CBSE Class 12 Mathematics Probability Notes Set 03

Important Notes and Formulas

1. Introduction

The theory of probability has been originated from the game of chance and gambling. In old days, gamblers used to gamble in a gambling house with a die to win the amount fixed among themselves. They were always desirous to get the prescribed number on the upper face of a die when it was thrown on a board. Shakuni of Mahabharat was perhaps one of them. People started to study the subject of probability from the middle of the seventeenth century. The mathematicians Huygens, Pascal, Fermat and Bernoulli contributed a lot to this branch of Mathematics. A.N. Kolmogorov proposed the set theoretic model to the theory of probability.
Probability gives us a measure of likelihood that something will happen. However, probability can never predict the number of times that an occurrence actually happens. But being able to quantify the likely occurrence of an event is important because most of the decisions that affect our daily lives are based on likelihoods and not on absolute certainties.

2. Definitions

(a) Experiment: An action or operation resulting in two or more well defined outcomes. e.g. tossing a coin, throwing a die, drawing a card from a pack of well shuffled playing cards etc.

(b) Sample space: A set S that consists of all possible outcomes of a random experiment is called a sample space and each outcome is called a sample point. Often, there will be more than one sample space that can describe outcomes of an experiment, but there is usually only one that will provide the most information.
e.g. in an experiment of "throwing a die", the following sample spaces are possible:
(i) {even number, odd number}
(ii) {a number less than 3, a number equal to 3, a number greater than 3}
(iii) {1, 2, 3, 4, 5, 6}
Here the 3rd sample space is the one which provides the most information.
If a sample space has a finite number of points, it is called a finite sample space and if it has an infinite number of points, it is called an infinite sample space. e.g. (i) "in a toss of a coin" either a head (H) or a tail (T) comes up, therefore the sample space of this experiment is S = {H, T}, which is a finite sample space. (ii) "Selecting a number from the set of natural numbers", the sample space of this experiment is S = {1, 2, 3, 4, ……}, which is an infinite sample space.

(c) Event: An event is defined as an occurrence or situation, for example
(i) in a toss of a coin, it shows head,
(ii) scoring a six on the throw of a die,
(iii) winning the first prize in a raffle,
(iv) being dealt a hand of four cards which are all clubs.
In every case it is a set of some or all possible outcomes of the experiment. Therefore an event (A) is a subset of the sample space (S). If the outcome of an experiment is an element of A, we say that the event A has occurred.
An event consisting of a single point of S is called a simple or elementary event.
ϕ is called the impossible event and S (sample space) is called the sure event.
Note: Probability of occurrence of an event A is denoted by P(A).

(d) Compound event: If an event has more than one sample point, it is called a compound event. If A and B are two given events, then A ∩ B is called a compound event and is denoted by A ∩ B or AB or A & B.

(e) Complement of an event: The set of all outcomes which are in S but not in A is called the complement of the event A and is denoted by \( \overline{A} \), \( A^c \), A′ or 'not A'.

(f) Mutually exclusive events: Two events are said to be mutually exclusive (or disjoint or incompatible) if the occurrence of one precludes (rules out) the simultaneous occurrence of the other. If A and B are two mutually exclusive events, then P(A ∩ B) = 0.
Consider, for example, choosing numbers at random from the set {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.
If event A is the selection of a prime number,
event B is the selection of an odd number,
event C is the selection of an even number,
then A and C are mutually exclusive as none of the numbers in this set is both prime and even. But A and B are not mutually exclusive as some numbers are both prime and odd (viz. 3, 5, 7, 11).

(g) Equally likely events: Events are said to be equally likely when each event is as likely to occur as any other event. Note that the term 'at random' or 'randomly' means that all possibilities are equally likely.

(h) Exhaustive events: Events A, B, C, …… N are said to be exhaustive events if no event outside this set can result as an outcome of an experiment. For example, if A and B are two events defined on a sample space S and A and B are exhaustive ⇒ A ∪ B = S ⇒ P(A ∪ B) = 1.

Note: Playing cards: A pack of playing cards consists of 52 cards of 4 suits, 13 in each, as shown in the figure.

