Probability

A coin can either give a head or a tail

So, when 1 coin is tossed once the sample space=2

when the coin is tossed 3 times sample space =2³=8

The sample space can be found by relative combination of the events

Let H denote the event of a head and T denote the event of a tail

We define the possible outcomes by an ordered set (x, y, z)

where

x denotes the outcomes when the coin is tossed for the first time

y denotes the outcomes when the coin is tossed for the second time

z denotes the outcomes when the coin is tossed for the third time

The sample spaces are:

S={(H,H,H) , (H,H,T) , (H,T,H) , (T,H,H) , (H,T,T) , (T,H,T) , (T,T,H) , (T,T,T)}

Let 1,2,3,4,5,6 denote the event the respective numbers comes when the die is thrown

The total number of sample space =(6×6)=36

We define the possible outcomes by an ordered set (x, y)

where

x denotes the outcome of the first time the die is thrown

y denotes the outcome of the second time the die is thrown

The sample spaces

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)}

Let H denote the event of a head and T denote the event of a tail

Total number of sample space=(2⁴)=16

We define the possible outcomes by an ordered set (w,x, y, z)

w denotes the event the coin is tossed for the first time

x denotes the event the coin is tossed for the second time

y denotes the event the coin is tossed for the third time

z denotes the event the coin is tossed for the fourth time

Sample space

S={(H,H,H,H),(H,H,H,T),(H,H,T,H),(H,T,H,H),(T,H,H,H),(H,H,T,T),(H,T,H,T),(T,H,H,T),(T,H,T,H),(T,T,H,H),(H,T,T,H),(H,T,T,T),(T,H,T,T),(T,T,H,T),(T,T,T,H),(T,T,T,T)}

Let H denote the event of a head and T denote the event of a tail,

1,2,3,4,5,6 denote the event the respective numbers come when the die is thrown

Total Number of space =(2×6)=12

We define the possible outcomes by an ordered set (x, y)

x denotes the first event when the coin is tossed

y denotes the second event when the die is thrown

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

Let H denote the event of a head and T denote the event of a tail ,

1,2,3,4,5,6 denote the event the respective numbers comes when the die is thrown

The problem can be solved by breaking it into two cases

Case 1: Head is encountered

We define the possible outcomes by an ordered set (x, y)

x denotes the first event when the head is encountered

y denotes the second event when the die is thrown

Sample Space S₁={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)}

Case 2: Tail is encountered

Sample Space S₂={(T)}

The overall sample space S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T)}

Let X denote the event Room X is selected, Y denote the event Room Y is selected

B1,B2 denote the event a boy is selected from room X ,B3 denote the event a boy is selected from Room Y and G1,G2 denote the event a girl is selected from Room X andG3,G4,G5 denote the event a girl is selected among the three girls from Room Y

The problem is solved by dividing into two cases

Case 1: Room X is selected

We define the possible outcomes by an ordered set (x , y)

x denotes the event Room X is chosen

y denotes the second event a student is selected

Sample Space S₁={(X,B1),(X,B2),(X,G1),(X,G2)}

Case 2: Room Y is selected

We define the possible outcomes by an ordered set (x , y)

x denotes the event Room Y is chosen

y denotes the second event a student is selected

Sample Space S₂={(Y,B3),(Y,G3),(Y,G4),(Y,G5)}

The overall sample space

S={(X,B1),(X,B2),(X,G1),(X,G2),(Y,B3),(Y,G3),(Y,G4),(Y,G5)}

Let R denote the event the red die is chosen W denote the event the white die is chosen B denote the event the Blue die is chosen

1,2,3,4,5,6 denote the event the respective numbers come when the die is thrown

We define the possible outcomes by an ordered set (x , y)

x denotes the first event a die is chosen

y denotes the second event the die is thrown and a number comes up

Total number of possible outcomes=(6×3)=18

The sample space of the event is

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

Let B be the event a boy is born, G be the event a girl is born

(i) We define the possible outcomes by an ordered set (x , y)

x denotes the event the first child is born

y denotes the event the second child is born

Sample space S={(G,G),(G,B),(B,G),(B,B)}

(ii)When there are two child in the family then only three possible cases are possible, 2 Girl child, 1 Girl Child or no Girl Child

Sample Space S= {2, 1, 0}

Let R be the event a red ball is drawn and W be the event a white ball is drawn

Since all the white balls are identical so the event of drawing any one of the three-white ball is same.

