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For a finite sample space S with equally likely We have n(A) = 13, n(S) = 52 outcomes, probability of an event A is denoted by P(A) = n(A)/n(S),

Where n(A) = number of elements in the set A n(S) = number of elements in the set S. 

Note: 

(i) If A and B are any two events, then 

P(A or B) = P(A) + P(B) – P(A and B) 

i.e., P( A ∪ B) P(A) + P(B) – P(A ∩ B) 

(ii) If A and B around mutually exclusive, then 

P (‘A or B’) = P(A) + P(B) 

i.e., P(A ∪ B) = P(A) + P(P) ∵ A ∩B = ϕ 

(iii) If A is any event, then 

P(‘not A’ ) = P(A’) = 1 – P(A)

Correct option is C. 17/19

The product of 3 randomly chosen integers are even only if there are even nos.

P(even) = 1 - P(odd)

P(odd) = \(\cfrac{10}{20}\times\cfrac9{19}\times\cfrac8{18}\)

= \(\cfrac2{19}\)

P(even) = \(\cfrac{17}{19}\)

3 integers out of 20 can be chosen in ²⁰C₃ ways. 

As in first 20 integers, 10 are even. 

To have the product of 3 integer even the chosen integers must be even.

∴ 3 integers that are even can be chosen from first 20 integers in ¹⁰C₃ ways. 

∴ Probability =  \(\frac{^{10}C_2}{^{20}C_2}\) = \(\frac{10\times9\times8}{20\times19\times18} = \frac{2}{19}\) 

Our answer matches with option (a) 

∴ Option (a) is the only correct choice

Probability of drawing an ace in the first draw = \(\frac{4}{52}.\)

Probability of drawing a queen of opposite shade in the second draw = \(\frac{2}{51}.\)

Probability of drawing a queen in the first draw = \(\frac{4}{52}.\)

Probability of drawing an ace of opposite shade in the second draw = \(\frac{2}{51}.\)

∴ Required probability = \(\frac{4}{52}\) x \(\frac{2}{51}\) + \(\frac{4}{52}\) x \(\frac{2}{51}\) = \(\frac{4}{663}.\)             

[‘AND’ and ‘OR’Theorems]

Sample space n(S) = 21 

Prime numbers from 1 to 21 are 2, 3, 5, 7, 11, 13, 17, 19. 

It ‘E’ be the event of getting a prime number, then n(E) = 8 

∴ p(E) = \(\frac{n(F)}{n(S)}\) = \(\frac{8}{21}\)

∴ The Probability that the number is prime = \(\frac{8}{21}.\)

Given, P(A) = 0.25 and P(B) = 0.5 

Also P(A ∩ B) = 0.14 

We have to find P(A’∩B’) 

By De Morgan’s theorem we know that: 

P(A’∩B’) = P(A∪B)’ 

We know that P(A∪B) = P(A) + P(B) – P(A∩B) 

∴ P(A∪B) = 0.25 + 0.5 – 0.14 = 0.61 

∴ P(A’∩B’) = P(A∪B)’ = 1 - P(A∪B) 

= 1 – 0.61 = 0.39 

As our answer matches with option (a) 

∴ Option (a) is the only correct choice.

0.2

The probability of happening of an events A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is 0.2.

Given that M and N are two events,

By General Addition Rule,

P(A ⋃ B) = P(A) + P (B) – P(A ⋂ B)

∴ P(M ⋃ N) = P(M) + P(N) – P(M ⋂ N).

Hence, the correct option is (B).

Correct option is A. 0.39

Given P(A∪B) = 0.14,P(A) = 0.25,P(B) = 0.50

By addition theorem,

P(A∪B) = P(A) + P(B) - P(A∩B) P(A∩B)

= 0.75 - 0.14 = 0.61

∴P(both A and B not happening) = 1 - P(A∩B) = 1 - 0.61 = 0.39

Correct option is A.0.0875

India wins = W = 2 pt

India loses = L = 0 pt

India draws = D = 1 pt

Total matches played by India = 4 

This means that the total points needs to be 7 at least

Either India wins all 4 matches or India wins 3 and draws 1

the probability of WWWW = (0.5)*(0.5)*(0.5)*(0.5) = 0.0625 

India wins 3 matches and draws 1: probability in that case = (0.5)*(0.5)*(0.5)*(0.05) = 0.00625 

