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The drawer contains 4 pens out of which one is red (R), one is green (G), one is blue (B) and the other one is purple (P). 

From this drawer, two pens are selected one after the other with replacement. 

∴ The sample space S is given by S = {RR, RG, RB, RP, GR, GG, GB, GP, BR, BG, BB, BP, PR, PG, PB, PP} 

(a) A : Select at least one red pen. At least one means one or more than one. 

∴ A = {RR, RG, RB, RP, GR, BR, PR}

(b) B : Two pens of the same colour are not selected. 

B = {RG, RB, RP, GR, GB, GP, BR, BG, BP, PR, PG, PB}

(c) \(\frac{4}{9}\)

n(S) = Total number of waysin which 2 socks can be drawn out of 9 socks (5 brown and 4 blue socks)

= ⁹C₂ = \(\frac{|\underline9}{|\underline7|\underline2}\) = \(\frac{9\times8}{2}\) = 36

Let A : Event of drawing 2 socks of same colour 

⇒ A = Drawing 2 brown socks out of 5 brown socks or Drawing 2 blue sock out of 4 blue socks

⇒ n(A) = ⁵C₂ + ⁴C₂ = \(\frac{|\underline5}{|\underline3|\underline2}\) + \(\frac{|\underline4}{|\underline2|\underline2}\) = \(\frac{5\times4}{2}\) + \(\frac{4\times3}{2}\) = 10 + 6 = 16

∴ Required probability P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{16}{36}\) = \(\frac{4}{9}\).

Let E : Event of drawing both the balls of same colour from the two urns 

E₁ : Getting 1 white ball from the first urn and 1 white ball from the second urn 

E₂ : Getting 1 blue ball from the first urn and 1 blue ball from the second urn

Then, P(E) = P(E) + P(E₂) 

(∵ The two events E₁ and E₂ are mutually exclusive) 

= {P(1 white from 1st urn) x P(1 white from 2nd urn)} + {P(1 blue from 1st urn) x P(1 blue from 2nd urn)}

= \(\frac{3}{8}\times\frac{4}{8}+\frac{5}{8}\times\frac{4}{8}\)  

(∵ Event of drawing a ball from one urn is independent of drawing a ball from other urn)

= \(\frac{12}{64}+\frac{20}{64}=\frac{32}{64}=\frac{1}{2}.\)

Let E₁ : Event that A speak the truth 

E₂ : Event that B speaks the truth 

Then, \(\bar{E}\)₁ : Event that A tells a lie 

\(\bar{E}\)₂ : Event that B tells a lie 

E₁, E₂ and so \(\bar{E}\)₁, \(\bar{E}\)₂, are independent events. 

∴ A and B will contradict each other in the following two mutually exclusive ways 

(i) A speaks the truth , B tells a lie 

(ii) A tells a lie, B speaks the truth

Now, P(\(\bar{E}\)₁) = \(\frac{75}{100}\) = \(\frac{3}{4}\) ⇒ P( \(\bar{E}\)₁) = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\)

 P(\(\bar{E}\)₂) = \(\frac{80}{100}\) = \(\frac{4}{5}\) ⇒ P(\(\bar{E}\)₂) = 1 - \(\frac{4}{5}\) = \(\frac{1}{5}\)

∴ Required probability = P(E₁) x  P(\(\bar{E}\)₂) +  P(\(\bar{E}\)₁) x P(E₂) = \(\frac{3}{4}\)x \(\frac{1}{5}\)+\(\frac{1}{4}\)x \(\frac{4}{5}\) = \(\frac{7}{20}\).

∴ % of cases in which A and B contradict each other = \(\bigg(\frac{7}{20}\times100\bigg)\)% = 35%

Given:

⇒ A speaks truth in 75% of cases

⇒ B speaks truth in 80% of cases

Now,

⇒ P(T_A) = P(A speaking truth) = 0.75

⇒ P(N_A) = P(A not speaking truth) = 1-0.75

⇒ P(N_A) = 0.25

⇒ P(T_B) = P(B speaking truth) = 0.80

⇒ P(N_B) = P(B not speaking truth) = 1-0.80

⇒ P(N_B) = 0.20

We need to find the probability for case in which A and B contradict each other for narrating an incident.