Comparative study of equally likely, mutually exclusive and exhaustive events:

3. Classical Definition of Probability

If n represents the total number of equally likely, mutually exclusive and exhaustive outcomes of an experiment and m of them are favourable to the happening of the event A, then the probability of happening of the event A is given by P(A) = m/n. There are (n − m) outcomes which are favourable to the event that A does not happen. 'The event A does not happen' is denoted by \( \overline{A} \) (and is read as 'not A').
Thus \( P(\overline{A}) = \frac{n - m}{n} = 1 - \frac{m}{n} \)
i.e. \( P(\overline{A}) = 1 - P(A) \)
Note:
(i) 0 ≤ P(A) ≤ 1
(ii) \( P(A) + P(\overline{A}) = 1 \), where \( \overline{A} \) = not A. This relationship is most useful in the 'at least one' type of problems.
(iii) If x cases are favourable to A and y cases are favourable to \( \overline{A} \), then \( P(A) = \frac{x}{(x + y)} \) and \( P(\overline{A}) = \frac{y}{(x + y)} \). We say that the odds in favour of A are x : y and the odds against A are y : x.

4. Addition Theorem

A ∪ B = A + B = A or B denotes the occurrence of at least A or B.
For 2 events A and B:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Note:
(a) P(A ∪ B) = P(A + B) = P(A or B) = P(occurrence of at least A or B)
= P(A) + P(B) − P(A ∩ B)   (This is known as the generalised addition theorem.)
= \( P(A) + P(B \cap \overline{A}) \)
= \( P(B) + P(A \cap \overline{B}) \)
= \( P(A \cap \overline{B}) + P(A \cap B) + P(B \cap \overline{A}) \)
= \( 1 - P(A^c \cap B^c) \)
= \( 1 - P(A \cup B)^c \)
(b) P(A\B) = P(A − B) = \( P(A \cap B^c) \) = P(A) − P(A ∩ B)
(c) The opposite of "at least A or B" is neither A nor B,
i.e. \( \overline{A + B} = 1 - (A \text{ or } B) = \overline{A} \cap \overline{B} \)
Note that \( P(A \cup B) + P(\overline{A} \cap \overline{B}) = 1 \).
(d) If A and B are mutually exclusive, then P(A ∪ B) = P(A) + P(B).
(e) For any two events A and B, P(exactly one of A, B occurs)
\( = P(A \cap \overline{B}) + P(B \cap \overline{A}) = P(A) + P(B) - 2P(A \cap B) \)
\( = P(A \cup B) - P(A \cap B) = P(A^c \cup B^c) - P(A^c \cap B^c) \)
(f) (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B)
(g) De Morgan's law: If A and B are two subsets of a universal set U, then
(i) \( (A \cup B)^c = A^c \cap B^c \)   and   (ii) \( (A \cap B)^c = A^c \cup B^c \)
(h) \( (A \cup B \cup C)^c = A^c \cap B^c \cap C^c \)   and   \( (A \cap B \cap C)^c = A^c \cup B^c \cup C^c \)
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)   and   A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

5. Conditional Probability and Multiplication Theorem

Let A and B be two events such that P(A) > 0. Then P(B|A) denotes the conditional probability of B given that A has occurred. Since A is known to have occurred, it becomes the new sample space replacing the original S. From this we are led to the definition
\( P(B|A) = \frac{P(A \cap B)}{P(A)} \)
which is called the conditional probability of B given A.
⇒ P(A ∩ B) = P(A) P(B|A), which is called compound probability or the multiplication theorem.
It says that the probability that both A and B occur is equal to the probability that A occurs times the probability that B occurs given that A has occurred.
Note: For any three events \( A_1, A_2, A_3 \) we have \( P(A_1 \cap A_2 \cap A_3) = P(A_1)\, P(A_2|A_1)\, P(A_3|(A_1 \cap A_2)) \)

