Probability
1.0 Basic Definitions
1.0 Basic Definitions
- Random Experiment
An experiment whose all possible outcomes are known in advance, but the result of any particular performance cannot be predicted before the completion of the experiment.
In other words, an experiment when repeated under identical conditions, do not produce the same outcome every time, but belongs to the set of all possible outcomes, is called a random experiment.
Illustration 1. Select the random experiments from the given set of experiments.
a. Tossing a fair coin.
b. Taking specific medicine for aches
c. Testing kinetics of hydrolysis of ester using alkali
d. Drawing a card from a pack of $52$ cards
e. Selecting a student for extempore by drawing lots
f. Jackpot lottery.
Solution:
| Experiment | Classification and Explanation |
| a.Tossing a fair coin. | Its a random experiment. A fair coin may either turn up head or tail on tossing.The result is known only after tossing the coin. And hence this goes well with the definition of a random experiment. |
| b.Testing specific medicine for aches on acidity | Its not a random experiment. Here, one knows for clear that the medicine which act specific to aches, do not have chemicals to cure the acidity. The result is that it does not work. As the result is known earlier, it does not fall under random experiments. |
| c.Testing kinetics of hydrolysis of ester using alkali | Its not a random experiment. Hydrolysis of ester may take first or second order depending upon the conditions. In that case, it will be random experiment. But since the condition is specified, hydrolysis of ester by alkali takes place only as a second order kinetics. So even before testing for the kinetics, the result of it being a second order kinetics is known. Hence it is not a random experiment. |
| d. Drawing a card from a pack of $52$ cards | It is a random experiment. There are $52$ different cards, out of which any one of these may turn up. We can not be sure of which card will show up. This can be decided only after we draw a card and see it. Hence it is a random experiment. |
| e. Selecting a student for extempore by drawing lots | It is a random experiment. Always, drawing lots is a random experiment, as the result is always unknown before drawing and seeing it. Any student's name can turn up on the chit. |
| f. Jackpot lottery | It is a random experiment. In a jackpot lottery, a person is declared winner by drawing lots. One is unaware of which number would show up before performing the experiment. Hence it is a random experiment. |
- Sample Space
The set of all possible outcomes of an experiment is said to be the sample space of the experiment. It is usually denoted by $S$.
For example, in tossing two coins, the sample space $S$ is {head-head , head-tail , tail-head , tail-tail}.
Illustration 2. Specify the sample space for the following experiments
a. Rolling a die once
b. Rolling two dice
c. Tossing a coin thrice.
d. Tossing a coin and a die together.
e. Forming two letter words from the vowels alone
Solution:
| Experiment | Sample space and Explanation |
| a. Rolling a die once | $S = \{1,2,3,4,5,6\}$ There are $6$ faces in a die. They are numbered from $1$ to $6$. Hence the sample space is ranged as above. |
| b. Rolling two dice | Each die has $6$ faces, i.e. $6$ numbers. Each number of the first die can have $6$ possible corresponding numbers showing up in the second die. Hence there are totally, $6 \times 6 = 36$ possible outcomes. Thus there are $36$ total possible outcomes. $S = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),$ $(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1),$ $(5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} $ |
| c. Tossing a coin thrice. | There are $3$ slots. They can be filled with $H$ or $T$. The first slot can be filled in two ways. The second slot can be filled in two ways. Similarly for the third. Hence $2 \times 2 \times 2 = {2^3} = 8$ ways. Thus there are $8$ total possible outcomes. $S = \{ (HHH),(HHT),(HTH),(THH),(HTT),(TTH),(THT),(TTT)\} $ |
| d.Tossing a coin and a die together | On tossing a coin, either head shows up or tail. If head shows up, the die can show $6$ possible numbers. Same is applicable for a tail showing up. Hence, there are $12$ ways. Thus there are $12$ total possible outcomes. $S = \{ (H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\} $ |
| e. Forming two letter words from the vowels alone. | There are total of $5$ vowels. There are $2$ places to be filled. Each place can be filled in $5$ ways, hence there are ${5^2} = 25$ ways of filling. Thus there are $25$ total possible outcomes. $S = \{ (AA), (AE), (AI), (AO), (AU), (EA), (EE), (EI), (EO), (EU),$ $(IA), (IE), (II), (IO), (IU), (OA), (OE), (OI), (OO), (OU),$ $(UA), (UE), (UI), (UO), (UU)\} $ |
- Event
Any subset of the sample space is called an event. If $E$ is an event, then $n(E)$ is the number of outcomes possible for that event.
