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P(A) = 0.8, P(A ∩ B) = 0.25 are given.

∴ P (A – B) = P(A) – P(A ∩ B)

= 0.8 – 0.25 = 0.55

• Selection with Replacement: In any trial if the selection of a unit from the population is done by replacing the unit selected in the previous trial back to the population then it is called the selection with replacement.
• Selection without Replacement: In any trial if the selection of a unit from the population is done by not replacing the unit selected in the previous trial back to the population, then it is called the selection without replacement.

A and B are any two events of a finite sample space U. The event B is called independent of the event A, if the probability of occurrence of event A does not affect the probability of occurrence of the event B. This means, if P(A|B) = P(A) and P(B|A) = P(B), then A and B are called mutually independent events.

If A and B are independent events, then

• A and B’ are independent events.
• A’ and B are independent events.
• A’ and B’ are also independent events.

Statement I. If two events A and B are independent, then probability that they will both occur is equal to the product of their individual probabilities. 

i.e. P (A and B) = P (A) × P (B)               [‘AND’ rule] 

In set notation, P (A ∩ B) = P (A) × P (B) 

Note. If A, B, C are independent events, then 

P (A and B and C) =P (A) × P (B) × P (C) 

In set notation, P (A ∩ B ∩ C) = P (A) . P (B) . P (C). 

In general, if A₁, A₂, ..... A_n are n independent events, then 

P (A₁ and A₂ and A₃ and ..... A_n) = P(A₁) × P(A₂) × P(A₃) × ..... × P(A_n) 

In set notation, P(A₁ ∩ A₂ ∩ A₃ ..... ∩ A_n) = P(A₁) × P (A₂) ..... × P (A_n) 

Finding probabilities of simultaneous occurrence of two independent events. 

Method. Use the relation P (A ∩ B) = P (A) . P (B).

If A and B are two events in a random experiment such that P(A) ≠ 0 and P(B) ≠ 0, then the probability of the simultaneous occurrence of the events A and B i.e., P(A ∩ B) is given by:

\(P(A\,\cap\,B)=P(A)\times P(B/A) \,or\,P(A\,\cap\,B) = P(B)\times P(A/B) \)

(This follows directly from the formula given for conditional probability in Key Fact No. 1) 

Thus, the above-given formulae hold true for dependent events. 

Corollary 1: In the case of independent events, the occurrence of event B does not depend on the occurrence of A, hence P(B/A) = P(B).

∴    \(P(A\,\cap\,B) = P(A)\times P(B)\)

Thus, we can say if \(P(A\,\cap\,B) = P(A)\times P(B)\), then the events A and B are independent.

Also, If A and B are two independent events associated with a random experiment having a sample space S, then

(a) \(\overline{A}\) and B are also independent events. So, 

\(P(\overline{A}\,\cap\,B)= P(\overline{A})\times P(B)\)

(b) A and \(\overline{B}\) are also independent events, so, 

\(P(A\,\cap\,\overline{B}) =P(A)\times P(\overline{B})\)

(c) \(\overline{A}\)and \(\overline{B}\) are also independent events, so, 

\(P(\overline{A}\,\cap\,\overline{B})= P(\overline{A})\times P(\overline{B})\)

Corollary 2: If A₁, A₂, A₃, ..., A_n are n independent events associated with a random experiment, then 

\(P(A_1\,\cap A_2 \cap A_3,...,\cap A_n)= P(A_1)\times P(A_2)\times P(A_3) ...\times P(A_n)\)

Corollary 3: If A₁, A₂, A₃, ..., A_n are n independent events associated with a random experiment, then 

\(P(A_1\cup A_2 \cup A_3\dotsb \cup A_n) = 1-P(\overline{A_1})\times P(\overline{A_2})\times P(\overline{A_3})\times ... \times P(\overline{A_n})\)

Corollary 4: If the probability that an event will happen is p, the chance that it will happen in any succession of r trials is p^r . 

Also for the r repeated non-occurrence of the event we have the probability = (1 – p)^r.

A and B are independent events. So A’ and B’ are also independent events. P(A) = 0.5, P(B) = 0.7 are given.

