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Probability of getting at most two heads \(=\frac{favourable\, outcome}{total\, outcome}\) \(=\frac{72+77+28}{200}=\frac{177}{200}\)

Correct option is: D) 1

Probability of getting at most two heads

= \(\frac {No \,of \,outcomes \,which \,favours \,at \,most \,two\, heads}{Total \,No \,of \,outcomes}\) = \(\frac 44 =1\)

Correct option is: D) 1

If two unbiased coins are tossed simultaneously, we obtain any one of the following as an out come :

HH, HT, TH, TT

So,  Total number of elementary events = 4

(i) Two heads are obtained if elementary event HH occurs.

So,  Favorable number of events = 1

Hence, P (Two heads) =1/4

(ii) At least one head is obtained if any one of the following elementary events happen: HH, HT, TH

So,  favorable number of events = 3

Hence P (At least one head) =4/6=2/3

(iii) If one of the elementary events HT, TH, TT occurs, than at most one head is obtained

So,  favorable number of events = 3 

Hence, P (At most one head) =3/6=1/2

Given: a pair of dice has been thrown, so the number of elementary events in sample space is 6² = 36

n (S) = 36

By using the formula,

P (E) = favourable outcomes / total possible outcomes

(i) Let E be the event that the sum 8 appears

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

n (E) = 5

P (E) = n (E) / n (S)

= 5 / 36

(ii) Let E be the event of getting a doublet

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

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(iii) Let E be the event of getting a doublet of prime numbers

E = {((2, 2) (3, 3) (5, 5)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(iv) Let E be the event of getting a doublet of odd numbers

E = {(1, 1) (3, 3) (5, 5)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(v) Let E be the event of getting sum greater than 9

E = {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(vi) Let E be the event of getting even on first die

E = {(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)}

n (E) = 18

P (E) = n (E) / n (S)

= 18 / 36

= 1/2

(vii) Let E be the event of getting even on one and multiple of three on other

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

n (E) = 11

P (E) = n (E) / n (S)

= 11 / 36

(viii) Let E be the event of getting neither 9 or 11 as the sum

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

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(ix) Let E be the event of getting sum less than 6

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

n (E) = 10

P (E) = n (E) / n (S)

= 10 / 36

= 5/18

(x) Let E be the event of getting sum less than 7

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

n (E) = 15

P (E) = n (E) / n (S)

= 15 / 36

= 5/12

(xi) Let E be the event of getting more than 7

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

n (E) = 15

P (E) = n (E) / n (S)

= 15 / 36

= 5/12

(xii) Let E be the event of getting neither a doublet nor a total of 10

E′ be the event that either a double or a sum of ten appears

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

n (E′) = 8

P (E′) = n (E′) / n (S)

= 8 / 36

= 2/9

So, P (E) = 1 – P (E′)

= 1 – 2/9

= 7/9

(xiii) Let E be the event of getting odd number on first and 6 on second

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

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(xiv) Let E be the event of getting greater than 4 on each die

E = {(5,5) (5,6) (6,5) (6,6)}

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 36

= 1/9

(xv) Let E be the event of getting total of 9 or 11

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

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 36

= 1/6

(xvi) Let E be the event of getting total greater than 8

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

n (E) = 10

P (E) = n (E) / n (S)

= 10 / 36

= 5/18

(i) 8 as the sum Total number of outcomes when a pair of die is thrown simultaneously is: 

Here the first number denotes the outcome of first die and second number the outcome of second die.

Total number of outcomes in the above table are 36 

Numbers of outcomes having 8 as sum are: (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)

Therefore numbers of outcomes having 8 as sum are 5 

Probability of getting numbers of outcomes having 8 as sum is 

= \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{5}{36}\) 

(ii) a doublet Total number of outcomes in the above table 1 are 36 

Numbers of outcomes as doublet are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) 

Therefore Numbers of outcomes as doublet are 6

Probability of getting numbers of outcomes as doublet is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\) 

Therefore Probability of getting numbers of outcomes as doublet is = \(\frac{1}{6}\)

(iii) a doublet of prime numbers Total number of outcomes in the above table 1 are 36 

Numbers of outcomes as doublet of prime numbers are: (1, 1), (3, 3), (5, 5)

Therefore Numbers of outcomes as doublet of prime numbers are 3 

Probability of getting numbers of outcomes as doublet of prime numbers is

= \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\) 

Therefore Probability of getting numbers of outcomes as doublet of prime numbers is = \(\frac{1}{12}\)

(iv) a doublet of odd numbers Total number of outcomes in the above table 1 are 36 

Numbers of outcomes as doublet of odd numbers are: (1, 1), (3, 3), (5, 5)

Therefore Numbers of outcomes as doublet of odd numbers are 3 

Probability of getting numbers of outcomes as doublet of odd numbers is

= \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

(v) a sum greater than 9 Total numbers of outcomes in the above table 1 are 36

Numbers of outcomes having sum greater than 9 are: (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6, 5) 