The 4 possible outcomes will be:

WWWD, WWDW, WDWW, DWWW

Total probability of India wins 3 matches and draws 1 = 4*(0.00625) = 0.025

P( total points have to be at least 7) = 0.0625 + 0.025 = 0.0875

Let E₁ = Event person will see the giraffe

and E₂ = Event person will see the bear

Then, we have

P(E₁) = 0.72 and P(E₂) = 0.84

P(person will see both giraffe and bear) = 0.52

By General Addition Rule,

P(A ⋃ B) = P(A) + P (B) – P(A ⋂ B)

⇒ P(E₁⋃ E₂) = P(E₁) + P(E₂) – P(E₁⋂ E₂)

= 0.72 + 0.84 – 0.52

= 1.04 ≠ 0.52

Hence, the given statement is False.

Let A = Event that student will pass examination

and B = Event that student will get compartment

Then, we have

P(A) = 0.73, P(B) = 0.13 and P(A ⋃ B) =0.96

Now, we have to find P(A ⋂ B)

By General Addition Rule, we have

P(A ⋃ B) = P(A) + P (B) – P(A ⋂ B)

⇒ P(A ⋃ B) = 0.73 + 0.13 – 0

[∵ A and B are independent events]

⇒ P(A ⋃ B) = 0.86

But P(A ⋃ B) = 0.96

Hence, the given statement is False.

Let X denote the number of doublets. Possible doublets are 

(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6) 

Clearly, X can take the value 0, 1, 2, or 3. 

Probability of getting a doublet = 6/36 = 1/6

Probability of not getting a doublet = 1 - 1/6 = 5/6

Now P(X = 0) = P (no doublet) = 5/6 x 5/6 x 5/6 = 125/216

P(X = 1) = P (one doublet and two non-doublets) 

= 1/6 x 5/6 x 5/6 + 5/6 x 1/6 x 5/6 + 5/6 x 5/6 x 1/6

= 3(1/6 x 5²/6²) = 75/216

P(X = 2) = P (two doublets and one non-doublet)

= 1/6 x 1/6 x 5/6 + 5/6 x 1/6 x 1/6 + 1/6 x 5/6 x 1/6

= 3(1/62 x 5/6) = 15/216

and P(X = 3) = P (three doublets) = 1/6 x 1/6 x 1/6 = 1/216

Thus, the required probability distribution is

Let the number of red ball be x.

total number of red balls = 4 + x.

probability of drawing a white ball is = \(\frac{2}{5}\)

\(\therefore \frac{4}{4+x}=\frac{2}{5}\)

2(4 + x) = 4 × 5

4 + x = 2 × 5 = 10

X = 10 - 4 = 6

X = 6

Thus the number of red balls are 6.

The probability of an event E is defined as the number of outcomes favorable to E divided by the total number of equally likely outcomes in the sample space S of the experiment.

In probability theory, an elementary event (also called an atomic event or simple event) is an event which contains only a single outcome in the sample space.

Using set theory terminology, an elementary event is a singleton.

Any particular performance of a random experiment is called a trial.

By Experiment or Trial in the subject of probability, we mean a Random experiment unless otherwise specified.

Each trial results in one or more outcomes.

Given,

Total number of workers = 30

(i) Probability that his wages are less than rs 150 \(=\frac{favourable\,outcome}{total\,outcome}\) \(=\frac{3+4}{30}=\frac{7}{30}\)

(ii) Probability that his wages are at least rs 210 \(=\frac{favourable\,outcome}{total\,outcome}\) \(=\frac{3+4}{30}=\frac{7}{30}\)

(iii) Probability that his wages are more than or equals to rs 150 but less than rs 200 \(=\frac{favourable\,outcome}{total\,outcome}\) \(=\frac{5+6+5}{30}=\frac{16}{30}=\frac{8}{15}\)

An Elementary Event is any single outcome or element of a sample space.

