Probability

Given word is ALGORITHM

⇒ Total number of letters in algorithm = 9

∴ Total number of words = 9!

So, n(S) = 9!

If ‘GOR’ remain together, then we consider it as one group.

∴ Number of letters = 7

Number of words, if ‘GOR’ remain together in the order = 7!

So, n(E) = 7!

[∵ n! = n×(n – 1)×(n – 2)…1]

Total new employees = 6

So, they can be arranged in 6! ways

∴ n(S) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Two adjacent desks for married couple can be selected in 5 ways i.e. (1, 2), (2, 3), (3, 4), (4, 5), (5, 6)

Married couple can be arranged in the two desks in 2! ways

Other four persons can be arranged in 4! ways

So, number of ways in which married couple occupy adjacent desks

= 5 × 2! × 4!

= 5 × 2 × 1 × 4 × 3 × 2 × 1

= 240

So, number of ways in which married couple occupy non – adjacent desks = 6! – 240

= (6 × 5 × 4 × 3 × 2 × 1) – 240

= 720 – 240

= 480 = n(E)

We have integers 1, 2, 3,… 1000

∴ Total number of outcomes, n(S) = 1000

Number of integers which are multiple of 2 are 2, 4, 6, 8, 10,… 1000

Let p be the number of terms

We know that,

a_p = a + (p – 1)d

Here, a = 2, d = 2 and a_p = 1000

Putting the value, we get

2 + (p – 1)2 = 1000

⇒ 2 + 2p – 2 = 1000

⇒ p = 500

Total number of integers which are multiple of 2 = 500

Let the number of integers which are multiple of 9 be n.

Number which are multiples of 9 are 9, 18, 27,…999

∴ n^th term = 999

We know that,

a_n = a + (n – 1)d

Here, a = 9, d = 9 and a_n = 999

Putting the value, we get

9 + (n – 1)9 = 999

⇒ 9 + 9n – 9 = 999

⇒ n = 111

So, the number of multiples of 9 from 1 to 1000 is 111.

The multiple of 2 and 9 both are 18, 36,… 990.

Let m be the number of terms in above series.

∴ m^th term = 990

We know that,

a_m = a + (m – 1)d

Here, a = 9 and d = 9

Putting the value, we get

18 + (m – 1)18 = 990

⇒ 18 + 18m – 18 = 990

⇒ m = 55

Number of multiples of 2 or 9

= No. of multiples of 2 + no. of multiples of 9

– No. of multiples of 2 and 9 both

= 500 + 111 – 55

= 556 = n(E)

= 0.556

Number of outcomes when die is thrown = 6

(i) Given that 2 appears on the k^th roll of the die.

So, first (k – 1)^th roll have 5 outcomes each and k^th roll results 2

∴ Number of outcomes = 5^k-1

(ii) If we consider that 2 appears not later than k^th roll of the die, then 2 comes before k^th roll.

If 2 appears in first roll, number of ways = 1 outcome

If 2 appears in second roll, number of ways = 5 x 1 (as first roll does not result in 2)

If 2 appears in third roll, number of ways = 5 x 5 x 1 (as first two rolls do not result in 2)

Similarly, if 2 appears in (k – 1)^th roll, number of ways

= (5 x 5 x 5 … (k- 1) times) x 1

= 5^k-1

Possible outcomes if 2 appears before k^th roll

= 1 + 5 + 5² + 5³+ … + 5^k-1

Here, we get the series

We know that,

So, here a = 1

and

Hence,

Given that probability of odd numbers

= 2 × (Probability of even number)

⇒ P (Odd) = 2 × P (Even)

Now, P (Odd) + P (Even) = 1

⇒ 2P(Even) + P(Even) = 1

⇒ 3P(Even) = 1

So,

Now, Total number occurs on a single roll of die = 6

and the number greater than 3 = 4, 5 or 6

So, P(G) = P(number greater than 3)

= P(number is 4, 5 or 6)

Here, 4 and 6 are even numbers and 5 is odd

∴ P(G) = 2 × P(Even) × P(Odd)

Hence, the required probability is

Let

E₁ = Event that a family owns colour television

E₂ = Event that the family owns black and white television

Given that P (E₁) = 0.87

P (E₂) = 0.36

and P (E₁ ⋂ E₂) = 0.30

Now, we have to find the probability that a family owns either anyone or both kinds of sets.