Total Number of Sample space =(2²-1)=3 (Since both the balls cannot be red)

We define the possible outcomes by an ordered set (x, y)

x denotes drawing the first ball

y denotes drawing the second ball

Sample space S={(W,W),(W,R),(R,W)}

Let H denote the event of a head and T denote the event of a tail,

1,2,3,4,5,6 denote the event the respective numbers come when the die is thrown

The problem can be solved by breaking it into two cases

Case 1: Head is encountered

We define the possible outcomes by an ordered set (x, y)

x denotes the first event when the head is encountered

y denotes the second event the coin is tossed again

Sample space S₁={(H,T),(H,H)}

Case 2: Tail is encountered

We define the possible outcomes by an ordered set (x, y)

x denotes the first event when the tail is encountered

y denotes the second event when the die is thrown

Sample Space S₂={(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

The Overall Sample space

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

Let D denote the event the bulb is defective, N denote the event the bulb is non-defective

Total Number of sample space=(2×2×2)=8

We define the possible outcomes by an ordered set (x, y, z)

where

x denotes the event first bulb is selected

y denotes the event second bulb is selected

z denotes the event third bulb is selected

Sample space S={(D,D,D), (D,D,N), (D,N,D), (N,D,D), (D,N,N), (N,D,N), (N,N,D), (N,N,N)}

Let H denote the event of a head and T denote the event of a tail,

1,2,3,4,5,6 denote the event the respective numbers come when the die is thrown

The problem can be solved by dividing it into 3 cases

Case 1: Head is encountered and the corresponding number on the die is Odd

Total number of sample space=(1×3)=3

We define the possible outcomes by an ordered set (x , y)

x denotes the first event the head is encountered

y denotes the second event the number on the die is Odd

Sample space S₁={(H,1),(H,3),(H,5)}

Case 2: Head is encountered and the corresponding number is even

Total number of sample space=(1×3×6)=18

We define the possible outcomes by an ordered set (x, y, z)

where

x denotes the outcomes when the coin is tossed and head is encountered

y denotes the outcomes when the die is tossed and the number encountered is even

z denotes the outcomes when the die is tossed for the second time

Sample Space

S₂={(H,2,1),(H,2,2),(H,2,3),(H,2,4),(H,2,5),(H,2,6),(H,4,1),(H,4,2),(H,4,3),(H,2,4),(H,4,5),(H,4,6), (H,6,1),(H,6,2),(H,6,3),(H,6,4),(H,6,5),(H,6,6)}

Case 3: Tail is encountered

Total number of sample space=1

Sample space S₃={(T)}

The overall sample spaces

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

Let 1, 2, 3, 4 denote the event 1, 2, 3, 4 numbered slip is drawn

When two slips are drawn without replacement the first event has 4 possible outcomes and the second event has 3 possible outcomes

So the total number of possible outcomes=(4×3)=12

We define the possible outcomes by an ordered event (x , y)

x denotes the first event the first slip is drawn

y denotes the second event the second slip is drawn

The sample space S={(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)}

Let 1,2,3,4,5,6 denote the event the respective numbers comes when the die is thrown

H denote the event of a head and T denote the event of a tail when coin is tossed