6. Independent Events

Two events A and B are said to be independent if the occurrence or non-occurrence of one does not affect the probability of the occurrence or non-occurrence of the other.
(a) If the occurrence of one event affects the probability of the occurrence of the other event, then the events are said to be dependent or contingent. For two independent events A and B:
P(A ∩ B) = P(A) . P(B). Often this is taken as the definition of independent events.
(b) Three events A, B and C are independent if and only if all the following conditions hold:
P(A ∩ B) = P(A) . P(B);   P(B ∩ C) = P(B) . P(C)
P(C ∩ A) = P(C) . P(A)   and   P(A ∩ B ∩ C) = P(A) . P(B) . P(C)
i.e. they must be pairwise as well as mutually independent.
Similarly, for n events \( A_1, A_2, A_3, \ldots, A_n \) to be independent, the number of these conditions is equal to \( {}^nC_2 + {}^nC_3 + \ldots + {}^nC_n = 2^n - n - 1 \).
Note: Independent events are not in general mutually exclusive and vice versa.
Mutual exclusiveness can be used when the events are taken from the same experiment, and independence can be used when the events are taken from different experiments.

7. Total Probability Theorem

Let an event A occur with one of the n mutually exclusive and exhaustive events \( B_1, B_2, B_3, \ldots, B_n \); then
\( A = AB_1 + AB_2 + AB_3 + \ldots + AB_n \)
\( P(A) = P(AB_1) + P(AB_2) + \ldots + P(AB_n) = \sum_{i=1}^{n} P(AB_i) \)
\( P(A) = P(B_1)\, P(A|B_1) + P(B_2)\, P(A|B_2) + \ldots + P(B_n)\, P(A|B_n) = \sum P(B_i)\, P(A|B_i) \)

8. Bayes' Theorem

If an event A can occur only with one of the n mutually exclusive and exhaustive events \( B_1, B_2, \ldots, B_n \) and the probabilities \( P(A/B_1), P(A/B_2), \ldots, P(A/B_n) \) are known, then
\( P(B_i/A) = \frac{P(B_i) \cdot P(A/B_i)}{\sum_{i=1}^{n} P(B_i) \cdot P(A/B_i)} \)

Explanation:
A ≡ event what we have;
\( B_i \) ≡ event what we want, and the remaining are alternative events.
Now, \( P(AB_i) = P(A) \cdot P(B_i/A) = P(B_i) \cdot P(A/B_i) \)
\( P(B_i/A) = \frac{P(B_i) \cdot P(A/B_i)}{P(A)} = \frac{P(B_i) \cdot P(A/B_i)}{\sum_{i=1}^{n} P(AB_i)} \)
\( P(B_i/A) = \frac{P(B_i) \cdot P(A/B_i)}{\sum_{i=1}^{n} P(B_i) \cdot P(A/B_i)} \)

Solved Examples

Illustration 1. A coin is tossed. If it shows head, we draw a ball from a bag consisting of 3 blue and 4 white balls; if it shows tail, we throw a die. Describe the sample space of this experiment. [1 Mark]
Answer:
Let us denote the blue balls by \( B_1, B_2, B_3 \) and the white balls by \( W_1, W_2, W_3, W_4 \). Then a sample space of the experiment is
S = {\( HB_1, HB_2, HB_3, HW_1, HW_2, HW_3, HW_4 \), T1, T2, T3, T4, T5, T6}.
Here \( HB_i \) means head on the coin and ball \( B_i \) is drawn, \( HW_i \) means head on the coin and ball \( W_i \) is drawn. Similarly, T1 means tail on the coin and the number 1 on the die. [1]

Illustration 2. Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space. [1 Mark]
Answer:
In the experiment head may come up on the first toss, or the 2nd toss, or the 3rd toss and so on. Hence, the desired sample space is S = {H, TH, TTH, TTTH, TTTTH, …} [1]