By General Addition Rule, we have

P(A∪B) = P(A) + P(B) – P(A∩B)

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

= 0.87 + 0.36 – 0.30

= 1.23 – 0.30

= 0.93

Hence, the required probability is 0.93

Given that P (A) = 0.35 and P (B) = 0.45

∵ the events A and B are mutually exclusive then P(A ⋂ B) = 0

To find: (a) P (A′)

We know that,

P (A) + P (A’) = 1

⇒ 0.35 + P(A’) = 1 [given]

⇒ P(A’) = 1 – 0.35

⇒ P(A’) = 0.65

To find: (b) P (B′)

We know that,

P (B) + P (B’) = 1

⇒ 0.45 + P(B’) = 1

⇒ P(B’) = 1 – 0.45

⇒ P(B’) = 0.55

To find: (c) P (A ⋃ B)

We know that,

P(A∪B) = P(A) + P(B) – P(A∩B)

⇒ P (A ⋃ B) = 0.35 + 0.45 – 0 [given]

⇒ P (A ⋃ B) = 0.80

To find: (d) P (A ⋂ B)

It is given that A and B are mutually exclusive events.

∴ P (A ⋂ B) = 0

To find: (e) P (A ⋂ B’)

P (A ⋂ B’) = P (A) – P (A ⋂ B)

= 0.35 – 0

= 0.35

To find: (f) P (A’ ⋂ B’)

P (A’ ⋂ B’) = P (A ⋃ B)’

= 1 – P (A ⋃ B)

= 1 – 0.8 [from part (c)]

= 0.2

Let

E₁ = event that surgeries are rated as very complex

E₂ = event that surgeries are rated as complex

E₃ = event that surgeries are rated as routine

E₄ = event that surgeries are rated as simple

E₅ = event that surgeries are rated as very simple

Given: P(E₁) = 0.15, P(E₂) = 0.20, P(E₃) = 0.31, P(E₄) = 0.26, P(E₅) = 0.08

(a) P(complex or very complex) = P(E₁ or E₂) = P(E₁⋃ E₂)

By General Addition Rule:

P (A∪B) = P(A) + P(B) – P(A∩B)

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

= 0.15 + 0.20 – 0 [given]

[∵ All event are independent]

= 0.35

(b) P(neither very complex nor very simple) = P(E₁’ ⋂ E₅’)

= P(E₁⋃ E₅)’

= 1 – P(E₁⋃ E₅)

[∵By Complement Rule]

= 1 – [P(E₁) + P(E₅) – P(E₁⋂ E₅)]

[∵ By General Addition Rule]

= 1 – [0.15 + 0.08 – 0]

= 1 – 0.23

= 0.77

(c) P(routine or complex) = P(E₃⋃ E₂)

= P(E₃) + P(E₂) – P(E₃⋂ E₂)

[∵ By General Addition Rule]

= 0.31 + 0.20 – 0 [given]

= 0.51

(d) P(routine or simple) = P(E₃⋃ E₄)

= P(E₃) + P(E₄) – P(E₃⋂ E₄)

[∵ By General Addition Rule]

= 0.31 + 0.26 – 0 [given]

= 0.57

Given that A is twice as likely to be selected as B

i.e. P(A) = 2 P(B) …(i)

and C is twice as likely to be selected as D

i.e. P(C) = 2P(D) …(ii)

Now, B and C are given about the same chance

∴ P(B) = P(C) …(iii)

Since, sum of all probabilities = 1

∴ P(A) + P(B) + P(C) + P(D) = 1

⇒ P(A) + P(B) + P(B) + P(D) = 1 [from (iii)]

[from (i) & (ii)]

[from (iii)]

[from (i)]

⇒ 9P(A) = 4

(a) P(C will be selected) = P(C)

= P(B) [from(iii)]

[from(i)]

(b) P(A will not be selected) = P(A’)

= 1 – P(A)

[By complement Rule]

Given Sample Space, S = John promoted, Rita promoted, Aslam promoted, Gurpreet promoted

Let

E₁ = events that John promoted

E₂ = events that Rita promoted

E₃ = events that Aslam promoted

E₄ = events that Gurpreet promoted

It is given that chances of John’s promotion is same as that of Gurpreet

P(E₁) = P(E₄) …(i)

It is given that Rita’s chances of promotion are twice as likely as John

P(E₂) = 2P(E₁) …(ii)

and Aslam’s chances of promotion are four times that of John

P(E₃) = 4P(E₁) …(iii)