The following problem can be divided in two cases

Case 1: Even number shows up in the die

We define the possible outcomes by an ordered set (x , y)

x denotes the first event even number shows up

y denotes the second event a coin is thrown

The sample space S₁={(2,H),(4,H),(6,H),(2,T),(4,T),(6,T)}

Case 2: Odd number shows up in the die

We define the possible outcomes by an ordered set (x, y, z)

x denotes the first event odd number shows up in the die

y denotes the second event the coin is thrown for first time

z denote the third event the coin is thrown for second time

The sample space

S₂={(1,H,H),(3,H,H),(5,H,H),(1,H,T),(3,H,T),(5,H,T),(1,T,H),(3,T,H),(5,T,H),(1,T,T),(3,T,T),(5,T,T)}

Therefore, the overall sample space for the problem= S₁+ S₂

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

Let H denote the event of a head and T denote the event of a tail when coin is tossed

R₁, R₂ denote the event the red balls are drawn B₁, B₂, B₃ denote the events black ball are drawn

1,2,3,4,5,6 denote the event the respective numbers come when the die is thrown

The following problem can be divided in two cases

Case 1: Tail turns up in the coin

We define the possible outcomes by an ordered set (x , y)

x denotes the first event the coin is tossed and tails turns up

y denotes the second event a ball is drawn

The sample space S₁={(T,R₁),(T,R₂),(T,B₁),(T,B₂),(T,B₃)}

Case 2: Head turns up in the coin

We define the possible outcomes by an ordered set (x , y)

x denotes the first event the coin is tossed and head turns up

y denotes the second event the die is thrown

The sample space S₂={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)}

Therefore, the overall sample space for the problem= S₁+ S₂

S={( T,R₁),(T,R₂),(T,B₁),(T,B₂),(T,B₃),(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)}

Let 1,2,3,4,5,6 denote the event the respective numbers comes when the die is thrown

Since this is a continuous event and doesn’t stop until six is found so the sample space is continuous in nature.

The six may come up on the very first throw or the second or the third and this goes on continuously until six comes

The sample space when 6 comes on very first throw={6}

The sample space when 6 comes on second throw ={(1,6),(2,6),(3,6),(4,6),(5,6)}

This event can go on for infinite times hence the sample space is infinitely defined

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

When the die is rolled the sample space will consist of following outcomes

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

According to the given conditions

E = (4) and

(F) = (2, 4, 6)

E ∩ F = (4) ∩ (2, 4, 6)

⇒ (4) ≠ ϕ as there is a common element between E & F

Hence E and F are not mutually exclusive event.

When the die is rolled the sample space will consist of following outcomes

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

According to the given conditions

(i) A: a number less than 7

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

(ii) B: a number greater than 7

B= (ϕ) as there is no number greater than 7 on the die.

(iii) C: a multiple of 3

C= (3, 6)

Only two numbers are multiple of 3 in the given sample space.

(iv) D: a number less than 4

D= (1, 2, 3)

(v) E: an even number greater than 4

E = (6)

(vi) F: a number not less than 3

F= (3, 4, 5, 6)

Since a pair of dice is thrown, all the possible outcomes or sample space (S) will be as follows

A: the sum is greater than 8

Possible sum greater than 8 are 9, 10, 11 & 12

∴ A=

B: 2 occurs on either die

Here possibilities are there that the number 2 will come on first die or second die or both the die simultaneously

B=

C: The sum is at least 7 and multiple of 3

So the sum can be only 9 or 12

C=

For 2 elements to be mutually exclusive, then there should not be any common element amongst them

(i) A∩ B =ϕ

As there is no common element in A and B

So A& B are mutually exclusive

(ii) B ∩ C = ϕ

Since there is no common element between

So B and C are mutually exclusive.

(iii) A ∩ C

⇒ ≠ ϕ

Since A and C has common elements,

∴ A and C are mutually exclusive.

