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Correct option is (B) equally likely

When an unbiased coin is tossed then

Probability of getting head is P(H) \(=\frac12\) and

Probability of getting tail is P(T) \(=\frac12.\)

Hence, getting a head when an unbiased coin is tossed is equally likely outcome.

B) equally likely

We know that the coin has two sides head (H) and tail (T)

So the possible outcomes are X^m. (where x is the number of outcomes when a coin is tossed and m is number of coins)

When there are 2 coins

∴ 2²= 4 i.e. head and tail

∴ The possible outcomes are HH, HT, TH, TT.

We know that the die has 6 faces 

So, they are 1, 2, and 3,4,5,6

∴ The possible outcomes are 1, 2, 3,4,5,6.

Correct option is (D) impossible

Possible outcome when a die is thrown are 1, 2, 3, 4, 5 and 6.

Hence, favourable outcome of event of getting a number more than 6 is n(E) = 0.

\(\therefore\) Probability of getting a number more than 6 is

\(P(E)=\frac{n(E)}{n(S)}\)

\(=\frac06=0\)

An event whose probability is 0 is known as an impossible event.

Hence, when a die is thrown, getting a number more than 6 is an impossible event.

D) impossible

(i) A coin has two sides a head(H) and a tail(T).

∴ There are X^m possible outcomes.

[Where X is number of outcomes when a coin is tossed and m is number of coins.]

That is 2¹ = 2 and they are H, T.

∴ All possible outcomes are H, T.

(ii) A coin has two sides a head(H) and a tail(T), and there are two such coins.

∴ There are X^m possible outcomes.

[Where X is number of outcomes when a coin is tossed and m is number of coins.]

That is 2² = 4 and they are HH, HT, TH, TT

∴ All possible outcomes are HH, HT, TH, TT.

(iii) A die has 6 faces and they are 1, 2, 3, 4, 5, 6

∴ All possible outcomes are 1, 2, 3, 4, 5, 6.

(iv) A deck of cards have a total of 52 cards.

∴ Number of possible outcomes are 52.

Correct option is (C) 6

Possible outcomes when a die is thrown are 1, 2, 3, 4, 5 and 6.

\(\therefore\) Number of possible outcomes when a die is thrown is 6.

Correct option is  C) 6

Total outcomes= 36

(i) Favourable outcomes are (2,2) (2,3) (2,5) (3,2)(3,3) (3,5) (5,2)(5,3) (5,5)  i.e., 9 outcomes

P(a prime number on each die)=9/36 or 1/4

(ii) Favourable outcomes are (3,6) (4,5) (5,4) (6,3) (5,6) (6,5) i.e.,  6 outcomes

P(a total of 9 or 11)=6/36 or 1/6

Correct option is (B) 2

When a coin is tossed there is only two outcomes are possible, either head occurs or tail occurs.

Hence, total outcomes = 2

Correct option is  B) 2

Total no. of possible outcomes when 2 dice are thrown = 6×6=36 which 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) }

E ⟶ event of getting sum on 2 dice as 2

No. of favourable outcomes = 1{(1, 1)}

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 1/36

E ⟶ event of getting sum as 3

No. of favourable outcomes = 2 {(1, 2) (2, 1)}

P(E) = 2/36

E ⟶ event of getting sum as 4

No. of favourable outcomes = 3 {(3, 1) (2, 2) (1, 3)}

P(E) = 3/36

E ⟶ event of getting sum as 5

No. of favourable outcomes = 4 {(1, 4) (2, 3) (3, 2) (4, 1)}

P(E) = 4/36

E ⟶ event of getting sum as 6

No. of favourable outcomes = 5 {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}

P(E) = 5/36

E ⟶ event of getting sum as 7

No. of favourable outcomes = 6 {(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)}

P(E) = 6/36

E ⟶ event of getting sum as 8

No. of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

P(E) = 5/36

E ⟶ event of getting sum as 9

No. of favourable outcomes = 4 {(3, 6) (4, 5) (5, 4) (6, 3)}

P(E) = 4/36

E ⟶ event of getting sum as 10

No. of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}

P(E) = 3/36

E ⟶ event of getting sum as 11

No. of favourable outcomes = 2 {(5, 6) (6, 5)}

P(E) = 2/36

E ⟶ event of getting sum as 12

No. of favourable outcomes = 1 {(6, 6)}

P(E) = 1/36

No, the outcomes are not equally likely from the above table we see that, there is different probability for different outcome.

Correct option is: D) 36

When two dice are thrown, the sample space is 

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)} 

∴ n(S) = 36

(a) A: Sum of the numbers on two dice is divisible by 3 or 4. 