Illustration 3. A coin is tossed three times. Consider the following events.
A: 'no head appears'
B: 'exactly one head appears'
C: 'at least two heads appear'
Do they form a set of mutually exclusive and exhaustive events? [1 Mark]
Answer:
The sample space of the experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Events A, B and C are given by
A = {TTT}
B = {HTT, THT, TTH}
C = {HHT, HTH, THH, HHH}
Now, A ∪ B ∪ C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S
Therefore A, B and C are exhaustive events. Also, A ∩ B = ϕ, A ∩ C = ϕ and B ∩ C = ϕ. Therefore, the events are pairwise disjoint, i.e., they are mutually exclusive.
Hence, A, B and C form a set of mutually exclusive and exhaustive events. [1]

Illustration 4. A coin is tossed successively three times. Find the probability of getting exactly one head or two heads. [1 Mark]
Answer:
Let S be the sample space and E be the event of getting exactly one head or exactly two heads; then
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
and E = {HHT, HTH, THH, HTT, THT, TTH}
∴ n(E) = 6 and n(S) = 8.
Now the required probability, \( P(E) = \frac{n(E)}{n(S)} = \frac{6}{8} = \frac{3}{4} \). [1]

Illustration 5. Words are formed with the letters of the word PEACE. Find the probability that the 2 E's come together. [1 Mark]
Answer:
Total number of words which can be formed with the letters P, E, A, C, E \( = \frac{5!}{2!} = 60 \)
Number of words in which the 2 E's come together = 4! = 24   ∴ required probability \( = \frac{24}{60} = \frac{2}{5} \) [1]

Illustration 6. A bag contains 5 red and 4 green balls. Four balls are drawn at random. Find the probability that two balls are red and two balls are green. [4 Marks]
Answer:
n(S) = the total number of ways of drawing 4 balls out of a total of 9 balls = \( {}^9C_4 \) [1]
A: drawing 2 red and 2 green balls; n(A) = \( {}^5C_2 \times {}^4C_2 \) [1]
∴ \( P(A) = \frac{n(A)}{n(S)} = \frac{{}^5C_2 \times {}^4C_2}{{}^9C_4} = \frac{\frac{5 \times 4 \times 4 \times 3}{2 \times 2}}{\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2}} = \frac{10}{21} \) [2]

Illustration 7. If the letters of INTERMEDIATE are arranged, then find the odds in favour of the event that no two E's occur together. [4 Marks]
Answer:
I → 2, N → 1, T → 2, E → 3, R → 1, M → 1, D → 1, A → 1   (3 E's, rest 9 letters)
First arrange the rest of the letters \( = \frac{9!}{2!\,2!} \) [1]
Now the 3 E's can be placed in \( {}^{10}C_3 \) ways, so favourable cases \( = \frac{9!}{2!\,2!} \times {}^{10}C_3 = 3 \times 10! \) [1]
Total cases \( = \frac{12!}{2!\,2!\,3!} = \frac{11}{2} \times 10! \); non-favourable cases \( = \left(\frac{11}{2} - 3\right) \times 10! = \frac{5}{2} \times 10! \) [1]
Odds in favour of the event \( = \frac{3}{5/2} = \frac{6}{5} \), i.e. 6 : 5 [1]

Illustration 8. From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. Find the probability that the selection contains at least one of each category. [4 Marks]
Answer:
n(S) = \( {}^{10}C_4 \) = 210. [1]
n(E) = \( {}^5C_2 \times {}^3C_1 \times {}^2C_1 + {}^5C_1 \times {}^3C_2 \times {}^2C_1 + {}^5C_1 \times {}^3C_1 \times {}^2C_2 \) = 105 [2]
∴ \( P(E) = \frac{105}{210} = \frac{1}{2} \). [1]

Illustration 9. If four cards are drawn at random from a pack of fifty-two playing cards, find the probability that at least one of them is an ace. [4 Marks]
Answer:
If A is a combination of four cards containing at least one ace (i.e. either one ace, or two aces, or three aces or four aces), then \( \overline{A} \) is a combination of four cards containing no aces.
Now \( P(\overline{A}) = \frac{\text{Number of combinations of four cards with no aces}}{\text{Total number of combinations of four cards}} \)
\( = {}^{48}C_4 / {}^{52}C_4 = 0.72 \) (approx.) [2]
Using \( P(A) + P(\overline{A}) = 1 \), we have \( P(A) = 1 - P(\overline{A}) = 1 - 0.72 = 0.28 \) [2]