Here, three coins are tossed once so the possible outcomes or sample space (S) consist of

S=

Now,

A: ‘three heads’

A= (HHH)

B: “two heads and one tail”

B= (HHT, THH, HTH)

C: ‘three tails’

C= (TTT)

D: a head shows on the first coin

D= (HHH, HHT, HTH, HTT)

(i) Mutually exclusive

A ∩ B = (HHH) ∩ (HHT, THH, HTH)

= ϕ

Since there is no common element between A&B,

they are mutually exclusive

A ∩ C = (HHH) ∩ (TTT) = ϕ

Since here is no common element,

So, A and C are mutually exclusive.

A ∩ D = (HHH) ∩ (HHH, HHT, HTH, HTT)

= (HHH) ≠ ϕ

Since there is a common element in A and D

So they are not mutually exclusive

B ∩ C = (HHT, HTH, THH) ∩ (TTT) = ϕ

Since there is no common element in B & C, so they are mutually exclusive.

B∩ D = (HHT, THH, HTH) ∩ (HHH, HHT, HTH, HTT)

= (HHT, HTH) ≠ ϕ

Since there are common elements in B & D,

so, they not mutually exclusive.

C ∩ D = (TTT) ∩ (HHH, HHT, HTH, HTT) = ϕ

Since there is no common element in C & D,

So they are not mutually exclusive.

(ii) Simple event

An event is said to be simple if it has only one sample point in its sample space.

Here A = (HHH)

C = (TTT)

Both A & C have only one element,

so they are simple events.

(iii) Compound events

Is an event has more than one sample point of a sample space , it is called as compound event.

Here B= (HHT, HTH, THH)

D= (HHH, HHT, HTH, HTT)

Both B & D have more than one element,

So, they are compound events.

since 3 coins are tossed, the sample space will consist of following possibilities

S= (HHH, HHT, HTH, HTT, THH,THT, TTH, TTT)

(i) Two events which are mutually exclusive.

Let A be the event of getting only head

A= (HHH)

And let B be the event of getting only Tail

B=(TTT)

So A ∩ B = ϕ

Since there is no common element in A& B so these two are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive

Now,

Let A be the event of getting exactly two tails

A= (HTT, TTH,THT)

Let B be the event of getting at least two heads

B = (HHT, HTH, THH, HHH)

Let C be the event of getting only one tail

C= (TTT)

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

= ϕ

Since there is no common element in A and B hence they are mutually exclusive

B ∩ C = (HHT, HTH, THH, HHH) ∩ (TTT)

= ϕ

Since there is no common element in B and C

So they are mutually exclusive.

A ∩ C =( HTT, TTH,THT) ∩ (TTT) = ϕ

Since there is no common element in A and C,

So they mutually exclusive

Now, since A & B, B& C and A& C are mutually exclusive

∴ A, B, and C are mutually exclusive.

Also,

A ⋃ B ⋃ C = (HTT, TTH,THT, HHT, HTH, THH, HHH, TTT) = S

Hence A, B and C are exhaustive events.

(iii) Two events , which are not mutually exclusive

Let a be the event of getting at least two heads

A = ( HHH, HHT, THH, HTH)

Let B be the event of getting only head

B= (HHH)

Now A ∩ B = (HHH) ≠ ϕ

Since there is a common element in A & B,

So they are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive

Let A be the event of getting only Head

A= (HHH)

Let b be the event of getting only tail

B = (TTT)

A ∩ B = ϕ

Since there is no common element in A & B,

These are mutually exclusive events.

But

A ⋃ B = (HHH) ∪ (TTT)

= ≠ S

Since A ∪ B ≠ S these are not exhaustive events.

(v) Three events which are mutually exclusive but not exhaustive

Let A be the event of getting only head

A = (HHH)

Let B be the event of getting only tail

B = (TTT)

Let C be the event of getting exactly two heads

C= (HHT, THH, HTH)

Now,

A ∩ B = (HHH) ∩ (TTT) = ϕ

A ∩ C = (HHH) ∩ (HHT, THH, HTH) = ϕ

B ∩ C = (TTT) ∩ (HHT, THH, HTH) = ϕ

⇒ A ∩ B = ϕ, A ∩ C = ϕ and B ∩ C = ϕ

i.e; they are mutually exclusive

also

A ∪ B ∪ C = (HHH TTT, HHT, THH, HTH) ≠ S

Since A ∪ B ∪ C ≠ S,

So, A, B and C are not exhaustive.