∴ P(A’ ∩ B’) = P(A’) ∙ P(B’)

= [1 – P(A)] [1 – P(B)]

= [1 – 0.5] [1 – 0.7]

= [0.5 × 0.3]

= 0.15

A and B are any two events of a finite sample space U. The rule of obtaining the probability of the event AnB which corresponds to the occurrence of two events A and B simultaneously, is called the rule of multiplication of probability. This rule is written as under:

• P (A ∩ B) = P (A) . P (B | A)
  OR
  P (A ∩ B) = P (B) . P (A|B)
• For two independent events law of multiplication of probability is written as under:
• P(A ∩ B) = P(A) P(B)
• P (A’ ∩ B’) = P (A’) . P (B’)
• P(A ∩ B’) = P(A) – P(B’)
• P(A’ ∩ B) = P (A’) . P (B)

Probability = (number of favorable outcomes)/(Total number of outcomes)

A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls.

Total numbers of balls = 4 + 5 + 2 + 4 = 15

(i) Total numbers of black balls = Numbers of favorable outcomes= 2

P(getting a black ball) = 2/15

(ii)

Total numbers of green balls = 4

Numbers of non-green balls = 15 – 4 = 11

P(not green ball) = 11/15

(iii)

Total numbers of red and white balls = 5 + 4 = 9

P(red ball or white ball) = 9/15 = 3/5

(iv)

Total numbers of red and green balls = 5 + 4 = 9

Number of balls which are neither red nor green = 15 – 9 = 6

P (getting neither red nor green ball) = 6/15 = 2/5

We know that, 

If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is \(\frac{a}{a+b}\), similarly, if odds are not in the favor of the occurrence an event are a:b, then the probability of not occurrence of the event is \(\frac{a}{a+b}\) 

We also know that,

Probability of occurring = 1 - the probability of not occurring

 =  \(1-\frac{a}{a+b}\) 

=  \(\frac{b}{a+b}\)  

Given a = 4 and b = 7 

Probability of occurrence = \(\frac{7}{4+7}\) = \(\frac{7}{11}\)

Conclusion: Probability that the event occurs is   \(\frac{7}{11}\)

Total outcomes are (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) 

Total cases where sum will be 6 is (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) i.e. 5 

Probability of getting sum 6 = \(\frac{5}{36}\)

We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is \(\frac{a}{a+b}\)

Now we got   \(\frac{a}{a+b}\) =   \(\frac{5}{36}\)

So, a = 5 and a + b = 36 i.e. b = 31 

Therefore odds in the favor of getting the sum as 6 is 5:31

Conclusion: Odds in favor of getting the sum as 6 is 5:31

(ii) Total outcomes are (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) 

Total cases where sum will be 7 is (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) i.e. 6

Probability of getting sum 6 = \(\frac{6}{36}\) = \(\frac{1}{6}\) 

We know that, 

If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is \(\frac{a}{a+b}\)

Now we got \(\frac{a}{a+b}\) = \(\frac{1}{6}\)

So, a = 1 and a+b = 6 i.e. b = 5 

Therefore odds in the favor of getting the sum as 7 is 1:5 

Odds against getting the sum as 7 is b:a i.e. 5:1 

Conclusion: Odds against getting the sum as 7 is 5:1

We know that, 

If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is  \(\frac{a}{a+b}\) , which indirectly came from 

Probability of the occurrence of an event

= \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\)

Where, Total no. of desired outcomes = a, and total no. of outcomes = a + b 

Given a = 8, b= 13 

The probability that the event occurs 

= \(\frac{8}{8+13}\) 

= \(\frac{8}{21}\)

Conclusion: Probability that the event occurs is  \(\frac{8}{21}\) 

(i) We know that, 

If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is \(\frac{a}{a+b}\)

Given, probability 

= \(\frac{5}{14}\)

We know, probability =  \(\frac{a}{a+b}\)

So, \(\frac{a}{a+b}\)  = \(\frac{5}{14}\)

a = 5 and a+b = 14 i.e. b = 9

odds in favor of its occurrence = a:b = 5:9 

Conclusion: Odds in favor of its occurrence is 5:9 

(ii) As we solved in part (i), a = 5 and b = 9 

As we know, odds against its occurrence is b:a = 9:5 

Conclusion: Odds against its occurrence is 9:5

Let n(s) = 90 

A = {12,15 ……………… 99} 

n(A) = 30

P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{30}{90}\) = \(\frac{1}{3}\)

Using a + (n — 1)d = T_n 

12+ (n – 1)³ = 99 

12 + 3x – 3 = 99 

3x = 99 – 9 

3x = 90; x = 30