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The law of multiplication of probability for two independent events A and B in a sample space is as follows:

P(A ∩ B) = P(A) × P(B)

Correct option is (a) 0

Correct option is (c) 0.28

P (A) = 0.7 and P(A ∪ B) = 0.45 is not possible because P(A ∪ B) < P(A).

A and B are independent events.

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – [P(A) ∙ P(B)]

1. Conditional Events: A and B are any two events of a finite sample space U. Under the condition ‘event A has occurred’ if event B occurs, then that event B is called the conditional event. It is denoted by the symbol B | A. Similarly, under the condition ‘event B has occurred’ if event A will occur then that event A is called the conditional event. It is denoted by the symbol A|B.

2. Law of Conditional Probability: A and B are any two events of a finite sample space U and P (A) > 0. The rule to obtain the probability of event B | A, the probability of occurrence of event B given that event A has already occurred, is called the law of conditional probability. This rule is written as under:

\(P(B|A) = \frac{P(A∩B)}{P(A)} P(A)> 0\)

Similarly, probability of conditional event A | B is obtained by following formula:

\(P(A|B) =\frac{ P(A∩B)}{P(B)}, P (B) > 0\)

The number of elements in the sample space of this random experiment of drawing two cards one by one with replacement from 52 cards is

n = ⁵²C₁ × ⁵²C₁ = 52 × 52 = 2704

P(B) = 2P(A|B) = 0.4

∴ P(B) = 0.4, 2P(A|B) = 0.4

∴ P(A|B) = \(\frac{0.4}{2}\) = 0.2

Now, P(A|B) = \(\frac{P(A∩B)}{P(B)}\)

∴ P(A ∩ B) = P(A|B) ∙P(B)

= 0.2 × 0.4 = 0.08

Total number of tickets = 2000

No. of tickets do not have a prize = 1998

No. of tickets eligible for prize = 2000 – 1998
= 2

Total number of outcomes of selecting a ticket is n = ²⁰⁰⁰c₁ = 2000

A = Event that the selected ticket is eligible for prize

∴ m = ²c₁ = 2

∴ P(A) = \(\frac{m}{n}\)

=\(\frac{ 2}{2000}\)

= \(\frac{1}{1000}\)

When two coins are tossed once, there are four possible outcomes, i.e., S = {HH, HT, TH, TT} 

∴ Total number of outcomes = n(S) = 4 

Let A : Event of getting at least one head 

⇒ A = {HH, HT, TH} ⇒ n(A) = 3

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{4}.\)

Let E₁, E₂, E₃ be the eventsthat the problem issolved by A, B, C respectively and let p₁, p₂, p₃ be corresponding probabilities. Then,

p₁ = P(E₁) = \(\frac{1}{3}\), p₂ = P(E₂) = \(\frac{2}{7}\),  p₃ = P(E₃) = \(\frac{3}{8}\), q₁ = P(\(\bar{E}_1\)) = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\),

q₂ = P(\(\bar{E}_2\)) = 1 - \(\frac{2}{7}\) = \(\frac{5}{7}\), q₃ = P(\(\bar{E}_3\)) = 1 - \(\frac{3}{8}\) = \(\frac{5}{8}\).

The problem will be solved by exactly one of them if it happens in the following mutually exclusive ways: 

(1) A solves and B, and C do not solve; 

(2) B solves and A, and C do not solve; 

(3) C solves and A, and B do not solve; 

Required probability = p₁ q₂ q₃ + q₁ p₂ q₃ + q₁ q₂ p₃

= \(\frac{1}{3}\) x \(\frac{5}{7}\) x \(\frac{5}{8}\) + \(\frac{2}{3}\) x \(\frac{2}{7}\) x \(\frac{5}{8}\) + \(\frac{2}{3}\) x \(\frac{5}{7}\) x \(\frac{3}{8}\) = \(\frac{25}{168}\) + \(\frac{5}{42}\) + \(\frac{5}{28}\) = \(\frac{25}{56}.\)

Sample space, n(S) = 8 

(i) p = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{8}\)

(ii)  p = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

(iii)  p = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\) 

(iv)  p = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{8}\) 

P(A) = 0.3, P(A ∩ B) = 0.03 are given.