It`s is hence proved that A, B and C are mutually exclusive but not exhaustive.

Here 2 dice are thrown

So the sample space (S) will have following events

According to the given conditions

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5

Now

(ii) = A

(iv) A and B (A ∩ B) = ϕ

∴ A ∩ B’ ∩ C’ = A ∩ A ∩ C’ = A ∩ C’

since 2 dice are thrown the sample space will consist of following outcomes

Here

And

(i) A and B are mutually exclusive (A ∩ B) = ϕ

Since A and B has no common element,

So, A & B are mutually exclusive.

Thus, the given statement is true.

(ii) A and B are mutually exclusive and exhaustive

A ∪ B = = S

⇒ A ∪ B = S

And from (i) we have A and b are mutually exclusive

∴ A and B are mutually exclusive and exhaustive.

So, the statement is true.

(iii) A = B

Thus, the statement is true.

(iv) A and C are mutually exclusive

Here A ∩ C = ≠ ϕ

Since A and C have common element,

So, they are not mutually exclusive

So, the given statement is false.

(v) A and B’ are mutually exclusive.

A ∩ B’ = A ∩ A = A

∴ A ∩ B’ ≠ ϕ

So, A and B’ not mutually exclusive.

Hence, the given statement is false.

(vi) A’, B’, C are mutually exclusive and exhaustive.

Here A’ =

B=

And 𝐶 ={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(3,1),(3,2),(2,3),(4,1)}

A’ ∩ B’ = ϕ

Hence there is no common element in A’ and B’

So they are mutually exclusive.

B’ ∩ C = ≠ ϕ

Since there is common element between B’ and C

They are not mutually exclusive.

Now, since B’ and C are not mutually exclusive,

So A’, B’ and C are not mutually exclusive and exhaustive.

Hence the given statement is false.

a) Both the conditions of axiomatic approach hold true in the given assignment, that is

1) Each of the number p(w_i) is less than zero and is positive

2) Sum of probabilities is

0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

The given assignment is valid.

b) Both the conditions of axiomatic approach hold true in the given assignment, that is

1) Each of the number p (w_i) is less than zero and is positive

2) Sum of probabilities is

The given assignment is valid.

c) Both the conditions of axiomatic approach in the given assignment are

1) Each of the number p(w_i) is less than zero and is positive

2) Sum of probabilities is

0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1

So, the 2^nd condition is not satisfied

Which states that p(w_i) ≤ 1

The given assignment is not valid.

d) The conditions of axiomatic approach don’t hold true in the given assignment, that is

1) Each of the number p(w_i) is less than zero but also negative

To be true each of the number p(w_i) should be less than zero and positive

So, the assignment is not valid

e) Both the conditions of axiomatic approach in the given assignment are

1) Each of the number p(w_i) is less than zero and is positive

2) Sum of probabilities is

The second condition doesn’t hold true so the assignment is not valid.

Here the sample space is S= (TT, HH, TH, HT)

∴ Number of possible outcomes n (S) = 4

Let E be the event of getting at least one tail

∴ n (E) = 3

Here S = {1, 2, 3, 4, 5, 6}

∴n(S) = 6

(i) Let A be the event of getting a prime number,

A = {2, 3, 5} and n(A) = 3

(ii) Let A be the event of getting a number greater than or equal to 3,

Then A = (3, 4, 5, 6) and n(A) = 4

(iii) Let A be the event of getting a number less than or equal to 1,

Then A = (1) ∴ n (A) = 1

(iv) Let A be the event of getting a number more than 6, then

Then A = (0), ∴ n (A) = 0

(v) Let A be the event of getting a number less than 6, then

Then A= (1, 2, 3, 4, 5), ∴ n (A) = 5

(a) Number of points\events in the sample space = 52 (number of cards in deck)

(b) Let A be the event of drawing an ace of spades.

