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As 2 dice are thrown so there are 6×6 = 36 possibilities. 

Let E denote the event of getting a total score of 5. 

E = {(4,6),(5,5),(6,4)} 

∴ n(E)= 3 

Hence, 

P(E) = \(\frac{3}{36} = \frac{1}{12}\)

As our answer matches only with option (b) 

∴ Option (b) is the only correct choice.

As 2 dice are thrown so there are 6×6 = 36 possibilities. 

Let E denote the event of getting a total score of 5. 

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

∴ n(E)= 6 

Hence,

P(E) = \(\frac{6}{36}\)

As our answer matches only with option (b) 

∴ Option (b) is the only correct choice.

Correct option is: B) \(\frac{1}{6}\)

When two dice are thrown total possible outcomes = 6 \(\times\) 6 = 36.

Outcomes favourble to event of getting sum greater than 9 are {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}.

\(\therefore\) Total favourable outcomes = 6

\(\therefore\) Probability of getting sum greater than 9 is

P = \(\frac {Total \, favourable \, outcome}{Total\, possible \, outcomes}\) = \(\frac {6}{36} = \frac 16\)

Correct option is: B) \(\frac{1}{6}\)

In a leap year we have 366 days. So everyday of a week comes 52 times in 364 days. Now we have 2 days remaining. 

These 2 days can be 

S = {MT, TW, WTh, ThF, FSa, SaSu, SuM} 

Where S is the sample space. 

As there are total 7 possibilities. 

∴ n(S) = 7 

Let A denote the event of getting 53 Fridays and B denote event of getting 53 Saturdays. We have to find P(A∪B) 

Clearly, 

P(A) = \(\frac{2}{7}\) 

P(B) =  \(\frac{2}{7}\) 

And P(A∩B) =  \(\frac{1}{7}\) 

∴ P(A∪B) =  \(\frac{2}{7}\) +  \(\frac{2}{7}\)  –  \(\frac{1}{7}\)  =  \(\frac{3}{7}\) 

Correct option is B. 3/7

There are 366 days in a leap year.52 × 7

= 364 days. So, there are 7 possible combinations for the last 2 days (Sunday, Monday) (Monday, Tuesday) ...

(Saturday, Sunday). Of these 7, 3 give you either 53 Saturdays or Fridays. So, ans is 3/7.

In a leap year we have 366 days. So, every day of a week comes 52 times in 364 days. 

Now we have 2 days remaining. These 2 days can be

S = {MT, TW, WTh, ThF, FSa, SaSu, SuM} 

Where S is the sample space. 

As there are total 7 possibilities. 

∴ n(S) = 7 

Let A denote the event of getting 53 Fridays and B denote event of getting 53 Saturdays. 

We have to find P(A ∪ B) Clearly, 

P(A) = \(\frac{2}{7}\) 

P(B) =  \(\frac{2}{7}\) 

And P(A∩B) =  \(\frac{1}{7}\) 

∴ P(A∪B) =  \(\frac{2}{7}\)  +  \(\frac{2}{7}\)  –  \(\frac{1}{7}\) =  \(\frac{3}{7}\) 

As our answer matches only with option (b) 

∴ Option (b) is the only correct choice.

Let E and F be the two events such that one must occur 

Given, 

P(E) = \(\frac{2}{3}\) P(F) 

Also, P(E∪F) = 1 P(E) + P(F) = 1 

⇒ P(F){ \(\frac{2}{3}\) + 1} = 1 

∴ P(F) =   \(\frac{3}{5}\) 

And P(F’) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)

We have to find \(\frac{P(F)}{P(\bar F)}\) = \(\frac{3}{\frac{5}{\frac{2}{5}}}\) = \(\frac{3}{2}\)

∴ Odds in favour of F = \(\frac{3}{2}\)

As our answer matches only with option (d) 

∴ Option (d) is the only correct choice.