For example, in throwing a die, sample space is $\{ 1,2,3,4,5,6\} $ , and occurrence of $1$ , $2$ , $3$ etc. are said to be events.
Illustration 3. Find the size of the event for the given experiments.
a. Rolling a die and getting a number less than $5$.
b. Getting the sum of numbers to be odd in rolling two dice.
c. Getting number cards from a pack of cards.
d. Getting at least two heads in tossing a coin four times.
e. Getting the product of the numbers as $14$ in rolling a die twice.
Solution:
| Event | Size of the event and Explanation |
| a. Rolling a die and getting a number less than $5$. | $E = \{1,2,3,4\}$ There are $4$ possible outcomes. Thus $n(E) =4$ |
| b. Getting the sum of numbers to be odd in rolling two dice. | To get the sum to be odd, the combination of the numbers should be even-odd or odd-even. $E = \{ (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5), \} $ There are $18$ such combinations. Thus $n(E) = 18$ |
| c. Getting number cards from a pack of cards. | The number cards from a pack range from $2$ to $10$. There are $9$ such cards for each flower. There are four flowers. Hence there are $9 \times 4 = 36$ outcomes for the event. Thus $n(E) = 36$ |
| d. Getting at least two heads in tossing a coin four times. | On tossing a coin four times, there are sixteen total outcomes. From this the outcomes pertaining to the given event are, $E = \{ (HHHH),(HHHT),(HHTH),(HTHH),(HHTT),(TTHH),(THHT),(THHH),(HTHT),(THTH),(HTTH)\} $ Thus $n(E) = 11$ |
| e. Getting the product of the numbers as $14$ in rolling a fair die twice. | On factorizing $14$, we get $1 \times 14$ and $2 \times 7$. A die has only numbers ranging from $1$ to $6$. Thus no outcome from the sample space satisfy the equation. Thus, $E$ is a null set. $E = \emptyset $ $n(E) = 0$ |
- Equally likely events
In an experiment, two or more events are said to be equally likely, if they have the same chances associated with them, i.e. none of them has more chance of occurrence than the others.
For example, in rolling a die, each number has ${{1 \over 6}^{th}}$ chance of occurring.
Illustration 4. State whether the given set of events are equally likely or not.
a. Rolling a die once
Event A : Getting an odd number greater than one
Event B : Getting an even number greater than one
b. Tossing a coin thrice
Event A: Getting at least one head
Event B: Getting at the most two tails
c. From a pack of cards
Event A: Getting a red card.
Event B: Getting a black card
d. Single toss of die
Event A : Getting an even prime number
Event B: Getting an odd prime number
e. From the alphabets in a lot
Event A: Getting a vowel.
Event B: Getting a consonant.
Solution:
| Experiment | Classification and Explanation |
| a. Rolling a die once Event A : Getting an odd number greater than one Event B : Getting an even number greater than one | $S = \{1,2,3,4,5,6\}$ Event $A$ = Getting an odd number greater than one = $\{ 3,5\} $ Thus $n(A) = 2$ Event $B$ = Getting an even number greater than one = $\{ 2,4,6\} $ Thus $n(B) = 3$ Since, $n(A) \ne n(B)$, the events are not equally likely. |
| b. Tossing a coin thrice Event A: Getting at least one head Event B: Getting at the most two tails | $S = \{ (HHH),(HHT),(HTH),(THH),(HTT),(TTH),(THT),(TTT)\} $ Event A: Getting at least one head = $ \{ (HHH),(HHT),(HTH),(THH),(HTT),(TTH),(THT)\} $ At least one head would imply all three cannot be tails. Thus $n(A) = 7$ Event B: Getting at the most two tails =$ \{ (HHH),(HHT),(HTH),(THH),(HTT),(TTH),(THT)\} $ This would imply all three cannot be tails, in other words there can be one, two or no tails. Thus $n(B) = 7$ Since, $n(A) = n(B)$ , the events are equally likely |
| c. From a pack of cards Event A: Getting a red card. Event B: Getting a black card | A pack of cards has four flowers namely, hearts, clubs, diamonds and spades. Out of which hearts and diamonds are red, and clubs and spade are black. There are $13$ cards of each flower. Event A: Getting a red card, $ \Rightarrow n(A) = 26$ Event B: Getting a black card, $ \Rightarrow n(B) = 26$ Since, $n(A) = n(B)$ , the events are equally likely |
| d. Single toss of die, Event A: Getting an even prime number Event B: Getting an odd prime number | $S = \{1,2,3,4,5,6\}$ Event A : Getting an even prime number = $\{ 2\} $ $n(A) = 1$ Event B: Getting an odd prime number = $\{ 3,5\} $ $n(B) = 2$ Since, $n(A) \ne n(B)$, the events are not equally likely. |
| e. From the alphabets in a lot Event A: Getting a vowel. Event B: Getting a consonant. | There are $26$ alphabets, out of which $5$ are vowels, and $21$ consonants. Event A: Getting a vowel $n(A) = 5$ Event B: Getting a consonant $n(B) = 26$ Since, $n(A) \ne n(B)$, the events are not equally likely. |
- Mutually Exclusive events
In an experiment, two or more events are said to be mutually exclusive, if the occurrence of one of them rules out probability of the occurrence of other events, i.e. they cannot occur simultaneously.