Since, sum of all probabilities = 1

⇒ P(E₁) + P(E₂) + P(E₃) + P(E₄) = 1

⇒ P(E₁) + 2P(E₁) + 4P(E₁) + P(E₁) = 1

⇒ 8P(E₁) = 1

…(iv)

(a) P (John promoted) = P(E₁)

[from (iv)]

P (Rita promoted) = P(E₂)

= 2P(E₁) [from (ii)]

[from (iv)]

P (Aslam promoted) = P(E₃)

= 4P(E₁) [from (iii)]

[from (iv)]

P (Gurpreet promoted) = P(E₄)

= P(E₁) [from (i)]

(b) Given A = (John promoted or Gurpreet promoted)

∴, A = E₁⋃ E₄

P(A) = P(E₁⋃ E₄)

= P(E₁) + P(E₄) – P(E₁⋂ E₄)

[∵ By general addition rule]

= P(E₁) + P(E₁) – 0 [from (i)]

Given P(A ⋂ B) = 0.07

From the given Venn Diagram

(a) P(A)

P(A) = 0.13 + 0.7 = 0.20

(b)

= 0.07 + 0.10 + 0.15 – 0.15

= 0.07 + 0.10

= 0.17

(c) P(A ⋃ B)

By General Addition Rule,

P(A ⋃ B) = P(A) + P(B) – P(A ⋂ B)

⇒ P(A ⋃ B) = 0.20 + (0.07 + 0.10 + 0.15) – 0.07

⇒ P(A ⋃ B) = 0.20 + 0.25

⇒ P(A ⋃ B) = 0.45

(d)

We know that,

= 0.20 – 0.07 [from part (a)]

= 0.13

(e) P(B ⋂ C)

P(B ⋂ C) = 0.15

(f) Probability of exactly one of the three occurs

P(exactly one of the three occurs) = 0.13 + 0.10 + 0.28

= 0.51

Given that one urn contains two black balls and one white ball

and second urn contains one black ball and two white balls

It is also given that one of the two urns is chosen, then a ball is randomly chosen from the urn, then second ball is chosen at random from the same urn without replacing the first ball

(a) Sample Space S = {B₁B₂, B₁W, B₂W, B₂B₁, WB₁, WB₂, W₁W₂, W₁B, W₂B, W₂W₁, BW₁, BW₂}

Total number of sample space = 12

(b) If two black balls are chosen

Total outcomes = 12

Favourable outcomes are B₁B₂, B₂B₁

∴ Total favourable outcomes = 2

We know that,

(c) If two balls of opposite colours are chosen

Favourable outcomes are B₁W, B₂W, WB₁, WB₂, W₁B, W₂B, BW₁, BW₂

∴ Total favourable outcomes = 8

and Total outcomes = 12

We know that,

Given that:

Number of red balls = 8

Number of white balls = 5

∴ Total balls, n = 13

It is given that 3 balls are drawn at random

⇒ r = 3

∴ n(S) = ⁿC_r = ¹³C₃

(a) All the three balls are white

We know that,

Total white balls are = 5

(b) All the three balls are red

We know that,

Total red balls are = 8

(c) One ball is red and two balls are white

We know that,

Given word is ASSASSINATION

Total number of letters in ASSASSINATION is 13

In word ASSASSINATION, there are 3A’s, 4S’s, 2I’s, 1T’s and 1O’s

Total number of ways these letters can be arranged =

(a) Four S’s come consecutively in the word

If 4S’s come consecutively then word ASSASSINATION become.

So, now numbers of letters is 1 + 9 = 10

(b) Two I’s and two N’s come together

So, now numbers of letters is 1 + 9 = 10

(c) All A’s are not coming together

Firstly, we find the probability that all A’s are coming together

If all A’s are coming together then

So, now numbers of letters is 1 + 10 = 11

Now, P(all A’s does not come together)

= 1 – P(all A’s come together)

(d) No two A’s are coming together

First we arrange the alphabets except A’s

There are 11 vacant places between these alphabets.