∴ P(B|A) =\( \frac{P(A∩B)}{P(A)} = \frac{0.03}{0.3} \)= 0.1

According to the law of addition of probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{1}{3}+\frac{2}{3}-\frac{1}{6}\)

= \(\frac{2+4−1}{6}\)

= \(\frac{5}{6}\)

Now, P(A’ ∩ B’) = P(A ∪ B)’

= 1 – P(A ∪ B)

= \(1 – \frac{5}{6}\)

= \(\frac{6−5}{6}\)

= \(\frac{1}{6}\)

The limitations of mathematical definition of probability are as follows:

• The probability of an event cannot be found if the outcomes are infinite.
• If the total number of outcomes is not known, the probability of an event cannot be found.
• If the elementary outcomes in the sample space are not equi-probable, the probability of an event cannot be found.

If there is no apparant reason to believe that out of one or more events of a random experiment, any one event is more or less likely to occur than the other events, then those events are called equi-probable.

Illustration: In the random experiment of tossing a balance coin, two events of getting head (H) and getting tail (T) are equi-probable, because P (H) = P (T) = \(\frac{1}{2}\)

Certain Event: The special subset U of the sample space of a random experiment is called a certain event.

Probability (Mathematical Definition): If out of n outcomes of the finite sample space of a random experiment which are mutually exclusive, exhaustive and equi-probable, m outcomes are favourable for an event A, then the probability of the event A, P(A) = \(\frac{m}{n}\).

Impossible Event: The special subset Φ or { } of the sample space of a random experiment is called an impossible event.

Let A, B and C be any two events associated with a random experiment whose sample space is S. Then, 

(i) A ∪ B. (Union of A and B) is the event that occurs if A occurs or B occurs or both A and B occur. 

(ii) A ∩ B. (Intersection of A and B). It is the event set which contains all sample points or outcomes which the two events A and B have in common.

Ex: (In a throw of a die), 

A : Event of getting an odd number 

B : Event of getting a prime number 

⇒ A = {1, 3, 5}, B = {2, 3, 5}. Then, A ∩ B = {3, 5}. 

(iii) \(\bar{A}\), (Complement of an event A). It is the set of all sample points of sample spaces that are not contained in A. 

In the toss of a coin, if A is getting a tail, then \(\bar{A}\) is getting a head. 

(iv) (A ∪ B ∪ C) is the event that occurs when at least one of the events A, B or C occurs 

(v) (A ∩ B ∩ C) is the event that occurs when all the three events A, B and C occur. 

(vi) As mutually exclusive events cannot occur together, if events A and B are mutually exclusive, then A ∩ B = ϕ since A and B have nothing in common. 

(vii) Mutually exclusive and exhaustive events. 

Let S be the sample space associated with a random experiment. If E₁, E₂, ..., En are mutually exclusive elementary events associated with the random experiment, then E_i ∩ E_j = ϕ for all i ≠ j and E₁ ∪ E₂ ∪ E₃... ∪ E_n = S 

(Since exhaustive means the total number of possible outcomes) Therefore, an event and its complementary event are both mutually exclusive and exhaustive since: 

A ∩ \(\bar{A}\) = ϕ and A ∪ \(\bar{A}\) = S. 

Ex. Let 1 ball be drawn from a bag containing 12 balls of which 4 balls are white, 4 are red and 4 are green. 

Let A : Event-ball drawn is white 

B : Event-ball drawn is red 

C : Event-ball drawn is green. 

It is obvious that one of the three events must occur as the ball drawn is either white or red or green. 

This means that A, B and C form a mutually exclusive and exhaustive set of events. 

∴ A∩B = ϕ, B∩C = ϕ, A ∩C = ϕ and A∪B∪C = S.

4 cards can be drawn from a pack of cards in ⁵²C₄ ways 

∴ Exhaustive number of cases = n(S) = ⁵²C₄ 

(a) There are 4 suits, each containing 13 cards. 