Illustration 10. If A and B are two events such that \( P(A \cup B) = \frac{3}{4} \), \( P(A \cap B) = \frac{1}{4} \) and \( P(A^c) = \frac{2}{3} \), then find
(i) P(A)    (ii) P(B)    (iii) \( P(A \cap B^c) \)    (iv) \( P(A^c \cap B) \) [4 Marks]
Answer:
(i) \( P(A) = 1 - P(A^c) = 1 - \frac{2}{3} = \frac{1}{3} \) [1]
(ii) \( P(B) = P(A \cup B) + P(A \cap B) - P(A) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = \frac{2}{3} \) [1]
(iii) \( P(A \cap B^c) = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \) [1]
(iv) \( P(A^c \cap B) = P(B) - P(A \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{5}{12} \) [1]

Illustration 11. Three dice are thrown at the same time. Find the probability of getting three 2's, if it is known that the sum of the numbers on the dice was six. [Exemplar] [4 Marks]
Answer:
Let events A and B be as defined below:
A: getting three 2's in a throw of three dice at the same time
B: sum of the numbers on the dice is 6
Now A = {222}, B = {114, 123, 132, 141, 213, 222, 231, 312, 321, 411}
and A ∩ B = {222} [1]
Hence \( P(A) = \frac{1}{216},\ P(B) = \frac{10}{216},\ P(A \cap B) = \frac{1}{216} \) [1]
\( P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{216}}{\frac{10}{216}} = \frac{1}{10} \) [2]

Illustration 12. A committee of 4 students is selected at random from a group consisting of 8 boys and 4 girls. Given that there is at least one girl on the committee, calculate the probability that there are exactly 2 girls on the committee. [Exemplar] [4 Marks]
Answer:
Let events A and B be as defined below:
A: event of exactly 2 girls on the committee
B: event of at least one girl on the committee
Now \( P(B') = \frac{{}^8C_4}{{}^{12}C_4} = \frac{70}{495} = \frac{14}{99} \)
\( P(B) = 1 - \frac{14}{99} = \frac{85}{99} \) [1]
Now \( P(A \cap B) = P(\text{2 boys and 2 girls}) = \frac{{}^8C_2 \cdot {}^4C_2}{{}^{12}C_4} = \frac{6 \times 28}{495} = \frac{56}{165} \) [1]
Hence \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{56}{165} \times \frac{99}{85} = \frac{168}{425} \) [2]

Illustration 13. A coin is tossed once. If it shows head, it is tossed again and if it shows tail, then a die is thrown. Let \( E_1 \) be the event 'the first throw of the coin shows tail' and \( E_2 \) the event 'the die shows a number greater than 4'. Find \( P(E_2/E_1) \). [4 Marks]
Answer:
The sample space of the experiment is
S = {HH, HT, T1, T2, T3, T4, T5, T6} [½]

This means that \( P(HH) = P(HT) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \) [½]
and \( P(T1) = P(T2) = P(T3) = P(T4) = P(T5) = P(T6) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \). [1]
Here, \( E_1 \) = {T1, T2, T3, T4, T5, T6} and \( E_2 \) = {T5, T6}
Hence, \( E_2 \cap E_1 \) = {T5, T6}. [1]
So, \( P(E_1) = 6 \times \frac{1}{12} = \frac{1}{2} \) and \( P(E_2) = 2 \times \frac{1}{12} = \frac{1}{6} \).
Also, \( P(E_2 \cap E_1) = 2 \times \frac{1}{12} = \frac{1}{6} \)
Now \( P(E_2/E_1) = \frac{P(E_2 \cap E_1)}{P(E_1)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3} \) [1]