Let E be the event of happening of rain 

P(E) is given as 0.85

Bar E ⟶not happening of rain

P(Bar E) = 1 – P(E) = 1 – 0.85 = 0.15 

∴ P(not happening of rain) = 0.15

Given: a leap year which includes 52 weeks and two days

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

so, we have to determine the probability of that remaining two days are Sunday and Monday 

S= {MT, TW, WT, TF, FS, SSu, SuM}

Therefore n(S)=7 

E= {SuM} 

n(E)=1 

P(E) = \(\frac{n(E)}{n(S)}\) 

P(E) = \(\frac{1}{7}\) 

Given: an ordinary year which includes 52 weeks and one day 

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

so, we have to determine the probability of that one day being Sunday 

Total number of possible outcomes are 7 

Therefore n(S)=7 

E= {M, T, W, T, F, S, SU} 

n(E) = 1

P(E) = \(\frac{n(E)}{n(S)}\) 

P(E) = \(\frac{1}{7}\) 

These are 4 kings, 4 queens, and 4 aces. These are removed.

Thus, remaining number of cards now = 52 - 4 - 4 = 40

(i) Number of black face cards now = 2 (only black jacks).

Therefore, P(getting a black face card) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{2}{40}\) = \(\frac{1}{20}\)

Thus, the probability that the drawn card is a black face card is \(\frac{1}{20}\).

(ii) Number of red cards now = 26 - 6 = 20.

therefore, P(getting a red card) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{20}{40}\) = \(\frac{1}{2}\)

Thus, the probability that the drawn card is a red card is \(\frac{1}{2}\).

Answer : (d) \(\frac{1}{3}\)

Given, 

P(B/A) =\(\frac{P(A\,\cap\,B)}{P(A)}\) 

\(P(A\,\cap\,B)= P(B/A)\times P(A)=\frac{2}{3}\times \frac{1}{4} = \frac{1}{6}\)
 

 Now P(A/B) = \(\frac{P(A\,\cap\,B)}{P(B)}\)

= \(\frac{1}{2} = \frac{1}{6}\times \frac{1}{P(B)} \implies P(B)=\frac{2}{6}= \frac{1}{3}\)

Answer: (C)  \(\frac{2}{3}\)

\( P\big(\frac{A'}{B'}\big) = \frac{P(A'\,\cap\,B')}{P(B')} = \frac{P(A\,\cap\,B)'}{P(B')}\)

= \(\frac{1-P(A\,\cup\,B)}{1-P(B)} = \frac {1-[P(A)+P(B)-P(A\,\cap\,B)]}{1-P(B)}\)

\(= \frac{1-[0.5+0.4-0.3]}{1-0.4}=\frac{0.4}{0.6}=\frac{2}{3}\)

Answer: (b)

P(X/Y) = \(\frac{1}{2}\) P (Y/X) = \(\frac{1}{3}\) P(X ∩ Y) = \(\frac{1}{6}\)

\(\therefore P(\frac{X}{Y}) =\frac{P(X\,\cap\,Y)}{P(Y)}\)

\(\implies\frac{1}{2} =\frac{1/6}{P(Y)}\implies P(Y)=\frac{1}{3}\) ... (i)

Now \(P(\frac{Y}{X}) =\frac{P(Y\,\cap\,X)}{P(X)} =\frac{P(X\,\cap\,Y)}{P(X)}\)

\(\implies \frac{1}{3}=\frac{1}{6}\times\frac{1}{P(X)}\implies P(X)= \frac{1}{2}\)  ...     (ii)

\(\therefore P(X\,\cup\,Y)= P(X)+P(Y) -P(X\,\cap\,Y) =\frac{1}{2}+\frac{1}{3}-\frac{1}{6} = \frac{2}{3}\) ... (iii)

\({P(X\,\cap\,Y)}=\frac{1}{6}\)  and P(X). P(Y) = \(\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)

⇒\({P(X\,\cap\,Y)}\) = P(X).P(Y)

⇒X and Y are independent events

Now  \({P(X^c\,\cap\,Y)}\) =P(Y) - \({P(X\,\cap\,Y)}\) = \(\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\)

Sample space of nos. 1 to 17.