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

(b) B: Sum of the numbers on two dice is 7. 

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

(c) C: Odd number on the first die. 

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

(d) D: Even number on the first die. 

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

(e) A and B are mutually exclusive events as A ∩ B = Φ. 

A ∪ B = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S 

∴ A and B are not exhaustive events as A ∪ B ≠ S. 

(f) C and D are mutually exclusive events as C ∩ D = Φ. 

C ∪ D = {(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)} = S 

∴ C and D are exhaustive events.

given: odds in favour of event is 2:3

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

we have to find the probability of occurrence of this event 

total possible outcomes is 2k + 3k = 5k  

therefore n(S) = 5k 

let E be the event that it occurs 

n(E) = 2k 

probability of occurrence is

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{2k}{5k}=\frac{2}{5}\) 

given: two men and two women

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

committee of two persons is to be formed from two men and two women, therefore total possible outcomes of selecting two persons is ⁴C₂ 

therefore n(S)=6 

(i) let E be the event that no man is in the committee 

E= {W, W} 

n(E)= ²C₂ = 1 (only woman)

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{1}{6}\) 

(ii) let E be the event that one man is present in committee 

E= {M W} 

n(E)= ²C₁ ²C₁ = 4

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{4}{6}=\frac{2}{3}\) 

(iii) let E be the event that two men is in the committee 

E= {M, M} 

n(E)= ²C₂ =1 (only men)

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{1}{6}\)

Total number of students = 40

Number of boys = 15

Number of girls = 25

(i) Let E₁ be the event of getting a girl's name of the cards

Therefore, P(selecting the name of a girl) = P(E₁) = \(\cfrac{number\,of\,outcomes\,favorable\,to\,E_1}{number\,of\,all\,possible\,outcomes}\) = \(\frac{25}{40}\) = \(\frac{5}{8}\)

Thus, the probability that the name written on the card is the name of a girls is \(\frac{5}{8}\).

(ii) Let E₂ be the event of getting a boy's name of the cards

Therefore, P(selecting the name of a boy) = P(E₂) = \(\cfrac{number\,of\,outcomes\,favorable\,to\,E_2}{number\,of\,all\,possible\,outcomes}\) = \(\frac{15}{40}\) = \(\frac{3}{8}\)

Thus, the probability that the name written on the card is the name of a boys is \(\frac{3}{8}\).

There are 40 students, and only one name card has to be chosen.

(i) The number of all possible outcomes is 40

The number of outcomes favourable for a card with the name of a girl = 25

Therefore, P (card with name of a girl) = P(Girl) =25/40=5/8

(ii) The number of outcomes favourable for a card with the name of a boy = 15 

Therefore, P(card with name of a boy) = P(Boy) =15/40=3/8

Note: We can also determine P(Boy), by taking

P(Boy) = 1 – P(not Boy) = 1 – P(Girl) =1-5/8=3/8

Total = 5 + 15 = 20, 

n(S) = 20c₂ = \(\frac{20\times19}{2\times1} = 190\)

Let A = both bulbs are non-defective 

n(A)= 15c₂= \(\frac{14\times15}{2\times 1} = 105\)

P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{105}{190}\) = \(\frac{21}{38}\)

As there are 3 + 2 + 4 = 9 persons 

So, 4 persons out of 9 can be drawn in ⁹C₄ ways = 126

Let E denote the event that there are exactly 

2 children in the selection. 2children out of 4 can be selected in ⁴C₂ = 6 ways 

And rest two persons can be male or female. So we will select 2 persons out of remaining 5.

∴ P(E) = \(\frac{4C_2\times5C_2}{126}\) = \(\frac{6\times10}{126}\) = \(\frac{10}{21}\)

Hence,

P(E) = \(\frac{10}{21}\)

As our answer matches only with option (c) 

∴ Option (c) is the only correct choice.

Sample Space = 40

Number of Girls = 25

Number of Boys = 15

(i) P(a girl) = 25/40

(ii) P(a boy) = 15/40

Probability of winning a game = 0.7

Let us say, P(A) = 0.7

Let B be probability of losing the game. Find: P(B)

We know that, P(A) + P(B) = 1

=> P(B) = 1 - 0.7 = 0.3

Total numbers of lottery tickets = 10 + 25 = 35

P(getting a prize) = 10/35 = 2/7

Total numbers of students = 35

Number of girls = 15

Number of boys = 20

(i) Numbers of favorable outcome = 20

P(choosing a boy) = 20/35 = 4/7

(ii) Numbers of favorable outcome are= 15

P(choosing a girl) = 15/35 = 3/7