This happens only when A not telling truth while B is telling truth and vice-versa.

⇒ P(C_AB) = P(A and B contradict each other)

⇒ P(C_AB) = P(A tells truth and B doesn’t) + P(B tells truth and A doesn’t)

Since the speaking of A and B are independent events their probabilities will multiply each other.

⇒ P(C_AB ) = (P(T_A)P(N_B)) + (P(N_A)P(T_B))

⇒ P(C_AB) = (0.75 × 0.20) + (0.25 × 0.80)

⇒ P(C_AB) = 0.15 + 0.20

⇒ P(C_AB) = 0.35

∴ The required probability is 0.35.

 A and B will agree on a certain statement if both speak the truth or both tell a lie. Now, let us define the events as follows: 

E₁: A and B both speak the truth ⇒ P(E1) = xy 

E₂: A and B both tell a lie ⇒ P(E₂) = (1 – x) (1 – y) 

E: A and B agree on a certain statement. 

Clearly, P(E/E₁) = P(E/E₂) = 1 

Now, we are required to find P(E₁/E).

P(E₁/E) = \(\frac{P(E_1).P(E/E_1)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)}\) = \(\frac{xy.1}{xy.1+(1-x)(1-y).1}\) = \(\frac{xy}{1-x-y+2xy}\)

Total number of possible outcomes, n(S) = 5 + 8 + 4 = 17 

(i) Number of favorable outcomes, 

n(E) = 5

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{17}\)

(ii) Number of favorable outcomes, 

n(E) = 8

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{17}\)

(iii) Number of events of drawing a green marble = 4

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{17}\)

Probability of not drawing green marble = 1 – P(E)

= 1 - \(\frac{4}{17}\) = \(\frac{13}{17}\)

Correct option is: A) 20

Total No. of marbles in the box is 28.

There are x green marbles and rest are white.

\(\therefore\) Total No. of white marbles = 28 - x

Probability of getting a green marble = \(\frac 27\)

\(\therefore\) \(\frac x{28} = \frac 27\)

\(\Rightarrow\) x = \(\frac 27 \times 28 = 2 \times 4 = 8\) 

\(\therefore\) Total No. of white marbles = 28- x = 28-8 = 20 

Hence, there are 20 white marbles in the box.

Correct option is: A) 20

Given : let H be head, and T be tails where as 1,2,3,4,5,6 

be the numbers on the dice which are thrown when a head comes up or else coin is tossed again if its tail.

According to the question ,

sample space S = {(TH),(TT) ,(H1),(H2),(H3),(H4),(H5),(H6)}

To Find: 

(i) the probability of obtaining two tails

From sample space, it is clear that the probability of obtaining two tails is\(\frac{1}{8}\)

i.e., {TT} with total no of elements in sample space as 8.

(ii) the probability of obtaining a head and the number 6

From sample space, it is clear that the probability of obtaining a head and the number 6 is\(\frac{1}{8}\)

i.e., {H6} with total no of elements in sample space as 8.

(iii) the probability of obtaining a head and an even number

From sample space, it is clear that the probability of obtaining a head and an even number is\(\frac{3}{8}\)

i.e, {H2,H4,H6} with total no of elements in sample space as 8.

Here the sample space is– S = {T, HT, HHT, HHHT, HHHHT,….}

In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2, 5, 6), … ,(5, 1, 6), (5, 2, 6), …}

Given: A coin is tossed repeatedly until comes up from the first time. 

To Find: Write the sample space for the given experiment. 

Explanation: In the given Experiment, a coin is tossed and if the outcome is tail the experiment is over 

And, if the outcome is Head then the coin is tossed again. 

In the second toss also if the outcome is tail then experiment is over, otherwise coin is tossed again. 

This process continues indefinitely 

So, The sample space for this experiment is 

S = {T, HT, HHT, HHHT, HHHHT…} 

Hence, S is the sample space for the given experiment.