Here A= 1, n (A) = 1

(c) Let A be the event of drawing an ace. There are four aces.

∴ n (A)= 4

(d) Let A be the event of drawing a black card. There are 26 black cards.

∴ n (A) = 26

The coin and die are tossed together.

Let S be the sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5),(6, 6)}

n(S) = 12

(i) Let A be the event having sum of numbers as 3.

A = {(1, 2)}, ∴ n (A) = 1

(ii) Let A be the event having sum of number as 12.

Then A = {(6, 6)}, n (A) = 1

Here total members in the council = 4 + 6 = 10, so the sample space has 10 points

∴ n (S) = 10

Number of women are 6

Let A be the event of selecting a woman

Then n (A) = 6

Here the sample space is,

S = (HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH,

TTTH, TTHT, THTT, HTTT, TTTT)

According to the given condition, a person will win or lose money depending up on the face of the coin so,

(i) For 4 heads = 1 + 1 + 1 + 1 = Rs. 4 = he wins Rs. 4

(ii) For 3 heads and 1 tail = 1 + 1 + 1 – 1.50 = Rs. 1.50 = he will be winning Rs. 1.50

(iii) For 2 heads and 2 tails = 1 + 1 – 1.50 – 1.50 = - Rs. 1 = he will be losing Re. 1

(iv)For 1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50 = - Rs. 3.50 = he will be losing Rs. 3.50

(v) For 4 tails = – 1.50 – 1.50 – 1.50 – 1.50 = - Rs. 6 = he will be losing Rs. 6

Now the sample space of amounts is

S= {4, 1.50, 1.50, 1.50, 1.50, – 1, – 1, – 1, – 1, – 1, – 1, – 3.50, – 3.50, – 3.50, – 3.50, – 6}

∴ n (S) = 16

When three coins are tossed then S = {HHH, HHT, HTH, THH, TTH, HTT, TTT, THT}

Where s is sample space and here n(S) = 8

Let A be the event of getting 3 heads

n(A)= 1

Let A be the event of getting 2 heads

n (A) = 3

Let A be the event of getting at least 2 head

n(A) = 4

•let A be the event of getting at most 2 heads

n(A) = 7

•Let A be the event of getting no heads

n(A) = 1

(vi) Let A be the event of getting 3 tails

n(A) = 1

(vii) Let A be the event of getting exactly 2 tails

n(A) = 3

(viii) Let A be the event of getting no tails

n(A) = 1

(ix) Let A be the event of getting at most 2 tails

n(A) = 7

Given:

∴ P (not A) = 1 – P(A)

There are 13 letters in the word ‘ASSASSINATION’ which contains 6 vowels and 7 consonants.

The sample space here will have all the letters of the word ASSASSINATION

So here n(S) = 13

(i) Let A be the event of selecting a vowel

n(A) = 6

(ii) Let A be the event of selecting the consonant

n(A)= 7

Total numbers of numbers in the draw = 20 and numbers to be selected = 6

∴ n (S) =

Let A be the event that six numbers match with the six numbers fixed by the lottery committee.

= 1

Probability of winning the prize

=

(i) Here P(A) = 0.5, P(B) = 0.7 and

Now

Therefore, the given probabilities are not consistently defined.

(ii) P(A) = 0.5, P(B) = 0.4,

Therefore, the given probabilities are consistently defined.

(i) here

(ii) here

0.6= 0.35 + P(B) – 0.25

P(B) = 0.5

(iii) Here

Given:

Since A and B are mutually exclusive events.

