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In a single throw of 2 die, we have total 36(6 × 6) outcomes possible. 

Say, n(S) = 36 where S represents Sample space 

Let A denotes the event of getting a doublet. 

∴ A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{36}=\frac{1}{6}\) 

And B denotes the event of getting a total of 9 

∴ B = {(3,6), (6,3), (4,5), (5,4)}

P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{4}{36} = \frac{1}{9}\) 

We need to find probability of the event of getting neither a doublet nor a total of 9. 

P(A’ ∩ B’) = ? 

As, P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s theorem} 

P(A’∩B’) = 1 – P(A∪B) 

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that: 

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A∪B) = \(\frac{1}{6} +\frac{1}{9}+0\) = \(\frac{5}{18}\) 

{As P(A ∩ B) = 0 since nothing is common in set A and B ⇒ n(A∩B) = 0} 

Hence, 

P(A’∩B’) = 1 – \(\Big(\frac{5}{18}\Big)\)  = \(\frac{13}{18}\)

Let event A: The student will pass in French. 

∴ P(A) = 0.64 

Let event B: The student will pass in Sociology. 

∴ P(B) = 0.45 

Also, P(A ∩ B) = 0.40 

∴ Required probability 

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

= 0.64 + 0.45 – 0.40 

= 0.69

Here, P(A) = 3/8, P (B) = 1/2, P(A ∪ B) = 5/8

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

∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

= 3/8 + 1/2 - 5/8

= 1/4

(b) P(A’ ∩ B’) = P(A ∪ B)’ 

= 1 – P(A ∪ B)

= 1- 5/8

= 3/8

(c) P(A’ ∪ B’) = P(A ∩ B)’ 

= 1 – P(A ∩ B)

= 1- 1/4

= 3/4

We know, when a pair of dice is thrown, total possible outcomes are = 36

A=No of outcomes for getting sum of 8 are

{(2,6),(3,5),(4,4),(5,3),(6,2)} = 5

Therefore, P(A) = \(\cfrac5{36}\)

B = No of outcomes for at least one die does not show 5

= {(1,1),(1,2),(1,3),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,6),(3,1),(3,2),(3,3),(3,4),(3,6),(4,1),(4,2), (4,3),(4,4),(4,6),(6,1),(6,2),(6,3),(6,4),(6,6) } = 25

Hence P(B) = \(\cfrac{25}{36}\)

(A B) = outcomes of getting sum as 8 with at least one die not showing 5

= {(2,6),(4,4),(6,2)= 3

Hence P(A B) = \(\cfrac3{36}\)

Therefore, \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac3{36}}{\frac{25}{36}}=\cfrac3{25}\)

(a) \(\frac{13}{32}\)

The three boxes B₁, B₂ and B₂ contain the different coloured balls as follows :

There can be three mutually exclusive cases of drawing 2 blue balls and 1 red balls in the ways as given :

∴ P(Drawing 2 blue and 1 red ball)

= P(B₁ (Blue), B₂ (Blue), B₃ (Red)) + P(B₁ (Blue), B₂ (Red), B₃ (Blue)) + P(B₁ (Red), B₂ (Blue), B₃ (Blue))

= \(\frac{3}{4}\) x \(\frac{2}{4}\) x \(\frac{3}{4}\) + \(\frac{3}{4}\) x \(\frac{2}{4}\) x \(\frac{1}{4}\) + \(\frac{1}{4}\) x \(\frac{2}{4}\) x \(\frac{1}{4}\)

= \(\frac{18}{64}\) + \(\frac{6}{64}\) + \(\frac{2}{64}\) = \(\frac{26}{64}\) = \(\frac{13}{32}\).

Let E₁, E₂, E₃, …, E_n be n mutually exclusive events associated with a random experiment. If A is an event which occurs as a result of the events (cases) E₁, E₂, E₃, …, E_n then

\(P(\frac{Ei}{A})\) = Probability of occurrence of event A as a result of a particular cause (event) E_i 

\(\frac{P(E_i).P(A)}{\displaystyle\sum_{i=1}^n P(E_i).P(\frac{A}{E_i})}\)  i = 1,2,..., n

(c) \(\frac{1}{4}\)

Let A : Event of getting a tail on the coin 

B : Event of getting an even number on the die. 

Then, P(A) = \(\frac{1}{2}\)

P(B) = \(\frac{3}{6}\) = \(\frac{1}{2}\) as B = {2,4,6}

A and B being independent events, 

P(Getting tail on the coin and even number on the die)

= P(A ∩ B) = P(A) × P(B) = \(\frac{1}{2}\)x\(\frac{1}{2}\) =  \(\frac{1}{4}\).

Let A, B, C, D be the events defined as: 

A: Selecting the first box 

B: Selecting the second box 

C: Selecting the third box 

D: Event of drawing a blue ball. 

Since there are three boxes and each box has an equally likely chance of selection, P(A) = P(B) = P(C) = \(\frac{1}{3}\)

♦ If the first box is chosen, i.e., A has already occurred, then 

Probability of drawing a blue ball from A = \(\frac{4}{10}\)⇒ P(D/A) =\(\frac{4}{10}\)

♦ If the second box is chosen, i.e., B has already occurred, then 

Probability of drawing a blue ball from B =\(\frac{5}{10}\) ⇒ P(D/B) = \(\frac{5}{10}\)

Similarly P(D/C) = \(\frac{6}{10}\)

Now we are required to find the probability (B/D), i.e., given that the ball drawn is blue, we need to find the probability that it is drawn from the second box. 

