Probability

Probability

Probability

The dictionary meaning of the word ‘probability’ is ‘likely though not certain to occur’. Thus when a coin is tossed a head is likely to occur but it may not occur. Here, the probability of occurrence of head or tail is ½ and it is also called 50/50 chance.

Probability is a concept which numerically measures the degree of certainty and therefore of uncertainty of the occurrence of events.

 

Definitions of basic terms on probability

There are some basic terms which are frequently used in probability. It is essentially important to have the proper concept of these terms.

1. Random experiments:

Any experiment whose outcome cannot be predicated or determined in advance is called a random experiment. Tossing a coin, rolling a dice, drawing a card from a well shuffled pack of 52 playing cards etc. are a few examples of random experiments.

 

2. Outcomes:

The results of a random experiment are called outcomes or events. For example, while tossing a coin, the occurrence of head or tail is the outcome.

 

3. Sample space:

Each performance in a random experiment is called a trial and an outcome of a trial is called a sample point. The set of all possible outcomes (i.e. sample points) of a random experiment is known as sample space. Usually a sample space is denoted by S. For example,

i.       The possible outcomes of a random experiment of throwing a dice are 1, 2, 3, 4, 5 or 6.

The sample space, S = {1, 2, 3, 4, 5, 6}

ii.      The possible outcomes of a random experiment of tossing a coin are head (H) or tail (T).

The sample space, S = {H, T}

iii.     When two coins are tossed simultaneously, the sample space, S = {HH, HT, TH, TT}

 

4. Event:

A sample space S of a random experiment is a universal set. Every non-empty subset of the sample space S is called an event. For example, while tossing a coin, the sample space, S = {H, T}.

Here, the subsets {H}, {T} and {H, T} are the events of S. The empty set φ is also an event of S, but it is the ‘impossible event’. S is called the ‘sure event’. An event containing only one element of S is called a simple or elementary event. For example,

If S = {H, T}, then {H} and {T} are the elementary events.

If S = {1, 2, 3, 4, 5, 6}, then {1}, {2}, {3}, {4}, {5} and {6} are the elementary events.

 

5. Exhaustive cases:

The total number of all possible outcomes of a random experiment is known as exhaustive cases. For example,

While tossing a coin, S = {H, T}

So, exhaustive cases = 2

While tossing two coins simultaneously, S = {HH, HT, TH, TT}

So, exhaustive cases = 4

While throwing a dice, S = {1, 2, 3, 4, 5, 6}

So, exhaustive cases = 6

 

6. Favourable and unfavourable cases:

The outcomes in an random experiment which are desirable (expected) to us are called favourable cases and all ther cases are unfavourable cases. For example:

While tossing a coin, S = {H, T}

Here, the favourable number of cases of head is 1 and tail is also 1.

While tossing two coins, S = {HH, HT, TH, TT}

The favourable number of cases of both of them head is 1 and tail is also 1. The favourable number of cases of at least one head is 3 and at least one tail is also 3.

 

7. Equally likely events:

Two or more events are said to be equally likely if the chance of occurring any one event is equal to the chance of occurring other cases. For example, while throwing a dice, the chance of coming up the numbers 1 to 6 is equal. Similarly, while tossing a coin, head (H) or tail (T) are equally likely events.

 

8. Independent and dependent events:

Two or more events are said to be independent if the occurrence of one of the events does not affect the occurrence of other events. For example, in the random experiment of tossing a coin twice or more, the occurrence of any one event in the first trial does not affect the occurrence of any event in the second trial.

On the other hand, two or more events are said to be dependent if the occurrence of one of the events affects the occurrence of the other events.

For example, while drawing a ball in two successive trials from a bag containing 2 red and 3 green without a replacement, getting any one coloured ball in the first trial affects to draw another ball in the second trial. So these are the dependent events.

 

9. Mutually exclusive events:

Two or more events of a sample space S are said to be mutually exclusive if the occurrence of any one event excludes the occurrence of the other events.

For example, while tossing a coin, the occurrence of head excludes the occurrence of tail or vice versa. So they are mutually exclusive events.

Furthermore, consider the experiment of throwing a dice. Let A be the event, ‘the number obtained is less than 5’. Then, A = {1, 2, 3, 4}. Again, B be the event, ‘the number obtained is at least 5’. Then, B = {5, 6}. Here AB = φ.

Mutually exclusive events

Thus, the joint occurrence of A and B is an impossible event. In this case the events A and B are called mutually exclusive events. In general, if A and B are any two events on a sample space S and AB = φ, the events A and B are said to be mutually exclusive.

