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Let E: First bag is selected

(a) P(E) = \(\frac{1}{2}\)

F: Second bag is selected G: Ball is red 

(b) P(F) = \(\frac{1}{2}\)

G: Ball is Red 

(c) P (G/E) = P (red ball from I bag) = \(\frac{3}{4}\)

(d) P(G/F)= P(red ball from Il bag) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Let E₁ : first bag is selected, E₂ : second bag is selected

Then, E₁ and E₂ are mutually exclusive and exhaustive. Moreover,

P(E₁) = P(E₂) = 1/2

Let E : ball drawn is red.

P(E/E₁) = P(drawing a red ball from first bag) = 4/8 = 1/2

P(E/E₂) = P(drawing a red ball from second bag) = 2/8 = 1/4

By using Baye’s theorem,

Required probability = P(E₁/E) 

= (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂))

= (1/2 x 1/2)/(1/2 x 1/2 + 1/2 x 1/4) = (1/4)/(1/4 + 1/8) 

= (1/4)/(2 + 1)/8 = (1/4)/(3/8) = 2/3

Let E₁ : the event that the student knows the answer and E₂ : the event that the student guesses the answer. 

Then, E₁ and E₂ are mutually exclusive and exhaustive. Moreover, 

P(E₁) = 3/4 and P(E₂) = 1/4 

Let E : the answer is correct. The probability that the student answered correctly, given that he knows the answer, is 1 i.e., P P(E/E₁) = 1 

Probability that the students answered correctly, given that the he guessed, is 1/4 

i.e., P(E/E₂) = 1/4

By using Baye’s theorem, we obtain

P(E₁/E) = (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂))

= (3/4 x 1)/(3/4 x 1 + 1/4 x 1/4) = (3/4)/(3/4 + 1/16) = (12/16)/(12 + 1)/16 = 12/13

Let E₁ : the event that the student is residing in hostel and E₂ : the event that the student is not residing in the hostel. 

Let E : a student attains A grade, 

Then, E₁ and E₂ are mutually exclusive and exhaustive. Moreover,

P(E₁) = 60% = 60/100 = 3/5 and P(E₂) = 40% = 40/100 = 2/5

Then P(E/E₁) = 30% = 30/100 = 3/10 and P(E/E₂) = 20% = 20/100 = 2/10

By using Baye’s theorem, we obtain

P(E₁/E) = (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂))

= (3/5 x 3/10)/(3/5 x 3/10 + 2/5 x 2/10) = 9/(9 + 4) = 9/13

(i) We have, 

P( A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 1/3 + 1/5 - 1/15 = (5 + 3 - 1)/15 = 7/15

Given that P(A) = 2/11 if A is an event. 

To find: P (‘not A’) = P(A’) 

Since P(A’) = 1 – P(A)

= 1 - 2/11 = 9/11

In the word ASSASSINATION,we have A – 3 ; 

S – 4; I – 2; N – 2, O – 1 and T- 1. 

Total number of letters = 13. 

Among them, the vowels are: A – 3, I – 2, 0 – 1 is vowels are 6 in number. 

The consonants are: S – 4 ; N – 2, T-1. 

i.e., co-consonants are 7 in number. 

(i) P (a vowel) =  6/13

(ii) P (a consonant) = 7/13

(i) P(A) = 0.6 and P(B) = 0.7 and P(A ∩ B) = 0.6

Here, p(A n B) > P(A) and P(A ∩ B)< P(B)

P(A ∩ B) is not less than or equal to P(A) and P(B)

∴ P(A) and P(B), P(A n B) are not consistently defined.

(ii) We have, P(A ∪ B) = P(A) + P(B) – p(A ∩ B)

P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

= 0.5 + 0.5 – 0.8

= 0.9 – 0.8  = 0.1

∴ P(A ∩ S) = 01

Here, P(A ∩ B) is less than P(A) and P(B)

P(A) and P(B)P(A ∪ B) are consistently defined.

(Note: P(A ∪ B)<P(A) + P(B))

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).
However, here, P(A ∩ B) > P(A).
Hence, P(A) and P(B) are not consistently defined.
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).
Here, it is seen that P(A ∪ B) > P(A) and P(A ∪ B) > P(B). 

Hence, P(A) and P(B) are consistently defined.

(c) Drawing a diamond from a pack of cards/ Drawing an ace from a pack of cards

As we have an ace of diamonds, so drawing a diamond from a pack of cards also includes the possibility of drawing an ace. Hence the events are not mutually exclusive.