Total no. of possible outcomes = 34 (18 girls, 16 boys)

(i) E ⟶ event of getting girl name

No. of favorable outcomes = 18 (18 girls)

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 18/34 = 9/17

(ii) E ⟶ event of getting boy name

No. of favorable outcomes = 16 (16 boys)

P(E) = 16/34 = 8/17

Given: In a class there are 18 girls and 16 boys, the class teacher wants to choose one name. The class teacher writes all pupils’ name on a card and puts them in basket and mixes well thoroughly. A child picks one card

Required to find: The probability that the name written on the card is

(i) The name of a girl

(ii) The name of a boy

Total number of students in the class = 18 + 16 = 34

(i) The names of a girl are 18, so the number of favourable cases is 18

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a name of girl on the card = 18/34 = 9/17

(ii) The names of a boy are 16, so the number of favourable cases is 16

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a name of boy on the card = 16/34 = 8/17

Given: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing number 1, 2, 3 ….12

Required to find: Probability of following

Total numbers on the spin is 12

(i) Favourable outcomes i.e. to get 10 is 1

So, total number of favourable outcomes i.e. to get 10 is 1

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a 10 = 1/12

(ii) Favourable outcomes i.e. to get an odd number are 1, 3, 5, 7, 9, and 11

So, total number of favourable outcomes i.e. to get a prime number is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number = 6/12 = 1/2

(iii) Favourable outcomes i.e. to get a multiple of 3 are 3, 6, 9, and 12

So, total number of favourable outcomes i.e. to get a multiple of 3 is 4

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting multiple of 3 = 4/12 = 1/3

(iv) Favourable outcomes i.e. to get an even number are 2, 4, 6, 8, 10, and 12

So, total number of favourable outcomes i.e. to get an even number is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an even number = 6/12 = 1/2

Given: One card is drawn from a well shuffled deck of 52 playing cards

Required to find: Probability of following

Total number of cards is 52

(i) Total number of cards which are king of red suit is 2

Number of favourable outcomes i.e. Total number of cards which are king of red suit is 2

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which is a king of red suit = 2/52 = 1/26

(ii) Total number of face cards are 12

Number of favourable outcomes i.e. total number of face cards is 12

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a face cards = 12/52 = 3/13

(iii) Total number of red face cards are 6

Number of favourable outcomes i.e. total number of red face cards is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a red face cards = 6/52 = 3/26

(iv) Total number of queen of black suit cards is 2

Total number of favourable outcomes i.e. total number of queen of black suit cards is 2

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which is a queen of black suit = 2/52 = 1/26

(v) Total number of jack of hearts is 1

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a card which is a jack of hearts = 1/52

(vi) Total number of spade cards are 13

Total numbers of favourable outcomes i.e. total number spade cards is 13

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a spade card = 13/52 = 1/4

Given: A bag contains 3 red, and 5 black balls. A ball is drawn at random

Required to find: Probability of getting a

(i) red ball

(ii) white ball

Total number of balls 3 + 5 = 8

(i) Total number red balls are 3

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a red ball = 3/8

(ii) Total number of black ball are 5

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a black ball = 5/8

Total number of possible outcomes, n(S) = 5 

(i) Number of events of drawing a queen, 

n(E) = 1

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

(ii) If a king is drawn first and put aside, total number of remaining cards, n(S) = 4 

a) Probability that second card is ace,

P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{4}\)

b) Since there was only a single king and when it is put aside, then

n(E) = 0

∴ P(E) = \(\frac{n(E)}{n(S)}\) = 0

Given: Five cards-ten, jack, queen, king and Ace of diamond are shuffled face downwards.

