Probability Problems or chance is a common term used in day-to-day life. For example, generally say, ‘it may rain today’. This statement has a certain uncertainty. In this, topics are related to experiment, random experiment, sample space, even point, discrete sample space, events, probability of occurrence of an event, odd in favour and odds of an event(E) etc.

**Random experiment**: An operation in which all possible outcomes are known and the exact output cannot be predicted in advance, is called random experiment.

Example: Coin contains heads or tails. Chance is either head or tail.

**Sample space or Event point**: while performing an experiment, each element of the sample space is called a sample point or an event point.

**Discrete sample space**: A sample space S is called a discrete sample space if S is a finite set.

**Events**: A subset of sample space S is called an event.

**Simple events or elementary events**: An event is called a simple event if it is a singleton subset of the sample space S.

**Compound events**: A subset of the simple space S which contains more than one element is called Compound event. It is also known as mixed event.

**Trial**: When an experiment is repeated under similar condition and it does not give the same result each time but may result in any one of the several possible outcomes are called cases. The number of times the experiment is repeated is called the number of trials.

**Mutually exclusive events**: Two or more events are said to be mutually exclusive if one of them occurs, other cannot occur.

Events \(A_{1}, A_{2},….,A_{n}\) are mutually exclusive if and only if \(A_{i} \cap A_{j}\) = \(\phi\) \(\forall i \neq\) j.

**Independent events**: Two or more events are said to be independent or mutually independent if the probability of occurrence or non- occurrence of any of them does not change by occurrence or non-occurrence of other events.

**Probability of occurrence of an event**: Let S be the sample space and E be an event. Then the probability of occurrence of the event E is denoted by P(E) and is defined as the ratio of number of cases favourable to be an event E & total number of cases.

**Odd in favour and odds of an event(E)**:

Odd in favour of an event is defined as the ratio between number of cases favourable to event(E) and the number of cases against.

Odds against an event is defined as the ratio between number of cases against E and the number of cases in favour of E.

The Probability of the occurrence of an event E is defined as:

- \(P(E)\) = \(\frac{(No. \ of \ ways \ E \ can \ occur)}{(Total \ no. \ of \ possible \ outcomes)}\)

**Example 1**:

When a single die is rolled, the sample space is {1, 2, 3, 4, 5, 6}. What is the probability of rolling a 5 when a die is rolled?

**Solution**:

- No. of ways it can occur = 1

Total no. of possible outcomes = 6

So the probability of rolling a particular number when a die is rolled = \(\frac{1}{6}\).

**Example 2**:

An unbiased diw is tossed. Find the probability of getting a multiple of 3.

**Solution**:

- Here Sample Space (S) = {1, 2, 3, 4, 5, 6}

Let E be the event of getting a multiple of 3.

Then, E = {3, 6}

Therefore, \(P(E)\) = \(\frac{(No. \ of \ ways \ E \ can \ occur)}{(Total \ no. \ of \ possible \ outcomes)}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\).

**Example 3**:

In a simultaneous throw of a pair of dice, ind the probability of getting a total more than 7.

**Solution**:

- Here, total number of posiible outcomes = 6 x 6 = 36.

Let E = Event of getting a total more than 7

= {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

Therefore, P(E) = \(\frac{15}{36}\) = \(\frac{5}{12}\).

**Compound probability**:

Compound probability is when the problem statement asks for the likelihood of the occurrence of more than one outcome.

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

Where \(A\) and \(B\) are any two events.

\(P(A \ or \ B)\) is the probability of the occurrence of atleast one of the events.

\(P(A \ and \ B)\) is the probability of the occurrence of both \(A\) and \(B\) at the same time.

**Mutually exclusive events**:

When two events cannot occur at the same time, they are considered mutually exclusive.

For a mutually exclusive event, P(A and B) = 0.

**Example 1**:

Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

**Solution**:

- We need to find out \(P(B \ or \ 6)\)

Probability of selecting a black card = \(\frac{26}{52}\)

Probability of selecting a 6 = \(\frac{4}{52}\)

Probability of selecting both a black card and a 6 = \(\frac{2}{52}\)

\(P(B \ or \ 6)\) = \(P(B)\) + \(P(6)\) – \(P(B \ and \ 6)\)

**Example 2**:

What is the probability of getting a 2 or a 5 when a die is rolled?

**Solution**:

- Taking the individual probabilities of each number, getting a \(2\) is \(\frac{1}{6}\) and so is getting a \(5\).

Applying the formula of compound probability,

Probability of getting a \(2\) or a \(5\),

\(P(2 \ or \ 5)\) = \(P(2)\) + \(P(5)\) – \(P(2 \ and \ 5)\)

=> \(\frac{1}{6}\) + \(\frac{1}{6}\) – 0

=> \(\frac{2}{6}\) = \(\frac{1}{3}\).

**Conditional probability**:

Conditional probability is calculating the probability of an event given that another event has already occured.

The formula for conditional probability \(P(A|B)\), read as \(P(A \ given \ B)\) is:

- \(P(A|B)\) =\(\frac{P(A \ and \ B)}{P(B)}\)

**Example**:

In a class, \(40%\) of the students study math and science. \(60%\) of the students study math. What is the probability of a student studying science given he/she is already studying math?

**Solution**:

- \(P(M \ and \ S)\) = 0.40

\(P(M)\) = 0.60

\(P(S|M)\) = \(\frac{P(M \ and \ S)}{P(S)}\) = \(\frac{0.40}{0.60}\) = \(\frac{2}{3}\) = 0.67

**Complement of an event**:

A complement of an event \(A\) can be stated as that which does NOT contain the occurrence of \(A\).