All numbers are different (given in question), this will be the same as picking r different objects from n objects which is ⁿc_r 

Here, n= 20 and r = 6(as we have to pick 6 different objects from 20 objects) 

Now we shall calculate the value of ²⁰C₆ = \(\frac{(20)!}{(20-6)!\times(6)!}\)as ⁿC_r = \(\frac{(n)!}{(n-r)!\times(r)!}\)  i.e. ²⁰C₆ = 38760 

Therefore, 38760 cases are possible, and in that only one them has prize, i.e. total no. of desired outcome is 1 

As we know,

Probability of occurrence of an event

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

Therefore, the probability of winning a prize is 

= \(\frac{1}{38760}\)

Conclusion: Probability of winning the prize in the game is \(\frac{1}{38760}\)

given: six numbers are chosen from 1-20

Formula: P(E) = \(\frac{favorable\ outcomes}{total\ possible\ outcomes}\) 

we have to find the probability of winning the prize 

total possible outcomes of selecting six numbers from 1-20 is ²⁰C₆ 

therefore n(S)= ²⁰C₆ = 38760 

let E be the event that all six numbers match with the given number (as winning number is fixed) 

n(E)=1 

probability of occurrence of the event is

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{1}{38760}\) 

Given, 

P(A) = 0.3, P(B) = 0.4, P(C) = 0.8 P(A ∩ B) = 0.08, P(A ∩ C) = 0.48, 

P(A ∩ B ∩ C) = 0. 09. 

We know– 

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

= 0.3 + 0.4 + 0.8 – 0.88 – p(B ∩ C) – 0.48 + 0.09 

= 0.23 – p(B ∩ C) 

Since 0 ≤ P(A ∪ B ∪ C) ≤ and P(A ∪ B ∪ C) ≥ 0.75

 ∴ P(B ∩ C) ≤ 0.48 

and P(B ∩ C) ≥ 0.23 

∴ P(B ∩ C) lies in the interval [0.23, 0.48]

We know that, 

Probability of occurrence of an event 

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

Total outcomes are 1, 2, 3, 4, 5, 6, and the desired outcomes are 2, 3 

Therefore, total no. of outcomes are 6, and total no. of desired outcomes are 2 

Probability of getting a 2 or 3 

= \(\frac{2}{6}\) 

= \(\frac{1}{3}\)

Conclusion: Probability of getting 2 or 3 when a die is thrown is \(\frac{1}{3}\)

Correct option is: A) \(\frac{1}{3}\)

When a dice is rolled, then possible outcomes are

{1, 2, 3, 4, 5, 6}

Total No of outcomes is n (S) = 6

Composite numbers from 1 to 6 are {4, 6}

Total No of composite numbers from 1 to 6 is 2

\(\therefore\) Probability of getting a composite number

= \(\frac {Total \,No\, of \,composite \,number}{Total \, No \, of\, Possible \, outcomes}\) 

= \(\frac 26 = \frac 13\)

Correct option is: A) \(\frac{1}{3}\)

We know that,

Probability of occurrence of an event

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

As 1, 3, 5 are odd numbers up to 6, so the desired outcomes are 1, 3, 5, and total outcomes are 1, 2, 3, 4, 5, 6 

Therefore, total no. of outcomes are 6, and total no. of desired outcomes are 3

Probability of getting an odd number

= \(\frac{3}{6}\) 

= \(\frac{1}{2}\) 

Conclusion: Probability of getting an odd number when a die is thrown is   \(\frac{1}{2}\) 

When we toss three coins, the sample space is– 

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

⇒ n(S) = 8. 

Let ‘A’ and ‘B’ be the event of getting almost two tails and atmost two heads respectively. 

∴ A = {HHH, HHT, HTH, HTT. THH, THT, TTH} 

⇒ n(A) = 7 

and B = {HHT, HTH, HTT, THH, THT, TTH, TTT} 

⇒ n(B) = 7

 A ∩ B = {HHT, HTH, HTT, THH, THT, TTH) 

⇒ n(A ∩ B) = 6

∴ P(A) = \(\frac{n(A)}{n(S)}= \frac{7}{8}\), P(B) = \(\frac{n(B)}{n(S)}= \frac{7}{8}\)

P(A ∩ B) = \(\frac{n(A∩B)}{n(S)}=\frac{6}{8}\)

We know– 

P(A ∪ B) = P(A) + P(B): P(A ∪ B)

= \(\frac{7}{8}+\frac{7}{8}- \frac{6}{8}\)

= \(\frac{8}{8}\)

= 1.

Correct option is: B)\(\frac 34\)

When two coins are tossed simultaneously, then total possible outcomes are

{HH, HT, TH, TT}

\(\therefore\) Total No of possible outcomes is n (S) = 4.

Fabourable outcomes to event of getting at least one head are {HH, HT, TH}.