Required to find: Probability of following

Total number of cards is 5

(i) Total number of cards which is a queen is 1

Number of favourable outcomes i.e. Total number of cards which is queen = 1

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which is a queen = 1/5

(ii) If a king is drawn first and put aside then

Total number of cards becomes 4

(a) Number of favourable outcomes i.e. Total number of ace card is 1

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an ace card = 1/4

(b) Number of favourable outcomes i.e. Total number of king cards is 0

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a king = 0

Total no. of possible outcomes = 5 {5 cards}

(i) E ⟶ event of drawing queen

No. favourable outcomes = 1 {1 queen card}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 1/5

(ii) When king is drawn and put aside, total no. of remaining cards = 4

Total no. of possible outcomes = 4

E ⟶ event of drawing ace card

No. favourable outcomes = 1 {1 ace card}

P(E) = 1/4

Let E₁ be the event that A speaks truth and E₂ be the event that B speaks truth. Then E and f are independent events such that

P(E₁) = 60/100 = 3/5, P(E₁) = 90/100 = 9/10

⇒ P(bar E₁) = 2/5, P(bar E₂) = 1/10

P (A and B contradict each other) = P(E₁)P(bar E₂) + P(bar E₁)P(E₂)

= 3/5 x 1/10 + 2/5 x 9/10 = (3 + 18)/50 = 21/50

Yes, the statement of B will carry more weight as the probability of B to speak truth is more than that of A.

A pack of 52 cards consists of 13 spade cards, 13 club cards, 13 heart cards and 13 diamond cards.

Let, S = Spade card,
C = Club card,
H = Heart card,
D = Diamond card,
J = Jack card,
Q = Queen card,
K = King card,
A = Ace card

Non-face card: Spade: S₂, S₃, S₄, …, S₁₀
Club: C₂, C₃, C₄, …, C₁₀
Heart: H₂, H₃, H₄, …, H₁₀
Diamond: D₂, D₃, D₄, …, D₁₀

Total number of ways of drawing a card randomly from a pack of 52 cards is ⁵²C₂ = 52

(1) U: The sample space of drawing a card randomly from a pack of 52 cards is expressed as follows:

u = {S_A, S₂, S₃, S₄, …….. S₁₀, S_J, S_Q, S_K
C_A, C₂, C₃, C₄, …….. C₁₀, C_J, C_Q, C_K
H_A, H₂, H₃, H₄, …….. H₁₀, H_J, H_Q, H_K
D_A, D₂, D₃, D₄, …….. D₁₀, D_J, D_Q, D_K}

(2) Event A = Drawing a spade card
S_A, S₂, S₃, S₄, …….. S₁₀, S_J, S_Q, S_K

(3) Event B = Drawing a card from ace to ten
= {S_A, S₂, S₃, …….. S₁₀, C_A, C₂, C₃, …….. C₁₀, H_A, H₂, H₃, …….. H₁₀, D_A, D₂, D₃, …….. D₁₀,}

(4) Event A ∪ B = Drawing a spade card or a card from ace to ten
= {S_A, S₂, S₃, S₄, …….. S₁₀, S_J, S_Q, S_K, C_A, C₂, C₃, …….. C₁₀, H_A, H₂, H₃, …….. H₁₀, D_A, D₂, D₃, …….. D₁₀}

(5 ) Event A ∩ B = Drawing a spade card and from ace to ten
= {S_A, S₂, S₃, …….. S₁₀}

(6) Event B’ = Drawing a face card (or not from ace to ten)
= {S_J, S_Q, S_K, C_J, C_Q, C_K, H_J, H_Q, H_K, D_J, D_Q, D_K}

We know that,

Probability of occurrence of an event

  = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Let B be Boy and G be Girl 

Total possible outcomes are BB, BG, GB, GG 

Our desired outcome is at least one boy. So, BB, BG, GB are desired outputs. 

Total no. of outcomes are 4, and the desired outcomes are 3 

Therefore, the probability of at least one boy = \(\frac{3}{4}\)

Conclusion: Probability of at least one boy is \(\frac{3}{4}\)

We know that,

Probability of occurrence of an event 

 = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) 

Total no. of outcomes = 10 + 25 = 35

Desired outcomes are prizes. Total no. of desired outcomes = 10 

Therefore, the probability of getting a prize = \(\frac{10}{35}\) = \(\frac{2}{7}\)

Conclusion: Probability of getting a prize is =  \(\frac{2}{7}\)