A complement of an event is denoted as \(P(A^{c})\) or \(P(A’)\).

- \(P(A^{c})\) = 1 – \(P(A)\)

or it can be stated, \(P(A)\) + \(P(A^{c})\) = 1

**Example**:

A single coin is tossed 5 times. What is the probability of getting at least one head?

**Solution**:

- Consider solving this using complement.

Probability of getting no head = \(P\)(all tails) = \(\frac{1}{32}\)

\(P\)(at least one head) = 1 – \(P\)(all tails) = 1 – \(\frac{1}{32}\) = \(\frac{31}{32}\)

- \(P(E_{1} \cup E_{2}) \leq (E_{1}) + P(E_{2})\)

**2.** If \(E_{1}, E_{2}, E_{3}, … E_{n}\) be n independent events with respective probabilities \(P_{1}, P_{2}, …..P_{n}\), then \(P\)(at least one of \(E_{1}, E_{2}, E_{3}, … E_{n}\) occurs).

- = \(P(E_{1} \cup E_{2} \cup E_{3} \cup ….. \cup E_{n}\))

= 1 – \(P({E_{1}}^{‘})P({E_{2}^{‘}})……P({E_{n}^{‘}})\)

= 1 – (1 – \(P_{1}\))(1 – \(P_{2}\))……(1 – \(P_{n}\))

**3.** If \(E_{1}, E_{2}, E_{3}\) be three events associated with an experiment, then

- \(P(E_{1} \cup E_{2} \cup E_{3})\) = \(P(E_{1}\)) + \(P(E_{2}\)) + \(P(E_{3}\)) – \(P(E_{1} \cap E_{2})\) – \(P(E_{2} \cap E_{3})\) – \(P(E_{3} \cap E_{1}) + \) + \(P(E_{1} \cap E_{2} \cap E_{3})\)

**4.** If \(E_{1}\) and \(E_{2}\) are independent events, then

- (a) \(P(E_{1} and E_{2}\) occurs) = \(P(E_{1} E_{2}\))

(b) \(P(E_{1} and E_{2}\) occurs)

= \(P\)(at least one of \(E_{1} and E_{2}\) occurs)

= \(P(E_{1}) + P(E_{2}) – P(E_{1} \cap E_{2})\)

= \(P(E_{1}) + P(E_{2}) – P(E_{1}) \cup P(E_{2})\)

(c) \(P(neither E_{1} nor E_{2} \) occurs)

= \(P{(E_{1} \cup E_{2}})^{‘}\) = \(P({E_{1}}^{‘} \cap {E_{2}}^{‘}\)) = \(P({E_{1}}^{‘})P({E_{2}}^{‘})\)

= (1 – \(P(E_{1})\))(1 – \(P(E_{2})\))

**5.** \(P(E)\) = \(\frac{n(E)}{n(S)}\) = \(\frac{Number \ of \ cases favourable \ to \ event \ E}{Total \ number \ of \ cases}\)

- Clearly, \( 0 \leq P(E) \leq 1 \)

\(P(E)\) = 0

⇒ \(E\) = \(\phi\) and \(P(E)\) = 1, \(E\) = \(S\)

- Here, n(s) = 6 x 6 = 36

Let E =event of getting a total more than 7

= {(2,6)(3,5)(3,6)(4,4)(4,5)(4,6)(5,3)(5,4)(5,5)(5,6)(6,2)(6,3)(6,4)(6,5)(6,6)}

Therefore, P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\)

**2. The probability that at least one of the events S and T occurs is 0.6. If S and T occur simultaneously with probability 0.2, then find \(P\bar{A} + P\bar{B}\)?**

**Solution**:

- Given,

\(P(A \cup B\)) = 0.6,

\(P(A \cap B\)) = 0.2

using the fact that,

\(P(A \cup B)\) = \(P(A) + P(B) – P(A \cap B)\)

⇒ 0.6 = [1 – \(P(\bar{A})\)] + [1 – \(P(\bar{B})\)] – 0.2

⇒ \(P(\bar{A})\) + \(P(\bar{B})\) = 1.2

**3. At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 -minute time interval. The probability that there is at the most one phone call during a 10 – minute time period is?**

**Solution**:

- Given,

P(X = 0) + P(X = 1)

= \(\frac{e^{-5}}{0!} * 5^0 + \frac{e^{-5}}{1!} * 5^1\)

= \(\frac{6}{e^{5}}\)

Therefore, the probability that there is at the most one phone call during a 10 – minute time period is \(\frac{6}{e^{5}}\)

**4. A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that are of same colour?**

**Solution**:

- Let S be the sample space. Then,

n(S) = Number of ways of drawing 2 balls out of (6 + 4) = \(10_{{c}_{2}}\) =\(\frac{10 * 9}{2 * 1}\)

Let E = Event of getting both balls of the same colour. Then,

n(E) = Number of ways of drawing (2 balls out of 6) or (2 balls out of 4)

= \(6_{{c}_{2}} + 4_{{c}_{2}}\)

= \(\frac{6 * 5}{2 * 1}\)

= 15 + 6

= 21.

P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{21}{45}\) = \(\frac{7}{15}\)

**5. Two unbiased coins are tossed. What is the probability of getting at most one head?**

**Solution**:

- Given,

S = {HH, HT, TH, TT}

Let E = event of getting at most one head.

Therefore, E = {TT, HT, TH}

Therefore, P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{4}\)