 

10. Non-mutually exclusive events:

If A and B are any two events on a sample space S and AB ≠ φ, the events are said to be non-mutually exclusive events. 

Non-mutually exclusive events

For example, let’s consider the experiment of throwing a dice. Let A be the event ‘the number obtained is even’ and B be the event ‘the number obtained is less than 4’. Then, A = {2, 4, 6} and B = {1, 2, 3} and AB = {2}. It means, ‘an even number from 1 to 6’ and ‘a number less than 4’ can occur simultaneously in a single throw of a dice. So A and B are said to be non-mutually exclusive events.

 

Some useful terms related to playing cards, coin and dice:

Playing cards:

·         There are 52 cards in a packed of playing cards.

·         There are 26 red and 26 black coloured cards in a packet.

·         13 Hearts () and 13 diamonds () are the red coloured cards.

·         13 Spade () and 13 clubs () are the black coloured cards.

·         There are 1/1 heart and diamond ace and 1/1 spade and club ace in a packet of playing cards.

·         There are 12 face cards in the packet. Among them 4 are Jacks (heart, diamond, spade and club), 4 are Queens and 4 are Kings.

Coin:

·         There are two faces of a coin, head (H) and tail (T).

Dice:

·         There are six faces in a dice which are numbered from 1 to 6.

 

Theoretical probability of an event:

Let n(S) be the total number of all possible outcomes (Exhaustive number of cases) of an experiment and n(E) be the favourable number of cases of the sample space S, then the theoretical probability of happening the event (E) is defined as,

P(E) = (Favourable number of cases)/(Exhaustive number of cases)  = (n(E))/(n(S))

Also, the probability of non-happening of the event is given by

P’(E) = 1 – P(E)

 

Empirical or experimental probability:

When a coin is tossed, theoretically the probability of getting head is ½. The probability obtained in this way is called theoretical probability. On the other hand, if a coin is tossed 20 times, theoretically the head should occur ½ × 20 = 10 times. However, in real experiment it may not happen, i.e. the occurrence of head may be 6, 9, 11 or 15 or any number of times. The probability of any event which is estimated on the basis of the number of actual experiments is known as empirical or experimental probability.

If n(S) be the total number of times, an experiment is repeated and n(E) be number of observed outcomes, then the empirical probability is defined as,

P(E) = (No.of observed outcomes)/(Total no.of times an experiment is repeated)  = (n(E))/(n(S))

 

Probability scale:

Probability is measured on a scale within 0 to 1. A probability of 0 means that there is no chance of an event happening. A probability or 1 means that it is certain the event will happen.

spinner

In the adjoining spinner, the probability of pointer to stop on red is ½. The probability of the pointer to stop on blue is ¼. The probability of the pointer to stop on yellow is ¼. The probability of the pointer to stop on black is 0, which means not possible. The probability of the pointer to stop on red, blue or yellow is 1 which means a sure event.

Thus, probability always lies in between 0 to 1, but it can never be less than 0 and more than 1.

 

Worked out probability examples:

Example 1: Among the number cards numbered from 1 to 20, a card is drawn at random. Find the probability of getting a card whose number is exactly divisible by 5.

Solution:

Let S be the sample space and A be the favourable event that the number is exactly divisible by 5. Then,

                   S = {1, 2, 3, 4, ……….., 20}

                  n(S) = 20

                   A = {5, 10, 15, 20}

                  n(A) = 4

          Now,

                    P(A) = n(A)/n(S) = 4/20 = 1/5

Hence, the probability of getting the card whose number is exactly divisible by 5 is 1/5.

 

Example 2: A card is drawn at random from a well shuffled pack of 52 cards. Find the probability of getting a face card.

Solution:

Let S be the sample space and F be the favourable event of getting a face card, then

                   n(S) = 52

                   n(F) = 12

          Now,

                   P(F) = n(F)/n(S) = 12/52 = 3/13

          Hence, the probability of getting a face card is 3/13.

 

Example 3: A bag contains a dozen of identical ball pens, out of which 5 are red colour and 7 are black colour. If a ball pen is taken out from the bag randomly, find the probability that it is of red colour.

Solution:

Let S be the sample space. R and B be the favourable events of getting red and black ball pens respectively. Then,

                   n(R) = 5

                   n(B) = 7

                   n(S) = 5+7 = 12

          Now,

                   P(R) = n(R)/n(S) = 5/12

          Hence, the probability of getting red ball pen is 5/12.

 

Example 4: What is the probability that Nisha celebrates her birthday on Tuesday?

Solution:

There are 7 days in a week. The birthday of a person must be one of these 7 days. Let S be the sample space and T be the favourable event that she celebrates her birthday on Tuesday. Then,

                   n(S) = 7

                   n(T) = 1 [ there is only one Tuesday in a week.]