Total A’s in the word ASSASSINATION are 3

∴ 3 A’s can be placed in 11 place in ¹¹C₃ ways

∴ Total number of words when no two A’s together

Total number of playing cards = 52

∴ n(S) = 52

Total number of king cards = 4

Total number of heart cards = 13

Total number of red cards = 13 + 13 = 26

∴ Favourable outcomes = 4 + 13 + 26 – 13 – 2

[Here, we subtract 13 and 2 cards because already these cards come in king cards and heart cards]

= 28

We know that,

Given that:

S = {e₁, e₂, e₃, e₄, e_5, e₆, e₇, e₈, e₉}

A = {e₁, e₅, e₈} and B = {e₂, e₅, e₈, e₉}

P(e₁) = P(e₂) = .08, P(e₃) = P(e₄) = P(e₅) = .1

P(e₆) = P(e₇) = .2, P(e₈) = P(e₉) = .07

(a) To find: P(A), P(B) and P(A ⋂ B)

A = {e₁, e₅, e₈}

P(A) = P(e₁) + P(e₅) + P(e₈)

⇒ P(A) = 0.08 + 0.1 + 0.07 [given]

⇒ P(A) = 0.25

B = {e₂, e₅, e₈, e₉}

P(B) = P(e₂) + P(e₅) + P(e₈) + P(e₉)

⇒ P(B) = 0.08 + 0.1 + 0.07 + 0.07 [given]

⇒ P(B) = 0.32

Now, we have to find P(A ⋂ B)

A = {e₁, e₅, e₈} and B = {e₂, e₅, e₈, e₉}

∴ A ⋂ B = {e₅, e₈}

⇒ P(A ⋂ B) = P(e₅) + P(e₈)

= 0.1 + 0.07

= 0.17

(b) To find: P(A ⋃ B)

By General Addition Rule:

P(A ⋃ B) = P(A) + P(B) – P(A ⋂ B)

from part (a), we have

P(A) = 0.25, P(B) = 0.32 and P(A ⋂ B) = 0.17

Putting the values, we get

P(A ⋃ B) = 0.25 + 0.32 – 0.17

= 0.40

(c) A = {e₁, e₅, e₈} and B = {e₂, e₅, e₈, e₉}

∴ A ⋃ B = {e₁, e₂, e₅, e₈, e₉}

⇒ P(A ⋃ B) = P(e₁) + P(e₂) + P(e₅) + P(e₈) + P(e₉)

= 0.08 +0.08 + 0.1 + 0.07 + 0.07

= 0.40

(d) To find:

By Complement Rule, we have

= 0.68

Given: B = {e₂, e₅, e₈, e₉}

= 0.08 + 0.1 + 0.1 + 0.2 + 0.2 [given]

= 0.68

(a) When a fair die is thrown, the possible outcomes are

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

∴ total outcomes = 6

and the odd numbers are 1, 3, 5

∴ Favourable outcomes = 3

We know that,

(b) When a fair coin is tossed two times, the sample space is

S = {HH, HT, TH, TT}

∴ Total outcomes = 4

If at least one head appears then the favourable cases are HH, HT and TH.

∴ Favourable outcomes = 3

We know that,

(c) When a pair of dice is rolled, total number of cases

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)}

Total Sample Space, n(S) = 36

If sum is 6 then possible outcomes are (1,5), (2,4), (3,3), (4,2) and (5,1).

∴ Favourable outcomes = 5

We know that,

In a non-leap year, there are 365 days and we know that there are 7 days in a week

∴ 365 ÷ 7 = 52 weeks + 1 day

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

∴ Total Outcomes = 7

If this day is a Tuesday or Wednesday, then the year will have 53 Tuesday or 53 Wednesday.

∴P(non-leap year has 53 Tuesdays or 53 Wednesdays)

Hence, the correct option is (B).

Since, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), ..., (18,19,20)

Considering 3 numbers as a single digit

∴ the numbers will be 18

Now, we have to choose 3 numbers out of 20. This can be done in ²⁰C₃ ways

∴ n(S) = ²⁰C₃

The desired event is that the 3 numbers are choose must consecutive. So,

Hence, the correct option is (B).

We know that, in a pack of 52 cards 26 are of red colour and 26 are of black colour.

It is given that 2 cards are accidentally dropped

So,

[Here, we take 51 because already one card is dropped. So, we left with 51 cards]

Similarly,

Probability of dropping a black card first

Probability of dropping a black card second

∴ P(both cards of different colour)

Hence, the correct option is (C).

Given that 7 persons are to be seated in a row.

If two persons sit next to each other, then consider these two persons as 1 group.

Now we have to arrange 6 persons.

∴ Number of arrangement = 2! × 6!

