Probability Problems, Shortcut Tricks and Examples

Probability In an experiment if ‘ n ‘ is the number of exhaustive cases and ‘m’ is the number of favourable cases of an event A. Then the probability of event A is denoted by P(A).

P(A) = $\frac{Number\; of\; Favourable\; Cases}{Number \; of \; Exhaustive\; Cases\; }\: =\:\;\frac{m}{n}\: =\,: \frac{n(A)}{n(S)}$

Probability Theory

Random Experiment : An experiment in which all possible out comes are known and the exact output cannot be predicated in advance is called Random Experiment.
EXAMPLE :
(i) Tossing a fair coin
(ii) Rolling an unbiased dice
(iii) Drawing a card from a pack of well shuffled cards
Trail : Conducting a Random Experiment once is known as a Trail
Outcome : The result of a Trail in the random experiment is called as Outcome
EXAMPLE : In tossing a single coin outcomes are H and T
Sample Space : A set of all possible outcomes of a random experiments is known as Sample Space
EXAMPLE :
( i ) In the experiment of tossing a coin the sample space S H,T
( ii ) If Two coins are tossed then S HH,HT,TH,TT
( iii ) In throwing a die S 1,2,3,4,5,6
Event : Any non-empty subset of a Sample Space is called an Event
EXAMPLE :
( i ). in Tossing a single coin getting Head or Tail is an Event
( ii )Getting an Ace (or) Diamond from a pack of 52 cards is an event
Exhaustive Events : The total number of possible outcomes of an experiment is known as Exhaustive Events
EXAMPLE : In the experiment of throwing a die the total number of possible outcomes = 6
Favourable Events : The number of event which favour the happing of the events are known as Favourable Cases (or) Events
EXAMPLE : In tossing two dice the number of cases favourable to getting the sum 3 is (2, 1), (1, 2) i.e., 2
Mutually Exclusive Events : If two events have no common outcomes then they are called Mutually Exclusive
EXAMPLE: In tossing a coin the events Head & Tail are mutually exclusive because Head & Tail cannot happen at the same time.
Independent Events : Two events are said to be independent if the happening of one event does not affect the happening of the other.
Dependent Events : Two events are said to be dependent if the happening of an event will affect the happening of the other event .
EXAMPLE: If we draw a card from a pack of 52 cards and replace it before we draw a second card. The second card is independent of first one. If we don’t replace the first card before the second draw. The second draw depends on the first one.
Axioms of Probability : Let S be the sample space and A be the event i.e., $A\: \sqsubseteq \: S$
1. $0\: \leq \: P(A)\: \leq \: 1\: ,\forall \: A\: \sqsubseteq \: S$
2. P(S) = 1
3. If A and B are mutually exclusive ( Disjoint i.e., $A\: \cap \: B\: =\: \Phi$ )
$P(A\cup B)$ = P(A)+P(B)
Results of Probability :
1. for any two events A & B , $P(A\cup B)\ =P(A)+P(B)-P(A\cap B)$
2. If $\bar{A}$ denotes compliment of event A then , $P\bar{A}$ = 1-P(A)
3. For any two events A & B , $P(\bar{A}\cap B)=P(B)-P(A\cap B)$
4. For any three events A, B & C , $(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$

Probability Examples and Probability Solutions

Question 1: A die is rolled, find the probability that an even number is obtained.
Solution : Let us first write the sample space S of the experiment.
S = {1,2,3,4,5,6}
Let E be the event "an even number is obtained" and write it down.
E = {2,4,6}
We now use the formula of the classical probability.
P(E) = $\frac{n(E)}{n(S)}=\frac{3}{6}\; =\; \frac{1}{2}$

Question 2: Two coins are tossed, find the probability that two heads are obtained.
Note: Each coin has two possible outcomes H (heads) and T (Tails).

Solution: The sample space S is given by.
S = {(H,T),(H,H),(T,H),(T,T)}
Let E be the event "two heads are obtained".
E = {(H,H)}
We use the formula of the classical probability.
P(E) = $\frac{n(E)}{n(S}\; =\; \frac{1}{4}$

Question 3: You toss a coin AND roll a die. What is the probability of getting a tail and a on the die?
Solution 3: Probability of getting a tail when a single coin is tossed = $\frac{1}{2}$
Probability of getting 4 when a die is thrown = $\frac{1}{6}$
Required probability = $\frac{1}{2}\times \frac{1}{6}\; =\: \frac{1}{12}$

Question 4: A number is selected from the numbers 1, 2, 3 and then a second number y is randomly selected from the numbers 1, 4, 9. What is the probability that the product xy of the two numbers will be less than 9 ways
Solution 4: Number X can be selected in three ways and corresponding to each such way there are three ways of selecting number y.
Therefore, two numbers can be selected in 9 ways.
The favourable number of elementary events for which the product xy of the two numbers will less than 9 = (1,1),(1,4),(2,1),(2,4),(3,1) = 5
Hence the required probability = $\frac{5}{9}$

Question 5: Find the probability of having 53 Sundays in
( i ) a leap year (ii) a non leap year
Solution 5: An ordinary year has 365 days i.e., 52 weeks and 1 odd day.
This day can be any one of the 7 days of the week,
$\therefore$ P(that this day is Sunday) = $\frac{1}{7}$
Hence, $\therefore$ P(an Ordinary Year has 53 Sundays) = $\frac{1}{7}$
A leap year has 366 days. i.e., 52 weeks and 2 odd days
This day can be any one of the 7 days of the week
$\therefore$ P(that this day is Sunday) = $\frac{2}{7}$
Hence $\therefore$ P(a leap Year has 53 Sundays) = $\frac{2}{7}$

Question 6: Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black and both are queen.
Solution 6: Let S be the Sample space. Then
n(S) = Number of ways of drawing 2 balls out of (6+4) = $10_{}^{C}\textrm{2}\; =\: \frac{10\times 9}{2\times 1}$
Let E = Event of getting both balls of the same colour . Then
N(E) = Number of ways of drawing ( 2 balls out 6 )or ( 2 balls out of 4)
= $(6_{}^{C}\textrm{2}+4_{}^{C}\textrm{2})\: =\: \frac{6\times 5}{2\times 1}$
P(E) = $\frac{n(E)}{n(S)}\; =\: \frac{21}{45}\; =\; \frac{7}{5}$