∴P(A∪B) or P(A or B) = P(A) + P(B)

(i) Now P(E∪F)=P(A) + P(B)-P(A∩B)

Given: P(not E and not F) = 0.25

⇒1-P(E∩F)=0.25

⇒P(E∩F)=1-0.25=0.75 ≠0

∴ E and F are not mutually exclusive events.

Given: P(A) = 0.42, P(B)= 0.48 and P(A and B) = 0.16

(i)

(ii)

= 0.42 + 0.48 – 0.16 = 0.74

Let A be the event that the student is studying mathematics and B be the event that the student is studying biology.

And , is probability of studying both mathematics and biology.

Here, Probability of studying mathematics or biology will be given by P (A∪B)

is the probability that the student will studying mathematics or biology

Let A be the event that the student passes the first examination and B be the event that the students passes the second examination.

P ( is probability of passing at least one of the examination

Then, P(A∪B)= 0.95 , P(A)=0.8, P(B)=0.7

∴ P(A∪B)=P(A) + P(B)-P(A∩B)

0.95 = 0.8 + 0.7 - P(A∩B)

P(A∩B)=1.5-0.95=0.55

0.55 is the probability that student will pass both the examinations

Let A be the event that the student passes English examination and B be the event that the students passes Hindi examination.

Here given

Now,

0.9 = 0.75 + P(B) - 0.5

P(B) = 0.9 – 0.25 = 0.65

Given: Total number of students = 60

So the sample space consist of n(S) = 60

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Here being number of students who have opted for both NCC and NSS

(i) P(Student opted for NCC or NSS)

P (A or B) = P(A) + P(B) –P(A and B)

(ii) P(student opted neither NCC nor NSS)

P(not A and not B) =

(by De Morgan’s law)

(iii) P(student opted NSS but not NCC)

n(B - A) = n(B) – n (AB)

⇒ 32 – 24 = 8

The probability that the selected student has opted for NSS and not NCC is

No. of red marbles = 10

No. of blue marbles = 20

No. of green marbles = 30

Total number of marbles = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from 60 marbles = ⁶⁰C₅

(i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.

Number of ways of drawing 5 blue marbles from 20 blue marbles = ²⁰C₅

Probability that all marbles will be blue = ²⁰C₅ / ⁶⁰C₅

(ii) Number of ways in which the drawn marble is not green = ^(20+10)C₅

Probability that no marble is green = ³⁰C₅ / ⁶⁰C₅

Probability that at least one marble is green = 1 - ³⁰C₅ / ⁶⁰C₅

Number of ways of drawing 4 cards from 52 cards = ⁵²C₄

In a deck of 52 cards, there are 13 diamonds and 13 spades.

Number of ways of drawing 3 diamonds and one spade = ¹³C₃ × ¹³C₁

Thus, the probability of obtaining 3 diamonds and one spade

= (¹³C₃ × ¹³C₁)/ ⁵²C₄

Total number of faces = 6

(i) Number faces with number ‘2’ = 3

(ii) P (1 or 3) = P (not 2) = 1 − P (2)

(iii) Number of faces with number ‘3’ = 1

P(not 3) = 1 – P(3) =

Total number of tickets sold = 10,000

Number prizes awarded = 10

(i) If we buy one ticket, then

P (getting a prize) =

P (not getting a prize) =

(ii) If we buy two tickets, then

Number of tickets not awarded = 10,000 − 10 = 9990

P (not getting a prize) = ⁹⁹⁹⁰C₂ / ¹⁰⁰⁰⁰C₂

(iii) If we buy 10 tickets, then

Number of tickets not awarded = 10,000 − 10 = 9990

P (not getting a prize) = ⁹⁹⁹⁰C₁₀ / ¹⁰⁰⁰⁰C₁₀

My friend and I are among the 100 students.

Total number of ways of selecting 2 students out of 100 students =

(a) Let S = The two of us will enter the same section if both of us are among 40 students or among 60 students.

