependent.
 Independent Events:

 Events that don't depend on each other (the outcome of one event does not affect the outcome of the second event) are called Independent events.
 Example: getting tails in a coin flip and rolling five on a sixsided die
 The probability of two independent events, A and B, can be found using the formula below.
$P\left(A\text{and}B\right)\mathit{=}P\left(A\right)\mathit{\xb7}P\left(B\right)$

 Thus to find the compound probability of two independent events, we simply multiply their individual probabilities.
 Dependent Events:

 Events that depend upon each other are called Dependent events.
 Example: drawing two cards from a standard without replacement
 The probability of two independent events, A and B, can be found using the formula below.
$P\mathit{(}A\mathit{}\mathit{\text{and}}\mathit{}B\mathit{)}\mathit{=}P\mathit{\left(}A\mathit{\right)}\mathit{\xb7}P\mathit{\left(}B\rightA\mathit{)}$

 Thus to find the compound probability of two dependent events, we multiply the probability of the first event $P\left(A\right)$ by the probability of the next event after the event has taken place $P\left(BA\right)$.
Mutually Exclusive and Mutually Inclusive Events
 Events can also be categorized as mutually exclusive and mutually inclusive.
 Mutually Exclusive Events:

 Compound events that cannot happen at the same time are called mutually exclusive events.
 Example: getting a number 4 and an odd number when rolling a sixsided die once
 For two mutually exclusive events, A and B,
$\mapsto P\left(A\text{and}B\right)=0$ because A and B cannot happen at the same time.
$\mapsto P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$
 Mutually Inclusive Events:

 Two events that can happen at the same time simultaneously are called mutually inclusive events.
 Example: getting a number less than four and an odd number when rolling a sixsided die once
 For two mutually inclusive events, A and B,
$\mapsto P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)P\left(A\text{and}B\right)$
 Note: When dealing with compound probabilities, we often encounter two symbols. The symbol '$\cup $' means "the union of" which is equivalent to $P\left(A\text{or}B\right)$ and '$\cap $' means "the intersection of" which is equivalent to $P\left(A\text{and}B\right)$.
Solved Examples
Question 1: A card is drawn randomly from a wellshuffled deck of 52 cards. Find the probability that the card drawn is a queen or an ace.
Solution: Drawing a queen and an ace card are mutually exclusive events, as they cannot happen simultaneously.
$P\left(A\right)=P\left(\text{drawingaqueen}\right)=\frac{13}{52}=\frac{1}{4}$
$P\left(B\right)=P\left(\text{drawinganacecard}\right)=\frac{4}{52}=\frac{1}{13}$
$P{\left(A\text{or}B\right)}_{mutuallyexclusive}=P\left(A\right)+P\left(B\right)$
$=\frac{1}{4}+\frac{1}{13}$
$=\frac{13+4}{52}$
$=\frac{17}{52}$
Question 2: A bowl contains ten green marbles, six red marbles, and four black marbles. Find the probability of drawing a green marble and then a red marble.
Solution: Total number of balls = 10 + 6 + 4 = 20
Probability of drawing green ball first, $P\left(\text{green}\right)=P\left(A\right)=\frac{10}{20}=\frac{1}{2}$
Probability of drawing a red ball after the green ball is taken out, $P\left(\text{red}\right)=P\left(B\right)=\frac{6}{19}$ Note that after drawing a green ball, 19 ball remains.
Required probability, $P\left(A\text{and}B\right)=P\left(A\right)\xb7P\left(AB\right)$$=\frac{1}{2}\times \frac{6}{19}=\frac{3}{19}$
Cheat Sheet
 $P\left(\text{event}\right)=\frac{\text{numberoffavourableoutcomes}}{\text{totalnumberofoutcomes}}$
 If A and B are two independent events, then $P\left(A\text{and}B\right)=P\left(A\right)\xb7P\left(B\right)$.
 If A and B are two dependent events, then $P\left(A\text{and}B\right)=P\left(A\right)\xb7P\left(BA\right)$.
 If A and B are two mutually exclusive events, then $P\left(A\text{and}B\right)=0$. Also, $P\left(Aor\text{}B\right)=P\left(A\right)+P\left(B\right)$.
 If A and B are two mutually inclusive events, then $P\left(Aor\text{}B\right)=P\left(A\right)+P\left(B\right)P\left(A\text{and}B\right)$.
Blunder Areas
 Sometimes a case may arise where there are a lot of favorable outcomes and fewer nonfavorable outcomes. In such cases, we should first find the probability of nonfavorable outcomes and then subtract it from the total number of outcomes.
 Abhishek Tiwari
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