Given,

(i) Probability that the life time of the selected bulb is less than 400 \(=\frac{favourable\,outcome}{total\,outcome}=\frac{14}{400}=\frac{7}{200}\)

(ii) Probability that the life time of the selected bulb is between 300 – 800 hours \(=\frac{favourable\,outcome}{total\,outcome}=\frac{14+56+60+86+74}{400}\) \(=\frac{290}{400}=\frac{29}{40}\)

(iii) Probability that the life time of the selected bulb is at least 700 hours \(=\frac{favourable\,outcome}{total\,outcome}=\frac{74+62+48}{400}=\frac{184}{400}=\frac{23}{50}\)

Given,

(i) The probability that the family is earning Rs. 10,000 – 13,000 per month and owning exactly 2 vehicles \(=\frac{favourable \,outcome}{total\,outcome}=\frac{29}{2400}\)

(ii) The probability that the family earning Rs. 16,000 or more per month and owning exactly 1 vehicles \(=\frac{favourable \,outcome}{total\,outcome}=\frac{579}{2400}\)

(iii) The probability that the family earning less than Rs. 7,000 per month and does not own any vehicles \(=\frac{favourable \,outcome}{total\,outcome}=\frac{10}{2400}=\frac{1}{240}\)

(iv) The probability that the family earning Rs. 13,000 – 16,000 per month and owning more than 2 vehicles \(=\frac{favourable \,outcome}{total\,outcome}=\frac{25}{2400}=\frac{1}{96}\)

(v) The probability that the family owning not more than 1 vehicle \(=\frac{favourable \,outcome}{total\,outcome}=\frac{10+0+1+2+1+160+305+535+469+579}{2400}\)

\(=\frac{2062}{2400}=\frac{1031}{1200}\)

(vi) The probability that the family owning at least one vehicle \(=\frac{favourable \,outcome}{total\,outcome}\)

\(=\frac{160+305+535+469+579+254+27+29+29+82+0+2+1+25+86}{2400}\)

\(=\frac{2356}{2400}=\frac{589}{600}\)

Number of total families surveyed = 10 + 160 + 25 + 0 + 0 + 305 + 27 + 2 + 1 +

535 + 29 + 1 + 2 + 469 + 59 + 25 + 1 + 579 + 82 + 88 = 2400

(i) Number of families earning Rs 10000 − 13000 per month and owning exactly 2 vehicles = 29

Hence, required probability, P = 29/2400

(ii) Number of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579

Hence, required probability, P = 579/2400

(iii) Number of families earning less than Rs 7000 per month and does not own any vehicle = 10

Hence, required probability, P = 10/2400 = 1/240

(iv) Number of families earning Rs 13000 − 16000 per month and owning more than 2 vehicles = 25

Hence, required probability, P = 25/2400 = 1/96

(v) Number of families owning not more than 1 vehicle = 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062

Hence, required probability, P = 2062/2400 = 1031/1200

Given: A coin is tossed and a die is rolled.

In the given experiment, coin is tossed and if the outcome is tail then, die will be rolled.

The possible outcome for coin is 2 = {H, T}

And, the possible outcome for die is 6 = {1, 2, 3, 4, 5, 6}

If the outcome for the coin is tail then sample space is S1= {(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)}

If the outcome is head then the sample space is S2 = {(H, H) (H, T)}

So, the required outcome sample space is S = S1 ⋃ S2

S = {(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6) (H, H) (H, T)}

∴ The sample space for the given experiment is {(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6) (H, H) (H, T)}

(a) Let event A: A yellow ball is drawn from each bag. 

Probability of drawing one yellow ball from total 8 balls of first bag and that of drawing one yellow ball out of total 10 balls of second bag is

P(A) = \(\frac {^{3}C_1}{^{8}C_1}\)x \(\frac {^{4}C_1}{^{10}C_1} = \frac {3} {8} \) x \(\frac {4} {10} = \frac {3}{20}\)

Let event B: A brown ball is drawn from each bag. Probability of drawing one brown ball out of total 8 balls of first bag and that of drawing one brown ball out of total 10 balls of second bag is

 P(A) = \(\frac {^{5}C_1}{^{8}C_1}\)x \(\frac {^{6}C_1}{^{10}C_1} = \frac {5} {8} \) x \(\frac {6} {10} = \frac {3}{8}\)

