1.Experiment: An operation which can produce some well-defined outcomes is called an experiment
Types of Experiment:
While studying probability theory, we will frequently use the term ‘experiment' which means an operation which can produce well defined outcome(s).
There are two types of experiments:
(i) Deterministic Experiment: The experiments whose outcome is same when done under exact conditions are called Deterministic experiment. E.g. all experiments are done in chemistry lab.
(ii) Random Experiment: The experiments whose outcomes are more than 1 when done under exact conditions are called Random Experiment. E.g. if a coin is tossed we may get a head or a tail.
Events in Probability:
When we perform any experiment, there are some outcomes which are called events. Let us study the different types of events can occur.
Trial and Elementary Events: If we repeat a random experiment under exact conditions, it is known as trial and all the possible outcomes are known as elementary events. E.g. if we throw a dice it is called a trial and getting 1, 2, 3, 4, 5 or 6 is called elementary event.
Compound Event: When two or more elementary events are combined it is known as compound event. When we throw a dice, getting a prime number is compound event as we can get 2, 3, 5 and all are elementary.
Exhaustive Number of Cases: It is the total possible outcome. When we throw a dice total number of cases are 6. When we throw a pair of dice exhaustive number of cases is 36.
Mutually Exclusive Events: It means simultaneous occurrence is not possible. In case of tossing a coin, either head will come or tail will come. So, both are mutually exclusive events.
Equally Likely Cases: It means chances are equal. When we throw a dice, each outcome has equal chance. So it is case of equally likely.
Total Number of Cases: As the name suggests, the total number of elementary events of a trial are known as total number of cases.
Favorable Events: Desired outcome of an elementary event is called Favorable event. E.g. when we throw a dice and it is asked that what is the probability of getting a multiple of 3? In this case favorable cases are 2 (3 and 6) and total cases are obviously 6.
Independent Events: Two events are called independent if outcome of one event is not affecting the outcome of other. If we toss a coin and throw a dice then outcome of coin is independent of outcome of coin, both are independent events.
Let us go through the Probability Formulas:
Probability in simple language is defined as ratio of favorable cases to the total number of cases.
Probability of happening of any event P(A) = fav. number of cases / Total no. of cases = n/N
If p is the probability of happening of an event A, then the probability of not happening of that event is P(Ā) = 1- p
Probability Equations: P (A) ≤ 1, P(A) + P(Ā) = 1.
Addition theorem: P(X or Y) = P(X) + P(Y) – P (X∩Y)
or P(X⋃Y) = P(X) + P(Y) – P(X∩Y)
Mutually exclusive events: Two events are mutually exclusive if they cannot occur simultaneously. For n mutually exclusive events, the probability is the sum of all probabilities of these events:
p = p1 + p2 + ... + p (n-1) + p (n)
or
P (A or B) = P (A) + P (B) where A and B denote mutually exclusive events.
Independent events: Two events are independent if the occurrence of one event does not influence the occurrence of other events. Therefore, for n independent events, the probability is the product of all probabilities of independent events:
p = p1 x p2 x ... x p (n-1) x p (n)
or P(X and Y) = P(X) x P(Y) , where X and Y denote independent events
Odds in favor of certain event = No. of successes: No. of failures
Odds against of an event = No. of failures: No. of successes
For solving the questions on probability, you are advised to revise the major probability formulas, go through 20 to 25 probability examples & solutions and solve around 100 probability sums. After doing this, you will feel confident to solve probability problems on your own.
2. Random Experiment: An experiment in which all possible outcomes are known and the exact output be predicted in advance, is called a random experiment.
Examples of Performing a Random Experiment
· Rolling an unbiased dice
· Tossing a fair coin
· Drawing a card from a pack of well-shuffled cards
· Picking up a ball of certain colour from bag containing balls of different colours
Details :
· When we throw a coin, then either a Head (H) or a Tail (T) appears.
· A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively, When we throw a die, the outcome is the number that appears on its upper face.
· A pack of cards has 52 cards. It has 13 cards of each suit, namely Spades, Clubs, Hearts and Diamonds.
Cards of spades and clubs are black cards.
Cards of hearts and diamonds are red cards.
There are 4 honours of each suit
These are Aces, Kings, Queens and Jacks.
These are called face cards.
3. Sample Space: When we perform an experiment, then the sets of all possible outcomes is called the Sample Space.
Examples of Sample Spaces:
· In tossing a coin, S = [H, T].
· If two coins are tossed, then S = (HH, HT, TH, TI).
· in In rolling a dice, we have, S = [1, 2, 3, 4, 5, 6].
4. Event: Any subset of a sample space is called an event.
5. Probability of Occurrence of an Event:
Let S be the sample space and let E be an event. Then, E ⊆S.
P(E)=n(E)/n(S)
P(E)=Number of out comes/Total out comes
6. Results on Probability:
· P(S) = 1
· 0<=P(E)<=1
· P(ϕ)=0
· For any events A and B, we have:
P(A⋃B )= P (A) + P (B) - P ( A⋂ B )
· If A bar denotes (not-A), then P(A bar) = 1-P (A).