Total number of arrangement of 7 persons = 7!

Hence, the correct option is (C).

We have digits 0, 2, 3, 5.

We know that, if unit place digit is ‘0’ or ‘5’ then the number is divisible by 5

If unit place is ‘0’

Then first three places can be filled in 3! ways = 3 × 2 × 1 × 1 = 6

If unit place is ‘5’

Then first place can be filled in two ways and second and third place can be filled in 2! ways = 2 × 2 × 1 × 1 = 4

∴ Total number of ways = 6 + 4 = 10 = n(E)

Total number of ways of arranging the digits 0, 2, 3, 5 to form 4 – digit numbers without repetition is 3 × 3 × 2 × 1 = 18

Hence, the correct option is (D).

Given that A and B are mutually exclusive events.

We know that,

When two events are mutually exclusive, then

P(A⋂B) = 0

By General Addition Rule:

P(A⋃B) = P(A) + P(B) – P(A⋂B)

⇒ P(A ⋃ B) = P(A) + P(B) – 0

⇒ P(A ⋃ B) = P(A) + P(B)

We know that, for all events A, B

0 ≤ P(A) ≤ 1

∴ P(A) + P(B) ≤ 1

⇒ P(A) ≤ 1 – P(B)

[By complement rule]

Hence, the correct option is (A).

We have, P(A ⋃ B) = P(A ⋂ B)

By General Addition Rule,

P(A) + P (B) – P(A⋂B) = P(A⋃B)

⇒ P(A) + P (B) – P(A ⋂ B) = P(A ⋂ B) [given]

⇒ [P(A) – P(A ⋂ B)] + [P(B) – P(A ⋂ B)] = 0

But P(A) – P(A ⋂ B) ≥ 0

and P(B) – P(A ⋂ B) ≥ 0

[∵ P(A ⋂ B) ≤ P(A) or P(B)]

⇒ P(A) – P(A ⋂ B) = 0

and P(B) – P(A ⋂ B) = 0

⇒ P(A) = P(A ⋂ B) …(i)

and P(B) = P(A ⋂ B) …(ii)

From (i) and (ii), we get

∴ P(A) = P(B)

Hence, the correct option is (A).

If all the girls sit together, then consider it as 1 group.

Total number of persons = 6 + 1 = 7 persons

∴ Total number of arrangements in a row of 7 persons = 7!

and the girls interchanges their seats in 6! ways.

Hence, the correct option is (C).

Total number of alphabet in the word probability = 11

Number of vowels in word Probability = 4 i.e. (O, A, I, I)

∴P(letter is vowel)

Hence, the correct option is (B).

Let E₁ be the event that A fails in an examination

and E₂ be the event that B fails in an examination

Then, we have

P (E₁) = 0.2 and P (E₂) = 0.3

Now, we have to find the P (either E₁ or E₂ fails)

∴ P(either E₁ or E₂ fails) = P(E₁) + P(E₂) – P(E₁⋂ E₂)

≤ P(E₁) + P(E₂)

≤ 0.2 + 0.3

≤ 0.5

Hence, the correct option is (C)

We have,

P( atleast one of the events A and B occurs) = 0.6

i.e. P(A ⋃ B) = 0.6

and P(A and B occur simultaneously) = 0.2

i.e. P(A ⋂ B) = 0.2

By General Addition Rule,

P(A) + P (B) – P(A⋂B) = P(A⋃B)

⇒ P(A) + P(B) – 0.2 = 0.6

⇒ P(A) + P(B) = 0.6 + 0.2

⇒ P(A) + P(B) = 0.8 …(i)

Now, by Complement Rule

P(A) = 1 – P(A’)

and P(B) = 1 – P(B’)

So, eq. (i) become

1 – P(A’) + 1 – P(B’) = 0.8

⇒ 2 – [P(A’) + P(B’)] = 0.8

⇒ 2 – 0.8 = P(A’) + P(B’)

⇒ P(A’) + P(B’) = 1.2

Hence, the correct option is (c)

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).

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

A be the event that typist will make 0 mistake in typing a report

B be the event that typist will make 1 mistake in typing a report

C be the event that typist will make 2 mistakes in typing a report

D be the event that typist will make 3 mistakes in typing a report

E be the event that typist will make 4 mistakes in typing a report

F be the event that typist will make 5 mistakes in typing a report

Then, we have

P(A) = 0.12, P(B) = 0.25, P(C) = 0.36, P(D) = 0.14, P(E) = 0.08 and P(F) = 0.11

We know that,

Sum of all probabilities = 1

∴ P(A) + P(B) + P(C) + P(D) + P(E) + P(F)

= 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11

= 1.06 > 1

Hence, the given statement is False.