Number of ways in which both of us enter the same section = P(S)

P(S) = (⁴⁰C₂ + ⁶⁰C₂) / ¹⁰⁰C₂

(b) P(we enter different sections) = 1 − P(we enter the same section)

P(we enter different sections) =

Let L1, L2, L3 be three letters and E1, E2, and E3 be their corresponding envelops respectively.

Let L_iE_i denote i^th letter is inserted in i^th envelope

∴ Sample space is

L₁E₁, L₂E₃, L₃E₂,

L₂E₂, L₁E₃, L₃E₁,

L₃E₃, L₁E₂, L₂E₁,

L₁E₁, L₂E₂, L₃E₃,

L₁E₂, L₂E₃, L₃E₁,

L₁E₃, L₂E₁, L₃E₂,

∴ there are 6 ways of inserting 3 letters in 3 envelops.

And there are 4 ways in which at least one letter is inserted in proper envelope. (first 4 rows of sample space)

Probability that at least one letter is inserted in proper envelope =

It is given that P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35

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

⇒ P (A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88

∴ P (A ∪ B) = 0.88

(ii) A′ ∩ B′ = (A ∪ B)′ [by De Morgan’s law]

So, P (A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12

∴ P (A’ ∩ B’) = 0.12

(iii) P (A ∩ B′) = P(A) − P(A ∩ B) = 0.54 − 0.35 = 0.19

∴ P (A ∩ B’) = 0.19

(iv) We know that: P (B ∩ A′) = P(B) − P(A ∩ B)

⇒ P (B ∩ A′) = 0.69 – 0.35

∴ P (B ∩ A′) = 0.34

Given Data

Total no. of person = 5

No. of spokesperson who are male = 3

No. of spokesperson who are 35 years of age = 2

Let E be the event in which the spokesperson will be a male and F be the event in which the spokesperson will be over 35 years of age.

Accordingly, P(E) = 3/5 and P(F) = 2/5

Since there is only one male who is over 35 years of age,

∴

We know that: P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

⇒

∴ P(E ∪ F) = 4/5

Thus, the probability that the spokesperson will either be a male or over 35 years of age is 4/5.

(i) When the digits are repeated

Since four-digit numbers greater than 5000 are formed,

The thousand’s place digit is either 7 or 5. The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed.

Total number of 4-digit numbers greater than 5000 = 2 × 5 × 5 × 5 − 1

= 250 − 1 = 249

[In this case, 5000 cannot be counted; so, 1 is subtracted]

A number is divisible by 5 if the digit at its unit’s place is either 0 or 5.

∴ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2 × 5 × 5 × 2 − 1 = 100 − 1 = 99

Thus, the probability of forming a number divisible by 5 when the digits are repeated is = P (number divisible by 5 when digits repeated)

⇒

(ii) When repetition of digits is not allowed

The thousands place can be filled with either of the two digits 5 or 7 i.e. by 2 ways.

The remaining 3 places can be filled with any of the remaining 4 digits.

Total number of 4-digit numbers greater than 5000 = 2 × 4 × 3 × 2 = 48

When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits. Here, number of 4-digit numbers starting with 5 and divisible by 5 = 1 × 3 × 2 × 1 = 6

When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and the tens and hundreds of places can be filled with any two of the remaining 3 digits. Here, number of 4-digit numbers starting with 7 and divisible by 5 = 1 × 2 × 3 × 2 = 12

Total number of 4-digit numbers greater than 5000 that are divisible by 5

= 6 + 12 = 18

Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed

= P (number divisible by 5 when digits are not repeated)

⇒

The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9.

Number of ways of selecting 4 different digits out of the 10 digits = ¹⁰C₄

Now, each combination of 4 different digits can be arranged in 4! Ways.

Number of four digits with no repetitions = 4! × ¹⁰C₄ = 5040

There is only one number that can open the suitcase.

Thus, the required probability is