Since both the events are mutually exclusive events, 

P(A ∩ B) = 0 

∴ P(both the balls are of the same colour) = P(both are of yellow colour) or P(both are of brown colour)

= P(A) + P(B)

= 3/20 + 3/8

= 21/40

(b) P(both the balls are of different colour) = 1 – P(both the balls are of the same colour)

= 1- 21/40

= 19/40

Sample space = S 

= {HH, HT, T1, T2,T3, T4, T5, T6}

Let R stands for red ball and W stands for white ball. Then required sample space = S 

= {RW, WR, WW} 

= {R,W),(W,R),(W,W)}

Total number of balls = 10 + 15 = 25 (a) 

(a) Let event A: First ball drawn is red.

∴ P(A) =\(\frac {^{10}C_1}{^{25}C_1} = \frac {10} {25} = \frac {2}{5}\)

Let event B: Second ball drawn is green. 

Since the first red ball is not replaced in the box, we now have 24 balls, out of which 15 are green. 

∴ Probability that the second ball is green under the condition that the first red ball is not replaced in the box = P(B/A) =

\(\frac {^{15}C_1}{^{24}C_1} = \frac {15} {24} = \frac {8}{5}\)

∴ Required probability = P(A ∩ B) = P(B/A) . P(A)

= 2/3 x 5/8

= 1/4

(b) To find the probability that one ball is red and the other is green, there are two possibilities: 

First ball is red and second ball is green. 

OR 

The first ball is the green and the second ball is red. 

From above, we get

P(First ball is red and second ball is green) = 1/4

Similarly, 

P(First ball is green and second ball is red) =

\(\frac {^{15}C_1}{^{24}C_1}\)x \(\frac {^{10}C_1}{^{24}C_1} = \frac {15} {25} \) x \(\frac {10} {24} = \frac {1}{4}\)

∴ Required probability = P(First ball is red and second ball is green) + P(First ball is green and second ball is red)

= 1/4 + 1/4

= 1/2

Given: There are three colored dice of red , white and black color. These dice are placed in bag. 

To Find: Write the sample space for the given experiment. 

Explanation: A dice has 6 faces containing numbers (1, 2, 3, 4, 5, 6) 

Let us Assume Red = R 

Let us Assume White = W 

Let us Assume Black = B 

According to question, when Dice is selected , then rolled. 

Firstly, selected Red Dice then, Possible samples are: 

S_R={(R, 1), (R, 2)(R, 3), (R, 4), (R, 5), (R, 6)} 

Then, White Dice will be selected , So the sample spaces are: 

S_w={(W, 1), (W, 2), (W, 3), (W, 4), (W, 5), (W, 6)} 

Lastly, Black Dice will be selected and rolled , So the sample spaces for Black die 

S_B= {(B, 1)(B, 2)(B, 3)(B, 4)(B, 5), (B, 6)} 

Thus, The total sample for the given experiment is 

S=S_R S_W S_B 

{(B, 1)(B, 2)(B, 3)(B, 4)(B, 5), (B, 6)(W, 1), (W, 2), (W, 3), (W, 4), (W, 5), (W, 6)(R, 1), (R, 2)(R, 3), (R, 4), (R, 5), (R, 6)}

Hence, B is the sample space for the given experiment.

It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.
When two balls are drawn at random in succession without replacement, the sample space is given by S = {RW, WR, WW}

(i) To Find: Write the sample space for the given experiment. 

Explanation: Here, The family has 2 children 

Let us Assume Boy = B 

Let us Assume Girl = G 

So, The number of sample spaces for 2 children is 2²=4 

Sample space are, S={(B₁, B₂), (G₁, G₂), (G₁, B₂), (B₁, G₂)} 

Where , number 1 and 2 are represent elder and younger. 

Hence, S is the sample space for given experiment. 

(ii) Explanation: Here, The family has 2 children, So the possibility of boys in a family is:

(a) No boys only girl 

(b) One boy and one girl 

(c) Two boys only 

So, The Sample space for the given condition is: 

S={0, 1, 2} 

Hence, S is the sample spaces for the given experiment.

(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S =
{GG, GB, BG, BB}
(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or
1 girl or no girl. Hence, the required sample space is S = {0, 1, 2}

A die has six faces that are numbered from 1 to 6, with one number on each face. Let us denote the red, white, and blue dices as R, W, and B respectively.
Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

Given: A coin is tossed and the die is rolled.