Illustration 14. An unbiased die is thrown twice. Let the event A be 'odd number on the first throw' and B the event 'odd number on the second throw'. Check the independence of the events A and B. [4 Marks]
Answer:
If all the 36 elementary events of the experiment are considered to be equally likely, we have
\( P(A) = \frac{18}{36} = \frac{1}{2} \) and \( P(B) = \frac{18}{36} = \frac{1}{2} \) [1]
Also P(A ∩ B) = P(odd number on both throws) [1]
\( = \frac{9}{36} = \frac{1}{4} \)
Now \( P(A)P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \) [1]
Clearly P(A ∩ B) = P(A) × P(B)
Thus, A and B are independent events. [1]

Illustration 15. A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B, C are defined as:
A = {The first bulb is defective}
B = {The second bulb is non-defective}
C = {The two bulbs are both defective or both non-defective}
Determine whether (i) A, B, C are pairwise independent, (ii) A, B, C are independent.
Answer:
We have \( P(A) = \frac{50}{100} = \frac{1}{2} \); \( P(B) = \frac{50}{100} = \frac{1}{2} \); \( P(C) = \frac{50}{100} \cdot \frac{50}{100} + \frac{50}{100} \cdot \frac{50}{100} = \frac{1}{2} \) [1]
A ∩ B is the event that the first bulb is defective and the second is non-defective.
∴ \( P(A \cap B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \) [1]
A ∩ C is the event that both bulbs are defective.
∴ \( P(A \cap C) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \) [1]
Similarly \( P(B \cap C) = \frac{1}{4} \) [1]
Thus we have P(A ∩ B) = P(A) . P(B); P(A ∩ C) = P(A) . P(C); P(B ∩ C) = P(B) . P(C)
∴ A, B and C are pairwise independent. [1]
There is no element in A ∩ B ∩ C
∴ P(A ∩ B ∩ C) = 0
⇒ P(A ∩ B ∩ C) ≠ P(A) . P(B) . P(C)
Hence A, B and C are not mutually independent. [1]

Illustration 16. A purse contains 4 copper and 3 silver coins and another purse contains 6 copper and 2 silver coins. One coin is drawn from any one of these two purses. Find the probability that it is a copper coin. [4 Marks]
Answer:
Let A ≡ event of selecting the first purse
B ≡ event of selecting the second purse
C ≡ event of drawing a copper coin
Then the given event has two disjoint cases: AC and BC
∴ P(C) = P(AC + BC) = P(AC) + P(BC) = P(A) P(C/A) + P(B) P(C/B) [2]
\( = \frac{1}{2} \cdot \frac{4}{7} + \frac{1}{2} \cdot \frac{6}{8} = \frac{37}{56} \) [2]

Illustration 17. Three groups A, B, C are contesting for positions on the Board of Directors of a company. The probabilities of their winning are 0.5, 0.3, 0.2 respectively. If the group A wins, the probability of introducing a new product is 0.7 and the corresponding probabilities for groups B and C are 0.6 and 0.5 respectively. Find the probability that the new product will be introduced. [4 Marks]
Answer:
Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2
∴ P(A) + P(B) + P(C) = 1 [1]
then events A, B, C are exhaustive.
If P(E) = probability of introducing a new product, then as given
P(E|A) = 0.7, P(E|B) = 0.6 and P(E|C) = 0.5 [1]
∴ P(E) = P(A) . P(E|A) + P(B) . P(E|B) + P(C) . P(E|C) [1]
= 0.5 × 0.7 + 0.3 × 0.6 + 0.2 × 0.5 = 0.35 + 0.18 + 0.10 = 0.63 [1]

Illustration 18. Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold? [6 Marks]
Answer:
Let \( E_1, E_2 \) and \( E_3 \) be the events that boxes I, II and III are chosen, respectively.
Then \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \) [1]
Also, let A be the event that 'the coin drawn is of gold'.
Then \( P(A/E_1) \) = P(a gold coin from box I) \( = \frac{2}{2} = 1 \) [1]
\( P(A/E_2) \) = P(a gold coin from box II) = 0 [1]
\( P(A/E_3) \) = P(a gold coin from box III) \( = \frac{1}{2} \) [1]
Now, the probability that the other coin in the box is of gold
= the probability that the gold coin is drawn from box I
= \( P(E_1/A) \)
By Bayes' theorem, we know that
\( P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \) [1]
\( = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times 0 + \frac{1}{3} \times \frac{1}{2}} = \frac{2}{3} \) [1]