S = {1,2, 3, 4,5,6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

E = (1, 3, 5, 7, 11, 13, 17}

Thus, n(S) = 17 and n(E) = 7

Thus required probability

P(E) = n(E)/n(s) = 7/17

Correct option is: (A) 5/14

Correct option is  A. \(\cfrac1{13}\times\cfrac1{13}\) 

M = Event that Queen is draw in in first draw.

P(M) = \(\cfrac4{52}\)

N = Event that Queen is drawn in second draw.

P(N) = \(\cfrac4{52}\)

Now, Probability that both the cards drawn are queen:

P = P(M) x P(N)

P = \(\cfrac4{52}\) x \(\cfrac4{52}\)

= \(\cfrac1{13}\times\cfrac1{13}\)

Correct option is D. none of the above is correct

Let A and B be two independent events, then,

P(A∩ B) = P(A) x P(B)

\(\therefore\) P(A∩B) ≠ 0 → Not mutually exclusive

Or,

P(A) + P(B) ≠ 1 → Their sum of probabilities is not 1

(i) 3/4        S = { HH, HT, TH, TT }

(ii) 3/4

(iii) 1/4

Total number of all possible outcomes = 52

(i) Number of space card = 13

Number of aces = 4 (including 1 of space)

Therefore, Number of space cards and aces = (13 + 4 - 1) = 16

Therefore, P(getting a space or an ace card) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) Number of red kings = 2

Therefore, P(getting a red king) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(iii) Total Number of kings = 4

Total Number of queens = 4

let E be the event of getting either a king or a queen 

Then, the favorable outcomes = 4 + 4 = 8

Therefore, P(getting a king or a queen) = P(E) = \(\frac{8}{52}\) = \(\frac{2}{13}\)

(iv) Let E be the event of getting either a king or a queen. Then,(not E) is the event that drawn card is neither a king nor a queen.

Then, P(getting a king or a queen) = \(\frac{2}{13}\)

Now, P(E) + P(not E) = 1

Therefore, P(getting a king nor a queen) = 1 - \(\frac{2}{13}\) = \(\frac{11}{13}\)

A family has 2 children, then Sample space = S = {BB, BG,GB,GG}, where B stands for Boy and G for Girl. 

(i) Let A and B be two event such that 

A = Both are girls = {GG} 

B = the youngest is a girl = {BG, GG}

Now, P(A/B) = (P(A ∩ B))/P(B) = (1/4)/(2/4) = 1/2

(ii) Let C be event such that 

C = at least one is a girl = {BG,GB,GG}

Now, P(A/C) = (P(A ∩ C))/P(C) = (1/4)/(3/4) = 1/3

Let E₁, E₂ and E be three events such that 

E₁ = six occurs

E₂ = six does not occurs 

E = man reports that six occurs in the throwing of the dice.

Now P(E₁) = 1/6, P(E₂) = 5/6

P(E/E₁) = 4/5, P(E/E₂) = 1 - 4/5 = 1/5

By using Baye’s theorem, we obtain

P(E₁/E) = (P(E₁)P(E/E₁))/(P(E₁)P(E/E₁) + P(E₂)P(E/E₂))

= (1/6 x 4/5)/(1/6 x 4/5 + 5/6 x 1/5) = 4/(4 + 5) = 4/9

Total number of marbles = 54

let x be the number of white marbles in the jar.

P (getting a white marble) = x/54

There are 3 types of marbles in the jar, blue, green and white.

P(getting a blue ball) + P(getting a green ball) + P(getting a white ball) = 1

(Because sum of probability = 1)

1/3 + 4/9 + x/54 = 1

(54 – x)/54 = 7/9

x = 12

There are 12 white marbles in the jar.

Total number of shirts = 100

Number of good shirts = 88

Number of shirts with minor defects = 8

Number of shirts with major defects = 100- 88 – 8 = 4

(i) P(shirt acceptable by Rohit) = 88/100 = 22/25

(ii) Given: Kamal only rejects shirt with major defects

So, there are 96 possible outcomes.

P(shirt acceptable by kamal) = 96/100 = 24/25

The total number of marbles = 54

It is given that,

P(getting a white marble) be x.