By Baye’s Theorem,

 \(P(B/D) = \frac {P(B)\times P(D/B)}{P(A)\times P(D/A)+P(B)\times P(D/B) +P(C)\times P(D/C)}\)

 = \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}+ \frac{1}{3}\times \frac{6}{10}}\) = \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{15}{10}}\)= \(\frac {1}{3}\)

The integers from 1 through 11 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Out of these, there are 5 even and 6 odd integers. 

Let A: Both numbers chosen are odd 

B: Sum of numbers is even at random 

S: Choosing 2 numbers from 11 numbers. 

Then, n(S) = ¹¹C₂ 

n(A) = ⁶C₂ (∴  There are 6 odd integers) 

As the sum of both chosen integers can be even if both are even or both are odd, so 

n(B) = ⁶C₂ + ⁵C₂ 

and n(A ∩ B) = ⁶C₂ 

∴                               P(A) = \(\frac{n(A)}{n(S)} = \frac {^6C_2}{^{11}C_2}= \frac{6\times 5}{11 \times 10} =\frac{3}{11}\)

  P(B) = \(\frac{n(B)}{n(S)} = \frac {^6C_2+^5C_2}{^{11}C_2}= \frac{6\times 5+ 5\times 4}{11 \times 10} =\frac{5}{11}\)

\(P(A\,\cap\,B) = \frac{n(A\,\cap\,B)}{n(S)} =\frac{^6C_2}{^{11}C_2} =\frac{5}{11}\)

∴ P(A/B) = \(\frac {P(A\,\cap\,B)}{P(B)}=\frac{\frac{3}{11}}{\frac{5}{11}} =\frac{3}{5}\)

Let A: Getting two heads 

B: At least one coin showing a head. 

S = {HH, HT, TH, HH} 

Then, A = {HH}, B = {HT, TH, HH} ⇒ A ∩ B = {HH} 

∴ P(A) = \(\frac{n(A)}{n(S)} =\frac{1}{4}\) , P(B) = \(\frac{n(B)}{n(S)} =\frac{3}{4}\),\(P(A\,\cap\,B) = \frac{n(A\,\cap\,B)}{n(S)} = \frac{1}{4}\)

Now, Required probability = P(A/B) = \(\frac{P(A\,\cap\,B)}{P(B)} = \frac {\frac{1}{4}}{\frac{3}{4}}\) = \(\frac{1}{3}\)

i) Even numbers in outcomes = 2, 4, 6 = 3 

∴ Required probability = 4/6 = 2/3

ii) Prime numbers = 1, 2, 3, 5 = 4 

∴ Required probability = 4/6 = 2/3

iii) Getting ‘7’ on top = 0 

∴ This is impossible event.

iv) Sum of the factors of that number is twice to that numbers = 6

∴ Required probability = 1/6

The sample space of an experiment is, 

5 = {1,2,3,4,5,6} 

Here, E = {4} and F = {2,4,6} 

Now, E ∩F = {4} ≠ ϕ 

⇒ E and F are not mutually exclusive.

Faces on the dice = 6 

∴ number of total possible outcomes = 6 

i) List of possible out comes less than five = 1, 2, 3, 4 

So number of favourable outcomes for less than five = 4 

∴ Probability of getting a number less than five = 4/6 = 2/3

ii) List of possible outcomes for getting more than five = (6) 

∴ Number of possible outcomes for getting more than five = 1 

∴ Probability of getting more than five = 1/6

Sample space, 

S = {10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44} 

∴ n(S) = 20 

i. Let A be the event that the number so formed is a prime number.

∴ A = {11,13,23,31,41,43} 

∴ n(A) = 6 

∴ P(A) = \(\frac{n(A)}{n(S)}\) = 6/20

∴ P(A) = 3/10

ii. Let B be the event that the number so formed is a multiple of 4. 

∴ B = {12,20,24,32,40,44}  

∴ n(B) = 6 

∴ P(B) = \(\frac{n(B)}{n(S)}\) = 6/20

∴ P(B) = 3/10

iii. Let C be the event that the number so formed is a multiple of 11. 

∴ C = {11,22,33,44} 

∴ n(C) = 4

∴ P(C) = \(\frac{n(C)}{n(S)}\) = 4/20

∴ P(C) = 1/5

∴ P(A) = 3/10, P(B) = 3/10, P(C) = 1/5

Sample space, 

S = {(0, 0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2), (3.0), (3,1), (3,2), (4.0), (4,1), (4,2), (5.0), (5,1), (5,2) }

∴ n(S) = 36 

Let A be the event that the product of digits on the upper face is zero. 

∴ A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1,0), (2, 0), (3,0), (4, 0), (5,0)} 

∴ n(A) = 11 

∴ P(A) = \(\frac{n(A)}{n(S)}\)

∴ P(A) = 11/36

∴ The probability that the product of the digits on the upper face is zero is 11/36.

Correct option is: A) \(\frac{7}{8}\)

When three coins are tossed simultaneously once.

Then total possible outcomes are

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

\(\therefore\) Total possible outcomes is n (S) = 8

Favourable outcomes to the even of getting at least one tail are

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

\(\therefore\) Total No of favourable outcomes = 7

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

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

= \(\frac 78\)

Hence, probability of getting at least one tail is \(\frac 78\)

Correct option is: A) \(\frac{7}{8}\)

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

No. of favourable outcomes = 7

(i) P(at least one tail) = 7/8

No. of favourable outcomes = 1

(ii) P(no tail) = 1/8