Shortcut way to remember probability of sum outcomes when die is rolled:
WHEN TWO DIES ARE ROLLED:
|
Number of Outcomes |
Probability of sum of outcomes |
|
2 |
1/36 |
|
3 |
2/36 |
|
4 |
3/36 |
|
5 |
4/36 |
|
6 |
5/36 |
|
7 |
6/36 |
|
8 |
5/36 |
|
9 |
4/36 |
|
10 |
3/36 |
|
11 |
2/36 |
|
12 |
1/36 |
WHEN THEREE DIES ARE ROLLED:
|
Number of Outcomes |
Probability of sum of outcomes |
|
3 |
1/216 |
|
4 |
3/216 |
|
5 |
6/216 |
|
6 |
10/216 |
|
7 |
15/216 |
|
8 |
21/216 |
|
9 |
25/216 |
|
10 |
27/216 |
|
11 |
27/216 |
|
12 |
25/216 |
|
13 |
21/216 |
|
14 |
15/216 |
|
15 |
10/216 |
|
16 |
6/216 |
|
17 |
3/216 |
|
18 |
1/216 |
Directions: Go through the following examples of probability. The examples cover variety of probability problems
Problems on Probability with solutions:
Example 1: A coin is thrown 3 times .what is the probability that atleast one head is obtained?
Sol: Sample space = [HHH, HHT, HTH, THH, TTH, THT, HTT, TTT]
Total number of ways = 2 × 2 × 2 = 8. Fav. Cases = 7
P (A) = 7/8
OR
P (of getting at least one head) = 1 – P (no head)⇒ 1 – (1/8) = 7/8
Example 2: Find the probability of getting a numbered card when a card is drawn from the pack of 52 cards.
Sol: Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36
P (E) = 36/52 = 9/13
Example 3: There are 5 green 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red.
Sol: P (G) × P (R) = (5/12) x (7/11) = 35/132
Example 4: What is the probability of getting a sum of 7 when two dice are thrown?
Sol: Probability math - Total number of ways = 6 × 6 = 36 ways. Favorable cases = (1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3) --- 6 ways. P (A) = 6/36 = 1/6
Example 5: 1 card is drawn at random from the pack of 52 cards.
(i) Find the Probability that it is an honor card.
(ii) It is a face card.
Sol: (i) honor cards = (A, J, Q, K) 4 cards from each suits = 4 × 4 = 16
P (honor card) = 16/52 = 4/13
(ii) face cards = (J,Q,K) 3 cards from each suit = 3 × 4 = 12 Cards.
P (face Card) = 12/52 = 3/13
Example 6: Two cards are drawn from the pack of 52 cards. Find the probability that both are diamonds or both are kings.
Sol: Total no. of ways = 52C2
Case I: Both are diamonds = 13C2
Case II: Both are kings = 4C2
P (both are diamonds or both are kings) = (13C2 + 4C2 ) / 52C2
Example 7: Three dice are rolled together. What is the probability as getting at least one '4'?
Sol: Total number of ways = 6 × 6 × 6 = 216. Probability of getting number ‘4’ at least one time
= 1 – (Probability of getting no number 4) = 1 – (5/6) x (5/6) x (5/6) = 91/216
Example 8: A problem is given to three persons P, Q, R whose respective chances of solving it are 2/7, 4/7, 4/9 respectively. What is the probability that the problem is solved?
Sol: Probability of the problem getting solved = 1 – (Probability of none of them solving the problem)
Probability of problem getting solved = 1 – (5/7) x (3/7) x (5/9) = (122/147)
Example 9: Find the probability of getting two heads when five coins are tossed.
Sol: Number of ways of getting two heads = 5C2 = 10. Total Number of ways = 25 = 32
P (two heads) = 10/32 = 5/16
Example 10: What is the probability of getting a sum of 22 or more when four dice are thrown?
Sol: Total number of ways = 64 = 1296. Number of ways of getting a sum 22 are 6,6,6,4 = 4! / 3! = 4
6,6,5,5 = 4! / 2!2! = 6. Number of ways of getting a sum 23 is 6,6,6,5 = 4! / 3! = 4.
Number of ways of getting a sum 24 is 6,6,6,6 = 1.
Fav. Number of cases = 4 + 6 + 4 + 1 = 15 ways. P (getting a sum of 22 or more) = 15/1296 = 5/432
Example 11: Two dice are thrown together. What is the probability that the number obtained on one of the dice is multiple of number obtained on the other dice?
Sol:Total number of cases = 62 = 36
Since the number on a die should be multiple of the other, the possibilities are
(1, 1) (2, 2) (3, 3) ------ (6, 6) --- 6 ways
(2, 1) (1, 2) (1, 4) (4, 1) (1, 3) (3, 1) (1, 5) (5, 1) (6, 1) (1, 6) --- 10 ways
(2, 4) (4, 2) (2, 6) (6, 2) (3, 6) (6, 3) -- 6 ways
Favorable cases are = 6 + 10 + 6 = 22. So, P (A) = 22/36 = 11/18
Example 12: From a pack of cards, three cards are drawn at random. Find the probability that each card is from different suit.
Sol: Total number of cases = 52C3
One card each should be selected from a different suit. The three suits can be chosen in 4C3 was
The cards can be selected in a total of (4C3) x (13C1) x (13C1) x (13C1)
Probability = 4C3 x (13C1)3 / 52C3
= 4 x (13)3 / 52C3
Example 13: Find the probability that a leap year has 52 Sundays.
Sol: A leap year can have 52 Sundays or 53 Sundays. In a leap year, there are 366 days out of which there are 52 complete weeks & remaining 2 days. Now, these two days can be (Sat, Sun) (Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed, Thur) (Thur, Friday) (Friday, Sat).
So there are total 7 cases out of which (Sat, Sun) (Sun, Mon) are two favorable cases. So, P (53 Sundays) = 2 / 7
Now, P(52 Sundays) + P(53 Sundays) = 1
So, P (52 Sundays) = 1 - P(53 Sundays) = 1 – (2/7) = (5/7)
Example 14: Fifteen people sit around a circular table. What are odds against two particular people sitting together?
Sol: 15 persons can be seated in 14! Ways. No. of ways in which two particular people sit together is 13! × 2!
The probability of two particular persons sitting together 13!2! / 14! = 1/7
Odds against the event = 6 : 1
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