Since, there are only 3 types of marbles in the jar, the sum of probabilities of all three marbles must be 1.

Therefore, \(\frac{1}3\) + \(\frac{4}9\) + x = 1

⇒ \(\frac{3+4}9\) + x = 1

⇒ x = 1 - \(\frac{7}9\)

⇒ x = \(\frac{2}9\)

Therefore, p(getting a white marble) = \(\frac{2}9\).......(1)

Let the number of white marbles be n.

then, P(getting a white marbles) = \(\frac{n}{54}\)........(2)

from (1) and (2).

\(\frac{n}{54}\) = \(\frac{2}9\)

⇒ n = \(\frac{2\times54}9\)

⇒ n = 12

Thus, there are 12 white marbles in the jar.

Total numbers of shirts = 100

The number of good shirts = 88.

The number of shirts with major defects = 100 - 88 = 4.

(i) P(the drawn shirts is acceptable to Rohit) = \(\frac{number\, of \,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\)= \(\frac{88}{100}\) = \(\frac{22}{25}\)

Thus, the probability that the drawn shirts is acceptable to Rohit is \(\frac{22}{25}\).

(ii) P(the drawn shirts is acceptable to Komal) = \(\frac{number\, of \,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\)= \(\frac{88+8}{100}\)= \(\frac{96}{100}\)= \(\frac{24}{25}\)

Thus, the probability that the drawn shirts is acceptable to Kamal is \(\frac{24}{25}\).

Total number of possible outcomes, n(S) = 12 

(i) Number of favorable outcomes, 

n(E) = 3

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{12}\)

(ii) Number of favorable outcomes, 

n(E) = 9

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{9}{12}\) = \(\frac{3}{4}\)

The total number of persons = 12.

The number of persons who are extremely patient = 3.

The number of persons who are extremely kind = 12 - 3 - 6 = 3.

(i) P(selecting a person who is extremely patient) = \(\frac{number\, of \,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Thus, the probability of selecting a person who is extremely patient is \(\frac{1}{4}\).

(ii) P(selecting a person who is extremely kind of honest) = \(\frac{number\, of \,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{6+3}{12}\) = \(\frac{9}{12}\)= \(\frac{3}{4}\)

Thus, the probability of selecting  persons who is extremely kind or honest  is \(\frac{3}{4}\).

from  the three given values, we prefer honesty more.

Total number of outcomes = 36

(i) cases where 5 comes up on at least one time are (1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) and (6,5).

the number of such cases = 11.

The number of cases where up either time = 36 - 11 = 25.

Therefore, P(5 will not come up either time) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{25}{26}\)

Thus, the probability that 5 will not come up either time is \(\frac{25}{26}\).

(ii) cases where 5 comes up on exactly one time are (1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) and (6,5).

the number of such cases = 10.

Therefore, P(5 will come up exactly one time) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{10}{36}\) = \(\frac{5}{18}\)

Thus, the probability that 5 will come up exactly one time is \(\frac{5}{18}\)

(iii) cases, where 5 comes up on exactly two times, is (5,5).

the number of such cases = 1.

Therefore, P(5 will come up both the time) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{1}{36}\)

Thus, the probability that 5 will come up both the time is \(\frac{1}{36}\).

S= {1, 2, 3, … 100} 

n(S) = 100 

Let A be the event of choosing a prime number 

∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89} 

n(A) = 25 

So P(A) = 25/100

Let B be the event of getting a number multiple of 8

B = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96} 

n(B)= 12 

So P(B) = 12/100

Also A ∩ B = ϕ 

⇒ A and B are mutually exclusive

∴ P(A ∪ B) = P(A) + P(B) = (25/100) + (12/100) = 37/100

Total numbers of elementary event are: 90 

Let E be the event of choosing a multiple of 3 

Favorable outcomes are: 3 × 4 =12 , 3 × 5 = 15, 3 × 6 = 18, 3 × 7= 21,….. 3 × 33 = 99 

Numbers of favorable outcomes are = 30 

P( multiple of 3) = P (E) = \(\frac{30}{90}\) = \(\frac{1}{3}\)