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Total numbers of outcomes = 17

(i) Odd numbers on the cards = { 1, 3, 5, 7, 9, 11, 13, 15, 17 }

Numbers of favorable odd numbers = 9

P (getting an old number) = 9/17

(ii) Numbers divisible by 5 form 1 to 17 = {5, 10, 15}

Number of favorable outcome = 3

P (getting a number divisible by 5) = 3/17

Total numbers of elementary events are: 25 

Let E be the event of getting number divisible by both 2 and 3 

Favorable outcomes are: { 6, 12, 18, 24} 

Numbers of favorable outcomes are: 4 

P (number divisible by 2 and 3) = P (E) = \(\frac{4}{25}\)

Correct option is: D) \(\frac 3{25}\)

Total number of outcomes = 50.

Number which is divisible by 8 from 1 to 50 are (8, 16, 24, 32, 40, 48)

\(\therefore\) Favourable outcomes = 6.

\(\therefore\) Required probability = \(\frac 6{50}\) = \(\frac 3{25}\)

Correct option is: D) \(\frac{3}{25}\)

Correct option is: C)\(\frac 3{25}\)

Total No which are divisible by 8 from 1 to 50 is 6.

\(\therefore\) Probability that drawn card is divisible by 8 = \(\frac 6{50}\) = \(\frac 3{25}\)

Correct option is: C) \(\frac{3}{25}\)

Total number cards kept in the box = 8

Number of cards having a number less than 4 on it = 3

Probability (P) = \(\frac{Number\,of\,favorable\,outcomes}{Total\,number\,of\,cards\,in\,a\,deck}\) = \(\frac{3}{8}\)

Here the total number of coins is N + 7. Therefore the total number of ways of choosing 5 coins out of N + 7 is N + ⁷C₅ Let E denotes the event that the sum of the values of the coins is less than one rupee and fifty paisa. 

Then E' denotes the event that the total value of the five coins is equal to or more than one rupee and fifty paisa. 

NOTE THIS STEP: 

The number of cases favorable to E' is 

= ²C₁ x ⁵C₄ x ^NC₀ + ²C₂ x ⁵C₃ x ^NC₀ + ²C₂ x ⁵C₂ x NC₁ 

= 2 x 5 + 10 + 10 N = 10 (N + 2)

∴ P (E) = 10 (N + 2)/^{n + 1} C₅

⇒ P(E) = 1 – P(E) = 1 – 10(N + 2)/^{N + 7}C₅

Total numbers of cards = 25

(i) The favorable outcome = numbers divisible by 2 = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}

The favorable outcome = number divisible by 3 = {3, 6, 9, 12, 15, 18, 21, 24}

Total number of favorable outcomes = 16

(Note: count 6,12,18 and 24 are common, so consider only one set)

Now,

P(drawn card is divisible by 2 or 3)= 16/25

(ii) The favorable numbers = prime number = 2, 3, 5, 7, 11, 13, 17, 19, 23

Numbers of favorable outcomes = 9

P(drawn card is prime number) = P (E) = 9/25

Total number of possible outcomes, n(S) = 50 

(i) Number of favorable outcomes, 

n(E) = 25

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{25}{50}\) = \(\frac{1}{2}\)

(ii) Number of favorable outcomes, 

n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{50}\) = \(\frac{2}{25}\)

(iii) Number of favorable outcomes, 

n(E) = 10

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{50}\) = \(\frac{1}{5}\)

(iv) Number of favorable outcomes, 

n(E) = 4

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{50}\) = \(\frac{2}{25}\)

The probability p₁ (say) of winning the best of three games is = the prob. of winning two games + the prob. of winning three games. 

= ³C₂ (0.6) (0.4)² + ³C₃ (0.4)³ 

 [Using Binomial distribution] 

Similarly the probability of winning the best five games is p₂ (say) = the prob. of winning three games + the prob. of winning 5 games.

 = ⁵C₃ (0.6)² (0.4)³ + ⁵C₃ (0.6) (0.4)³ + ⁵C₅ (0.4)⁵ 

We have p₁ = 0.288 + 0.064 = 0.352 

And p₂ = 0.2304 + 0.0768 + 0.01024 = 0.31744 

As p₁ > p₂ 

∴ A must choose the first offer i.e. best of three games. 

No. of all possible outcomes = 113

(i) Perfect square numbers between 11 to 123 area 16,

25, 36, 49, 64, 81, 100, 121

No. of all favourable outcomes = 8

P(Number drawn is Perfect square) = 8/113

(ii) No. of multiples of 7 from 11, to 123 = 16.