For example, on tossing a coin, when head occurs, occurrence of tail is ruled out.
Illustration 5. Sort the following events as mutually exclusive and mutually non-exclusive.
a. Rolling a die once,and getting an odd number and a number less than $5$.
b. Getting an even or odd number in a single throw of die.
c.Tossing a coin once, and getting a head and a tail.
d. Getting Ace and a card of hearts from a pack of cards in a single draw.
e. Getting a card of diamonds and a card of spade in a single draw.
Solution:
| Event | Classification and Explanation |
| a. Rolling a die once, and getting an odd number and a number less than $5$ | Here $1$,$3$ are odd and less than $5$. As an outcome satisfying both events can occur , they are mutually non-exclusive. |
| b. Getting an even or odd number in a single throw of die | In event of getting an even number, the possible outcomes are $2$,$4$,$6$ and the possible outcomes in getting odd number is $1$,$3$,$5$. There is not even a single common outcome. Hence they are mutually exclusive. |
| c. Tossing a coin once, and getting a head and a tail. | Clearly, getting a head, rules out the occurrence of a tail. Thus they are mutually exclusive events. |
| d. Getting Ace and a card of hearts from a pack of cards in a single draw | There are four flower types of Ace, i.e. Ace of hearts, Ace of diamond, Ace of spade and Ace of clubs. In the event of getting card of hearts there are $13$ possible outcomes out of which getting Ace is also one. Since there is an outcome in common to both the events they are mutually non-exclusive events. |
| e. Getting a card of diamonds and a card of spade in a single draw | Clearly occurrence of a diamond rules out occurrence of a spade and vice-versa. Thus they are mutually exclusive events. |
- Exhaustive set of events
A set of outcomes is said to be exhaustive if it covers each and every possible outcome of the sample space of the experiment, i.e. at least one of them has to occur in the performance of the experiment.
Illustration 6. Sort the given events as exhaustive and not exhaustive events.
a. Rolling a die once,and getting a number less than $7$.
b. Getting an even number in a single throw of die.
c. Getting a head or a tail in a single toss of coin.
Solution:
| Event | Classification and Explanation |
| a. Rolling a die once,and getting a number less than $7$. | The sample space is $S = \{1,2,3,4,5,6\}$. Since all the outcomes are less than $7$. This event is exhaustive. |
| b. Getting an even number in a single throw of die. | In event of getting an even number, the possible outcomes are $2$,$4$,$6$. But the sample space is $S = \{1,2,3,4,5,6\}$. Since the set of outcomes in not the complete sample space, the event is not exhaustive. |
| c. Getting a head or a tail in a single toss of coin. | Here there are two events. $A$ occurrence of head. $B$ occurrence of tail. There are only two total possible outcomes in the sample space. As happening of $A$ or $B$ would imply the total outcomes being covered. Hence, events $A$ and $B$ are set of exhaustive events. |
Note: Tossing a fair coin is equally likely, and mutually exclusive as occurrence of one rules out the other, and also exhaustive as either one has to occur.