Let A : Event of drawing one card from each suit 

⇒ Favourable number of cases = n(A) = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{^{13}C_1\times^{13}C_1\times^{13}C_1\times^{13}C_1}{^{52}C_4}\) = \(\frac{13\times13\times13\times13}{\frac{52\times51\times50\times49}{4\times3\times2\times1}}\)               ∵ \(\bigg[\,^nC_r = \frac{|\underline{n}}{|\underline{n-r}|\underline{r}}\bigg]\)

= \(\frac{2197}{20825}\)

(b) Let A : Drawing 4 spade cards of which one is king of spades. 

Then, Favourable number of cases = n(A) = ¹²C₃ x 1

(∵ There is only one king ofspades and the rest of the three spades we draw from remaining 12 spade cards) n(S) = ⁵²C₄

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{^{12}C_1\times1}{^{52}C_4}\) = \(\frac{\frac{12\times11\times10}{3\times2\times1}}{\frac{52\times51\times50\times49}{4\times3\times2\times1}}\) = \(\frac{12\times11\times10\times4}{52\times51\times50\times49}\) = \(\frac{44}{54145}\)

(c) Let A : Drawing at least one ace. 

Now since there are 4 aces in the pack of 52 cards, therefore, the number of ways of drawing 4 cards so that no card is an ace = ⁴⁸C₄ 

∴ Probability of drawing four cards so that none is an ace

P(\(\bar{A}\)) = \(\frac{^{48}C_4}{^{52}C_4}\) = \(\frac{48\times47\times46\times45}{52\times51\times50\times49}\) = \(\frac{38916}{54145}\)

[Here \(\bar{A}\) denotes the complement of event A, i.e, non-happening of event A]

∴ P(A) = 1 - P(\(\bar{A}\)) = 1 - \(\frac{38916}{54145}\) = \(\frac{15229}{54145}\)             

(∵ P(Event) + P(complement of event) = 1)

The two digit numbers can be formed by putting any of 5 digits at the one 's place and also one of the 5 digits at ten’s place. So, 

Total number of 2–digit numbers that can be formed using these 5–digits = 5 × 5 = 25 

The 2–digit numbers formed by 1, 2, 3, 4 and 5 that are divisible by 4 are {12, 24, 32, 44, 52}, i.e, 5 in number. 

∴ Required probability = \(\frac{5}{25}\) = \(\frac{1}{5}.\)

The chosen consists of players (6 + 5). 

∴ Number of ways of selecting 11 players out of 15 players = n(S) = ¹⁵C₁₁ 

Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers 

Then, n(A) = ⁸C₆ x ⁷C₅

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{^8C_6\times^7C_5}{^{15}C_{11}}\) = \(\frac{\frac{8\times7}{2}\times\frac{7\times6}{2}}{\frac{15\times14\times13\times12}{4\times3\times3\times1}}\)= \(\frac{8\times7}{2}\) x \(\frac{7\times6}{2}\) x \(\frac{4\times3\times2\times1}{15\times14\times13\times12}\) = \(\frac{28}{65}.\)

Let A : Event of getting at least 3 black balls 

Then n(A) = ⁵C₃ x ¹¹C₁ + ⁵C₄       

(∵ Besides 5 black balls, there are 11 other balls)

(3 black + others) (4 black)

= \(\frac{5\times4}{2}\) x 11 + 5 = 115

Total numbers of ways in which 4 balls can be drawn from (7 + 5 + 4) = 16 balls

n(S) = ¹⁶C₄ = \(\frac{16\times15\times14\times13}{4\times3\times2\times1}\) = 1820

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{115}{1820}\) = \(\frac{23}{364}.\)

Total number of ways in which 5 tickets can be drawn = n(S) = ³⁰C₅. 