          Now,

                   P(T) = n(T)/n(S) = 1/7

          Hence, the probability of celebrating her birthday on Tuesday is 1/7.

 

Example 5: What is the probability that a leap year selected at random has 53 Saturdays.

Solution:

The total number of days in a leap year is 366 and also there are 7 days in a week.

Now,

          366 ÷ 7 = 52+2/7

Clearly we can see that all the days of a week appear 52 times and remaining 2 days may appear as: {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)}. So there are 7 exhaustive cases and 2 favourable cases.

Let S be the sample space and E be the favourable event of getting 53 Saturdays, then

                   n(S) = 7

                   n(E) = 2

          Now,

                   P(E) = n(E)/n(S) = 2/7

          Hence, the probability of 53 Saturdays in a leap year is 2/7.

 

Example 6: A circular plate has eight equal sectors numbered from 1 to 8 and a fixed arrow headed needle at the centre. If the plate is rotated, find the probability that the needle will point at:

A circular plate

a)   2

b)  Lucky number 7

c)   An odd number

d)  A number greater than 5

Solution:

Let S be the sample space. Then,

                   n(S) = 8 [ There are 8 sectors]

a)   Let, A be the favourable event that the arrow points at 2, then

                   n(A) = 1

                   P(A) = n(A)/n(S) = 1/8               

b)  Let, B be the favourable event that the arrow points at 7, then

                   n(B) = 1

                   P(B) = n(B)/n(S) = 1/8

c)   Let, C be the favourable event that the arrow points at an odd number, then

                   n(C) = 4

                   P(C) = n(C)/n(S) = 4/8 = ½

d)  Let, D be the favourable event that the arrow points at number greater than 5, then

                   n(D) = 3

                   P(D) = n(D)/n(S) = 3/8

 

Example 7: The given table represents the number of people of different ages and sex in a live concert. If a person among them gets a car in a lucky draw, find the probability that the person is:

Table 

a)   of age 16 to 24

b)  male of age 25 to 40

c)   a female

d)  not a female

Solution:

Here, the total number of persons in the concert is

          = 500 + 1500 + 1800 + 1200

          = 5000

                  n(S) = 5000

a)   Let, A be the favourable event of the person being of age 16-24, then

                   n(A) = 500 + 1800 = 2300

                   P(A) = n(A)/n(S) = 2300/5000 = 23/50                 

b)  Let, B be the favourable event of the person being male of age 25-40, then

                   n(B) = 1500

                   P(B) = n(B)/n(S) = 1500/5000 = 3/10

c)   Let, F be the favourable event of the person being a female, then

                   n(F) = 1800 + 1200 = 3000

                   P(F) = n(F)/n(S) = 3000/5000 = 3/5

d)  Let, NF be the favourable event of the person not being a female, then

                   P(NF) = 1 – P(F) = 1 – 3/5 = 2/5

 

Example 8: If a person having alcohol drives a vehicle, then the probability of meeting an accident is 11/100. If an experiment is done where 1000 persons having alcohol drive vehicle, how many of them will:

a)   meet an accident

b)  not meet an accident

Solution:

Let S be the sample space and A be the favourable event of meeting an accident where 1000 persons having alcohol drive vehicle and the probability of meeting an accident is 11/100. So,

                   n(S) = 1000

                   P(A) = 11/100

                  P(A) = n(A)/n(S)

          or,     11/100 = n(A)/1000

          or,     100 × n(A) = 11000

          or,     n(A) = 11000/100

          or,     n(A) = 110

                  110 of them are likely to meet an accident and 1000-110 = 890 will not meet an accident.

 

Example 9: A cubical dice is thrown two times. Find the probability that the outcomes are 6 in the first time and odd number in the second time.

Solution:

Here,

36 pairs can be obtained as given in the table below.

Table

So, n(S) = 36

The number of pairs in which ‘6’ in the first and the odd number in the second are (6, 1), (6, 3) and (6, 5), so n(S) = 3

          We know,

                   P(A) = n(A)/n(S) = 3/36 = 1/ 12  

 

Example 10: Head occurs 10 times when a coin is tossed 25 times. Find the empirical probability of tail.

Solution:

Let S be the sample space i.e. total number of times an experiment is repeated and T be number of events in which tail occurred. Then,

                   n(S) = 20 (total number of experiments)

                   n(T) = 25-10 = 15

          Now,

                   P(T) = n(T)/n(S) = 15/20 = 3/5

          Hence, the empirical probability of tail = 3/5.

 

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