Illustration 19. A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, what is the probability that the missing card is black? [6 Marks]
Answer:
Let \( E_1 \) be the event that the missing card is a black card and \( E_2 \) the event that the missing card is red. Let E be the event that the first 13 cards which are examined are all red; then we are required to find \( P(E_1/E) \).
Now \( P(E_1) = P(E_2) = \frac{1}{2} \). [1]
Also \( P(E/E_1) = \frac{{}^{26}C_{13}}{{}^{51}C_{13}} \) and \( P(E/E_2) = \frac{{}^{25}C_{13}}{{}^{51}C_{13}} \). [2]
By Bayes' theorem, we get \( P(E_1/E) = \frac{P(E_1)P(E/E_1)}{P(E_1)P(E/E_1) + P(E_2)P(E/E_2)} \) [1]
\( = \frac{\frac{1}{2} \times \frac{{}^{26}C_{13}}{{}^{51}C_{13}}}{\frac{1}{2} \times \frac{{}^{26}C_{13}}{{}^{51}C_{13}} + \frac{1}{2} \times \frac{{}^{25}C_{13}}{{}^{51}C_{13}}} = \frac{{}^{26}C_{13}}{{}^{26}C_{13} + {}^{25}C_{13}} \) [1]
\( = \frac{\frac{26!}{13!\,13!}}{\frac{26!}{13!\,13!} + \frac{25!}{13!\,12!}} = \frac{26(25!)}{26(25!) + 13(25!)} = \frac{26}{39} = \frac{2}{3} \) [1]

Do Yourself

Do Yourself 1
(i) Two balls are drawn from a bag containing 2 red and 3 black balls. Write the sample space of this experiment.
(ii) A coin is tossed and if head comes up, a die is thrown. But if tail comes up, the coin is tossed again. Write the sample space of this experiment.

Do Yourself 2
(i) A bag contains 6 white, 5 black and 4 red balls. Find the probability of getting either a white or a black ball in a single draw.
(ii) In a class of 125 students, 70 passed in English, 55 in Mathematics and 30 in both. Find the probability that a student selected at random from the class has passed in (a) at least one subject (b) only one subject.

Do Yourself 3
(i) Three cards are drawn successively without replacement from a well shuffled pack of cards. Find the probability that the first two cards are kings and the third card drawn is an ace.
(ii) A bag contains 8 red, 3 white and 9 blue balls. If three balls are drawn at random without replacement, determine the probability that (a) all the balls are blue (b) all the balls are of different colours.

Do Yourself 4
(i) A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.
(ii) Events A and B are such that \( P(A) = \frac{1}{2},\ P(B) = \frac{7}{12} \) and \( P(\text{not A or not B}) = \frac{1}{4} \). State whether A and B are independent.

Do Yourself 5
(i) A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and both are found to be spades. Find the probability of the lost card being a spade.
(ii) Suppose that 6% of the persons with blood group O are left handed and 10% of those with other blood groups are left handed. It is given that 30% of the persons have blood group O. If a left handed person is selected at random, what is the probability that the person has blood group O? [Exemplar]

Answers for Do Yourself

Do Yourself 1
(i) {\( B_1R_1, B_2R_1, B_3R_1, B_1R_2, B_2R_2, B_3R_2, B_1B_2, B_2B_3, B_1B_3, R_1R_2 \)}
(ii) {H1, H2, H3, H4, H5, H6, TH, TT}

Do Yourself 2
(i) \( \frac{11}{15} \)    (ii) (a) \( \frac{19}{25} \)  (b) \( \frac{13}{25} \)

Do Yourself 3
(i) \( \frac{2}{5525} \)    (ii) (a) \( \frac{7}{95} \)  (b) \( \frac{18}{95} \)

Do Yourself 4
(i) A and B are independent    (ii) A and B are not independent

Do Yourself 5
(i) \( \frac{11}{50} \)    (ii) \( \frac{9}{44} \)