The tickets are arranged in the form T₁, T₂, T₃ (= 20), T₄, T₅ 

Where T₁, T₂ ∈{1, 2, 3, …, 19} and T₄, T₅ ∈{21, 22, …, 30} 

∴ Number of favourable cases = ¹⁹C₂ x 1 x ¹⁰C₂

∴ Required probability = \(\frac{^{19}C_2\times^{10}C_2}{^{30}C_5}\) = \(\frac{19\times18}{2}\times\frac{10\times9}{2}\) x \(\frac{5\times4\times3\times2\times1}{30\times29\times28\times27\times26}\) = \(\frac{285}{5278}.\)

Answer: (b) \(\frac{1}{37}\)

We define the given events as: 

A1: Student knows the answer

A2: Student does not know the answer 

E: He gets the correct answer.

P(A₁) =\(\frac{9}{10}\) , P(A₂) = 1-\(\frac{9}{10}\)  = \(​​\frac{1}{10}\)

\(\therefore\)   P(E/A1) = P(Student gets the correct answer when he knows the answer) = 1 

P(E/A2) = P(Student gets the correct answer when he does not know the correct answer) = 1/4 

\(\therefore\)  Required probability 

\(P(A_2/E) = \frac{P(A_2).P(E/A_2)}{P(A_1).P(E/A_1)+P(A_2).P(E/A_2)}\)

= \(\frac{\frac{1}{10}.\frac{1}{4}}{\frac{9}{10}.1+\frac{1}{10}.\frac{1}{4}}\) = \(\frac{\frac{1}{40}}{\frac{37}{40}}\) = \(\frac{1}{37}\)

Answer : (c) \(\frac{24}{29}\)

Let, 

E1: Examinee guesses the answer 

E2: Examinee copies the answer 

E3: Examinee knows the answer 

E : Event examinee answers correctly

Given, P(E₁) = \(\frac{1}{3}\) , P(E₂) = \(\frac{1}{6}\)

∴  P(E₃) = 1 – (P(E₁) + P(E₃)) = 1- \(\big(\frac{1}{3}+\frac{1}{6}\big)\) = \(\frac{1}{2}\)

Given, \(P(E/E_1) =\frac{1}{4},P(E/E_2) =\frac{1}{8},P(E/E_3)=1\)

∴ Required probability =  P(E₃/E)

= \(\frac{P(E_3).P(E/E_3)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)} \)

\(= \frac{\frac{1}{2}.1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}\)   = \(\frac{\frac{1}{2}}{\frac{1}{12}+\frac{1}{48}+\frac{1}{2}}\) = \(\frac{\frac{1}{2}}{\frac{4+1+24}{48}}\) = \(\frac{\frac{1}{2}}{\frac{29}{48}}\)  \(=\frac{24}{29}\)

Required sample space = S 

= {(x,y) : x,y ∈ {1,2,3,4,5,6}} 

= {(x,y) : x,y = 1,2,3,4,5,6} 

= {(1.1). (1,2)(1,3), (1,4), (1,5), (1,6), 

= (2,6), ……………… , (3,1), …………….. , (3,6), (4,1), ……………. , (4,6), (5,6), …………. (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
When a die is thrown two times, the sample space is given by
S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by:
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)}

Sample space = S 

= {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}

When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).
When a coin is tossed four times, the total number of possible outcomes is 2⁴= 16
Thus, when a coin is tossed four times, the sample space is given by:
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

A coin has two faces: head (H) and tail (T).
When a coin is tossed three times, the total number of possible outcomes is 2³= 8 Thus, when a coin is tossed three times, the sample space is given by:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and a die is thrown, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is given by: S = {H1, H2, H3, H4, H5, H6, T}

Let B₁, G₁, G₂ are in room A and B₂, B₃, B₄, G₃ are in room B. Then sample space is– 

S = {AB₁, AG₁, AG₂, BB₂, BB₃, BB₄, BG₃}.

Let B stands for boys and G stands for girl. Then 

(i) Required sample space = S 

= {BB, BG, GB, GG} 

Required sample space = {0, 1, 2}

Let R,W,B be the red die, white die, blue die respectively. 

∴ Required sample space = S 

= {R₁, R₂, R₃, R₄, R₅, R₆, W₁, W₂, W₃, W₄, W₅, W₆, B₁, B₂, B₃, B₄, B₅, B₆}