14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91,98, 105, 112, 119

P(number drawn card is multiple of 7) = 16/113

Answer: (a)\(\frac{1}{14}\)

Let S = {00, 01, 02, ... , 48, 49}. 

n(S) = 50 

Let A be the event that the sum of the digits on the selected ticket is 8, then 

A = {08, 17, 26, 35, 44} ⇒ n(A) = 5 

Let B be the event that the product of the digits is zero. 

Then, B = {00, 01, 02, ... , 08, 09, 10, 20, 30, 40} 

⇒ n(B) = 14 

∴  A ∩ B = {08} ⇒ n(A ∩ B) = 1

∴  Required probability = \(P\big(\frac{A}{B}\big)\) 

= \(\frac{P(A\,\cap\,B)}{P(B)} =\frac{1/150}{14/50}\) = \(\frac{1}{14}\)

The correct option is: (B) 10/49

Explanation:

Total 13 cards are present in suit of club if 3 cards are removed, then 10 cards of clubs are remaining. 

Total cards remaining = 4

. ^..  Probability of getting a club = 10/49

There are 52 playing cards. 

∴ n(S) = 52 i. Let A be the event that the card drawn is an ace. 

∴ n(A) = 4 

∴ P(A) = \(\frac{n(A)}{n(S)}\)= 4/52 

∴ P(A) = 1/13

ii. Let B be the event that the card drawn is a spade. 

∴ n(B) = 13 

∴ P(B) = \(\frac{n(B)}{n(S)}\) = 13/52 

∴ P(B) = 1/4

∴ P(A) = 1/13 ; P(B) = 1/4

After removing three face cards of spades (king, queen, jack) from a deck of 52 playing cards, there are 49 cards left in the pack. Out of these 49 cards one card can be chosen in 49 ways.

So, Total number of elementary events = 49

(i) There are 6 black face cards out of which 3 face cards of spades are already removed. So, out of remaining 3 black face cards one black face card ban be chosen in 3 ways.

So, Favorable number of elementary events = 3

Hence, P (Getting a black face card ) =3/49

(ii) There are 3 queens in the remaining 49 cards. So, out of these three queens, on queen can be chosen in 3 ways

So, Favorable number of elementary events = 3

Hence P (Getting a queen) =3/49

(iii) There are 23 black cards in the remaining 49 cards, So, out to these 23 black card, one black card can be chosen in 23 ways

So, Favorable number of elementary events = 23

Hence, P (Getting a black card) =23/49

Sample space, 

S = {1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15} 

∴ n(S) = 15 

i. Let A be the event that the ticket drawn shows an even number. 

∴ A = {2, 4, 6, 8, 10, 12, 14} 

∴ n(A) = 7 

∴ P(A) = \(\frac{n(A)}{n(S)}\) 

∴ P(A) = 7/15

ii. Let B be the event that the ticket drawn shows a number which is a multiple of 5.

∴ B = {5, 10, 15} 

∴ n(B) = 3 

∴ P(B) = \(\frac{n(B)}{n(S)}\)= 3/15

∴ P(B) = 1/5

∴ P(A) = 7/15 ; P(B) = 1/5

We know that, 

Probability of any event = (Number of favorable outcome) / (Total number of trials) 

Total number of trials = 95 + 290 + 115 = 500 

Now, 

P(Getting two heads) = \(\frac{95}{500}\) = 0.19 

P(Getting one tail) = \(\frac{290}{500}\) = 0.58 

P(Getting no head) = \(\frac{115}{500}\) = 0.23

Given,

Total number of trials = 500 times

Probability (E) \(=\frac{number\,of\,trials\, in\,which\,the\,event\,happens}{total\,number\,of\,trials}\)

P (getting two heads) = \(\frac{95}{500}=0.19\)

P (getting one tail) = \(\frac{290}{500}=0.58\)

P (getting no head) = \(\frac{115}{500}=0.23\)

(d) \(\frac{1}{12}\)

Total number of exhaustive cases in a single throw of two dice = 6 × 6 = 36

Doublets are obtained as (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) 

Number of doublets of odd numbers = 3

\(\therefore\) Required probability = \(\frac{3}{36}\) = \(\frac{1}{12}\)

Probability of occurrence of an event = (Total number of favorable outcomes) / (Total number of outcomes)

Possible outcomes are as follow:

(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 number of outcomes = 36

(i) getting a sum less than 6

Pick entries having sum less than 6:

(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)

Total number of favorable outcomes = 10

Probability (getting a sum less than 6) = 10/36 or 5/18

(ii) getting a doublet of odd numbers

Pick entries having doublet of odd numbers:

(1, 1), (3, 3), (5, 5)

Total number of favorable outcomes = 3

Probability (getting a doublet of odd numbers) = 3/36 or 1/12

(iii) getting the sum as a prime number

Pick entries having sum as a prime number:

(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 number of favorable outcomes = 15

Probability (getting the sum as a prime number) = 15/36 or 5/12