\(\therefore\) Total No of favourable outcomes = 3

\(\therefore\) Probability of getting at least one head 

= \(\frac {Total \,No\, of \,favourable \,outcomes}{Total \,Possible \, outcomes}\) = \(\frac 34\)

Correct option is: B) \(\frac{3}{4}\)

(d) \(\frac{7}{30}\)

Triangular numbers between 1 and 30 are 1, 3, 6, 10, 15, 21, 27, i.e., 7 in number.

\(\therefore\) Reqd. probability = \(\frac{7}{30}\)

(b) \(\frac{1}{4}\)

The sample space for a simultaneous toss of two coins is 

S = { HH, HT, TH, TT} 

Favourable cases = {TT} 

\(\therefore\) P ( getting two tails ) =\(\frac{1}{4}\)

(c) \(\frac{1}{256}\)

Probability of drawing a blue ball

= \(\frac{4}{4+5+7}\) = \(\frac{4}{16}\)= \(\frac{1}{4}\)

\(\therefore\) Required probability = \(\frac{1}{4}\) x \(\frac{1}{4}\) x \(\frac{1}{4}\) x \(\frac{1}{4}\) = \(\frac{1}{256}\).

We know that,

Probability of occurrence of an event

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

As 2, 3, 5 are prime numbers up to 6, so the desired outcomes are 2, 3, 5, and total outcomes are 1, 2, 3, 4, 5, 6 

Therefore, total no. of outcomes are 6, and total no. of desired outcomes are 3

Probability of getting a prime number 

= \(\frac{3}{6}\) 

= \(\frac{1}{2}\)

Conclusion: Probability of getting a prime number when a die is thrown is \(\frac{1}{2}\)

(c) \(\frac{3}{4}\)

The sample space for a simultaneous toss of three coins 

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

Favourable cases = {HHT, HTT, THH, TTH, THT, HTH} 

\(\therefore\) P ( getting at least one head, one tail) =\(\frac{6}{8}\) =\(\frac{3}{4}\).

Given: three dices are thrown

Formula: P(E) = \(\frac{favourable \ outcomes}{total \ possible \ outcomes}\)

Total number of possible outcomes are 6³=216 

Therefore n(S)=216 

Let E be the event of getting total of 17 or 18 

E= {(6,6,5) (6,5,6) (5,6,6) (6,6,6)} 

n(E)=4

P(E) = \(\frac{n(E)}{n(S)}\) 

P(E) = \(\frac{4}{216}\) = \(\frac{1}{54}\)  

The correct option is: (A) 3 

Explanation:

 If is given that, total number of balls = 12 

Number of white balls = x

.'. Probability of getting a white ball = x/12

Now, 6 white balls are added. 

.'. Total number of balls = 12 + 6 = 18

Number of white balls = x + 6

.'. Probability of getting a white ball = x + 6/18

According to the question, (x + 6)/18 = 2 x x/12

=> x + 6 = 3x => x = 3 

.'. Number of white balls = 3

We know that,

Probability of occurrence of an event

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

By permutation and combination, total no. of ways to pick r objects from given n objects is ⁿC_r 

Now, total no. of ways to pick a ball from 9 balls is ⁹c₁ = 9

Our desired output is to pick a white ball. So, no. of ways to pick a white ball from 4 white balls (because the white ball can be picked from only white balls) is ⁴c₁ = 4 

Therefore, the probability of picking a white ball = \(\frac{4}{9}\)

Conclusion: Probability of picking a white ball from 4 white balls and 5 white balls is  \(\frac{4}{9}\) 

(b) \(\frac{1}{18}\)

Total number of exhaustive cases = 6 × 6 = 36 A total of 11 may be obtained in 2 ways as (5, 6), (6, 5).

\(\therefore\) P (total of 11) = \(\frac{2}{36}\)=\(\frac{1}{18}\).

The correct option is: (C)

Explanation:

 (i) Total number of outcomes = 6 

Numbers greater than 2 are 3,4, 5, and 6. 

Favorable outcomes = 4

.'. Required probability = 4/6 = 2/3

(ii) Total number of outcomes = 52 

Number of red cards = 26

.'. Probability of drawing a red card = 26/52 = 1/2

Number of face cards = 12

.'. Probability of drawing a face card = 12/52 = 3/13

(iii) Since, there is no red marble in the bag, so probability of drawing a red marble is zero.