We know that, we have a coin and a die,

So, when coin is tossed there will be 2 events Head and tail,

According to question, If Head occurs on coin then Die will be rolled out otherwise not.

So, the sample spaces are:

S = {(T, (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

∴ The sample space is {T, (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)1}

S = {HHHH, HHHT, HHTH, HTHH,THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT,TTHT,TTTH ,TTTT}

Required sample space = S = {H₁, H₂, H₃, H₄, H₅, H₆, T}

Probability (Statistical Definition) :
Suppose, a random experiment is repeated n times under identical conditions. If an ‘event A occurs in m trials then the relative frequency\( \frac{m}{n} \)of the event A gives the estimate of the probability of the event A. When n tends to infinity, the limiting value of \( \frac{m}{n} \) is called the probability of the event A. Thus,

\(P(A)= \lim\limits_{n\to\infty}\frac{m}{n}\)
 

The limitations of statistical definition of probability are as follows:

• The infinite value of n cannot be taken in practice.
• The exact value of probability cannot be known.

If in a random experiment there are n mutually exclusive and equally likely elementary events and m of them are favourable to an event A, then the probability P of happening of A denoted by P(A) is defined as the ratio \(\frac{m}{n}.\)

i.e, P(A) = \(\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\) = \(\frac{n(A)}{n(S)},\)

where   n(A) = number of sample points in event A 

n(S) = number of sample points in sample space S. 

If P(A) = probability of occurrence of A, then 

P(\(\bar{A}\)) = probability of failure of A or non-occurrence of A 

= 1 – P(A) as P(A) + P(\(\bar{A}\)) = 1

Note : The probabilities of mutually exclusive and exhaustive events always adds up to 1.

If a random experiment is repeated n times under identical conditions and out of n such trials, m trials are favourable to the occurrence of some event A, then the relative frequency is called the estimate of the probability of occurrence of event A, which is denoted by P (A). If the value of n is taken larger and larger, i.e., as n tends to infinity, the limiting value of the ratio is taken as the probability of occurrence of the event A. Symbolically,

\(P(A)=\lim\limits_{n\to \infty}\,\frac{m}{n}\)

(a) For Two Events. If A and B are two events associated with a random experiment, then 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

⇒ P(A or B) = P(A) + P(B) — P(A and B) 

Corollary 1: If A and B are mutually exclusive events, then, 

P(A ∩ B) = 0, therefore 

P(A ∪ B) = P(A) + P(B) 

Corollary 2: P(A or B) ≤ P(\(\underline{A}\)) + P(B) 

Given, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

Since, P(A ∩ B) is greater than or equal to 0, 

P(A or B) < P(A) + P(B) 

Equality in the above result holds when A and B are mutually exclusive as P(A ∩ B) = 0

(b) For three events 

If A, B and C be any three events associated with a random experiment, then 

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C) 

Corollary 1: If A, B and C are mutually exclusive events, then 

P(A ∩ B) = P(B ∩ C) = P(A ∩ C) = P(A ∩ B ∩ C) = 0 

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C)

Note : (i) If A and B are two events such that A ⊆ B, then P(A) ≤ P(B) 

(ii) If E is an event associated with a random experiment, then 0 P(E) ≤ 1

P(B|A) = \(\frac{1}{2}\), P(A ∩ B) = \(\frac{1}{5}\), P(A) = ?

P(B|A) = \(\frac{P(A∩B)}{P(A)}\)

∴ 12 = \(\frac{\frac{1}{5}}{P(A)}\)

∴ \(\frac{1}{2}p(A) = \frac{1}{5}\)

∴ P(A) = \(\frac{1}{5}\times2=\frac{2}{5}\)

3P(A) = 2P(B) = 0.12

∴ 3P(A) = 0.12 and 2P(B) = 0.12

∴ P(A) = \(\frac{0.12}{3}\)

= 0.04

and

P(B) = \(\frac{0.12}{2}\)

= 0.06

Now, A and B are independent events.

∴ P(A ∩ B) = P(A) – P(B)

= 0.04 × 0.06

= 0.0024