Let E₁ be the event that A is selected in engineering college

and E₂ be the event that B is selected in engineering college

Then, we have

P(E₁) = 0.5, P(E₂) = 0.7 and P(E₁⋂ E₂) ≤ 0.3

Now, P(E₁) × P(E₂) ≤ 0.3

⇒ 0.5 × P(E₂) ≤ 0.3

⇒ P(E₂) ≤ 0.6

But P(E₂) = 0.7

Hence, the given statement is False.

Given A and B are two events.

We know that A ⋂ B ⊂ A

⇒ P(A ⋂ B) ≤ P(A)

Hence, the given statement is True.

Given:

P(occurrence of event A) = 0.7

P(occurrence of event B) = 0.3

and P(occurrence of both) = 0.4

We know that,

P(A⋂B) = P(A) × P(B)

[when A and B are independent events]

⇒ P(A ⋂ B) = 0.7 × 0.3

⇒ P(A ⋂ B) = 0.21

But given that P(A ⋂ B) = 0.4

Hence, the given statement is False.

Probability of each student getting distinction in their final examination is less than or equal to 1

Since, the two given events are not related to the given Sample Space

∴, The sum of the probabilities of two may be 1.2.

Hence, the given statement is True.

Let

A be the event that home team will win the football game

B be the event that home team will tie the game

and C be the event that they will lose the game

then, we have

P(A) = 0.77, P(B) = 0.08

Now, we have to find the P( losing the game) or P(C)

We know that,

Sum of all Probabilities = 1

∴ P(A) + P(B) + P(C) = 1

⇒ 0.77 + 0.08 + P(C) = 1

⇒ P(C) = 1 – (0.77 + 0.08)

= 0.15

Ans. The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is 0.15

Given: P(e₁) = 0.1, P(e₂) = 0.5, P(e₃) = 0.1

We know that,

Sum of all probabilities = 1

∴ P(e₁) + P(e₂) + P(e₃) + P(e₄) = 1

⇒ 0.1 + 0.5 + 0.1 + P(e₄) = 1

⇒ P(e₄) = 1 – 0.7

⇒ P(e₄) = 0.3

Ans. If e₁, e₂, e₃, e₄ are the four elementary outcomes in a sample space and P(e₁) =.1, P(e₂) = .5, P (e₃) = .1, then the probability of e₄ is 0.3

We have, S = {1, 2, 3, 4, 5, 6}

and E = {1, 3, 5}

Now, we have to find the

∴ {w: w Є S and w ∉ E}

= {2, 4, 6}

Ans. Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then is {2, 4, 6}

Given that: P(A) = 0.3, P(B) = 0.2 and P(A ⋂ B) = 0.1

We know that,

= 0.3 – 0.1

= 0.2

Given P(happening of an event A) = 0.5

and P(happening of an event B) = 0.3

It is also given that A and B are mutually exclusive events

⇒ P(A ⋂ B) = 0

Now, we have to find the P(neither A nor B)

= 1 – P(A ⋃ B)

= 1 – [P(A) + P(B)]

[Since, A and B are mutually exclusive]

= 1 – (0.5 + 0.3)

= 1 – 0.8

= 0.2

(a) 0.95 = very likely to happen, so it is close to 1

(b) 0.02 = very little chance of happening as the probability is very low

(c) -0.3 = an incorrect assignment because probability is never negative

(d) 0.5 = as much chance of happening as not because sum of chances of happening and not happening is one

(e) 0 = no chance of happening

(a) If E₁ and E₂ are the two mutually exclusive events

We know that,

When two events are mutually exclusive then

P(E₁⋂ E₂) = ϕ

Hence, (a) ↔ (iv)

(b) If E₁ and E₂ are the mutually exclusive and exhaustive events

Hence, (b) ↔ (iii)

(c) If E₁ and E₂ have common outcomes (E₁ – E₂) ⋃ (E₁⋂ E₂) = E₁

Hence, (c) ↔ (ii)

(d) If E₁ and E₂ are two events such that E₁⊂ E₂

⇒ E₁⋂ E₂ = E₁

Hence, (d) ↔ (i)