We know that, 

Probability of occurrence of an event 

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

As 3, 6 are multiples up to 6, so the desired outcomes are 3, 6, and total outcomes are 1, 2, 3, 4, 5, 6 

Therefore, total no. of outcomes are 6, and total no. of desired outcomes are 2 

Probability of getting multiple of 3

= \(\frac{2}{6}\) 

= \(\frac{1}{3}\)

Conclusion: Probability of getting multiple of 3 when die is thrown is \(\frac{1}{3}\)

We know that,

Probability of occurrence of an event

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

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) 

Desired outcomes are (3, 6), (4, 5), (5, 4), (6, 3), (6, 5) , (5, 6) 

Total no. of outcomes are 36 and desired outcomes are 6

Therefore, probability of getting total equal to 9 or 11 = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Conclusion: Probability of getting total equal to 9 or 11, when two dice are rolled is \(\frac{1}{6}\)

We know that,

Probability of occurrence of an event

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

As 4, 5 are two numbers between 3 and, so the desired outcomes are 3, 6, and total outcomes are 1, 2, 3, 4, 5, 6 

Therefore, total no. of outcomes are 6, and total no.of desired outcomes are 2 

Probability of getting a number between 3 and 6

= \(\frac{2}{6}\) 

= \(\frac{1}{3}\)

Conclusion: Probability of getting a number between 3 and 6 when a die is thrown is  \(\frac{1}{3}\)

We know that, 

Probability of occurrence of an event

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

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) 

Desired outcomes are (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) 

Total no. of outcomes are 36 and desired outcomes are 10

Therefore, probability of getting total greater than 8 = \(\frac{10}{36}\) = \(\frac{5}{18}\) 

Conclusion: Probability of getting total greater than 8, when two dice are rolled is  \(\frac{5}{18}\) 

(i) We know that,

Probability of occurrence of an event

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

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 no. of outcomes are 36 

In that only (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) are our desired outputs as there sum is less than 6 

Therefore no. of desired outcomes are 10 

Therefore, the probability of getting a sum less than 6

= \(\frac{10}{36}\) 

= \(\frac{5}{18}\) 

Conclusion: Probability of getting a sum less than 6, when two dice are rolled is  \(\frac{5}{18}\) 

(ii) We know that, 

Probability of occurrence of an event

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

In (a, b) if a=b then it is called a doublet 

Total doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) 

In (a, b) if a=b and if a, b both are odd then it is called a doublet 

Odd doublets are (1, 1), (3, 3), (5, 5) 

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 no. of outcomes are 36 and desired outcomes are 3 

Therefore, probability of getting doublet of odd numbers

= \(\frac{3}{36}\) 

= \(\frac{1}{12}\) 

Conclusion: Probability of getting doublet of odd numbers, when two dice are rolled is  \(\frac{1}{12}\) 

(iii) We know that,

Probability of occurrence of an event

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

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 no. of outcomes are 36 

Desired outputs are (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) 

Total no. of desired outputs are 15 

Therefore, probability of getting the sum as a prime number

= \(\frac{15}{36}\) 

= \(\frac{5}{12}\) 

Conclusion: Probability of getting the sum as a prime number, when two dice are rolled is  \(\frac{5}{12}\) 

We know that,

Probability of occurrence of an event 

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

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) 

Desired outcomes are (4, 6), (5, 5), (6, 4) 

Total no. of outcomes are 36 and desired outcomes are 3 

Therefore, the probability of getting a total of 10 

=   \(\frac{3}{36}\)

=   \(\frac{1}{12}\)

Conclusion: Probability of getting total sum 10, when two dice are rolled is \(\frac{1}{12}\)

We know that, 

Probability of occurrence of an event 

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

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) 

Desired outcomes are (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) 

Total no. of outcomes are 36 and desired outcomes are 9 

Therefore, probability of getting number greater than 3 on each die

= \(\frac{9}{36}\) 

= \(\frac{1}{4}\)

Conclusion: Probability of getting a number greater than 3 on each die, when two dice are rolled is \(\frac{1}{4}\)

We know that,

Probability of occurrence of an event

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

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) 

Desired outcomes are (1, 6), (3, 6), (5, 6) 

Total no. of outcomes are 36 and desired outcomes are 3 

Therefore, probability of getting odd on the first die and 6 on the second die

= \(\frac{3}{36}\) 

= \(\frac{1}{12}\) 

Conclusion: Probability of getting odd on the first die and 6 on the second die, when two dice are rolled is  \(\frac{1}{12}\) 

P(A/B) = P(A ∩ B)/P(B)

= 0.32/0.5 = 32/50 

= 16/25

We know that,

Probability of occurrence of an event

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

Let T be tails and H be heads 

Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH 

Desired outcomes are exactly one tail. So, desired outputs are THH, HTH, HHT 

Total no. of outcomes are 8 and desired outcomes are 3 

Therefore, the probability of getting exactly one tail = \(\frac{3}{8}\) 

Conclusion: Probability of getting exactly one tail is \(\frac{3}{8}\)