Therefore Numbers of outcomes having sum greater than 9 are 6 

Probability of getting numbers of outcomes having sum greater than 9 is

= \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\) 

Therefore Probability of getting numbers of outcomes having sum greater than 9 is = \(\frac{1}{6}\)

(vi) An even number on first 

Total numbers of outcomes in the above table 1 are 36 

Numbers of outcomes having an even number on first are: (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) and (6, 6) 

Therefore Numbers of outcomes having an even number on first are 18 

Probability of getting numbers of outcomes having An even number on first is

= \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{18}{36}\) = \(\frac{1}{2}\) 

(vii) an even number on one and a multiple of 3 on the other 

Total numbers of outcomes in the above table 1 are 36 

Numbers of outcomes having an even number on one and a multiple of 3 on the other are: (2, 3), (2, 6), (4, 3), (4, 6), (6, 3) and (6, 6) 

Therefore Numbers of outcomes having an even number on one and a multiple of 3 on the other are 6 

Probability of getting an even number on one and a multiple of 3 on the other is

= \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\) 

 (viii) neither 9 nor 11 as the sum of the numbers on the faces 

Total numbers of outcomes in the above table 1 are 36 

Numbers of outcomes having 9 nor 11 as the sum of the numbers on the faces are: (3, 6), (4, 5), (5, 4), (5, 6), (6, 3) and (6, 5) 

Therefore Numbers of outcomes having neither 9 nor 11 as the sum of the numbers on the faces are 6 

Probability of getting 9 nor 11 as the sum of the numbers on the faces is 

= \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

The probability of outcomes having 9 nor 11 as the sum of the numbers on the faces P(E) = \(\frac{1}{6}\)

Probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is given by P(\(\text{E}\) ) = \(1-\frac{1}{6}=\frac{6-1}{6}=\frac{5}{6}\)

Therefore probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces = P (\(\text{E}\)) = \(\frac{5}{6}\) 

Therefore Probability of getting neither 9 nor 11 as the sum of the numbers on the faces is = \(\frac{1}{6}\)

(ix) a sum less than 6 

Total numbers of outcomes in the above table 1 are 36 

Numbers of outcomes having a sum less than 6 are: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1) 

Therefore Numbers of outcomes having a sum less than 6 are 10

Probability of getting a sum less than 6 is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{10}{36}\) = \(\frac{5}{18}\) 

Therefore Probability of getting sum less than 6 is = \(\frac{5}{18}\)

(x) a sum less than 7 

Total numbers of outcomes in the above table 1 are 36 

Numbers of outcomes having a sum less than 7 are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1) 

Therefore Numbers of outcomes having a sum less than 7 are 15

Probability of getting a sum less than 7 is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\) 

Therefore Probability of getting sum less than 7 is = \(\frac{5}{12}\)

(xi) a sum more than 7 Total numbers of outcomes in the above table 1 are 36

Numbers of outcomes having a sum more than 7 are: (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) 

Therefore Numbers of outcomes having a sum more than 7 are 15

Probability of getting a sum more than 7 is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\)

Therefore Probability of getting sum more than 7 is = \(\frac{5}{12}\)

(xii) at least once 

Total numbers of outcomes in the above table 1 are 36 

Therefore Numbers of outcomes for atleast once are 11 

Probability of getting outcomes for atleast once is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{11}{36}\) 

Therefore Probability of getting outcomes for atleast once is = \(\frac{11}{36}\)

(xiii) a number other than 5 on any dice. 

Total numbers of outcomes in the above table 1 are 36 

Numbers of outcomes having 5 on any die are: (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5) 

Therefore Numbers of outcomes having outcomes having 5 on any die are 15

Probability of getting 5 on any die is = \(\frac{Total\,number}{Total\,number\,of\,outcomes}\) = \(\frac{11}{36}\) = \(\frac{11}{36}\) 

Therefore Probability of getting 5 on any die is = \(\frac{11}{36}\)

Probability of not getting 5 on any die P( ) = 1 –P (E)

P(\(\text{E}\)) = \(1-\frac{11}{36}\) = \(\frac{36-11}{36}\) = \(\frac{25}{36}\)

(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6 

Total number of outcome = 6 

Prime numbers are: 1, 3 and 5 

Total number of prime numbers = 3 

Probability of getting a prime number = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{3}{6}=\frac{1}{2}\)

Therefore probability of getting a prime number = \(\frac{1}{2}\)

(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6 

Total number of outcome = 6

Probability of getting 2 and 4 is = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\) 

Therefore probability of getting 2 and 4 is \(\frac{1}{3}\)

(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6

Multiples of 2 and 3 are = 2, 3, 4 and 6

Probability of getting a multiple of 2 or 3 is = \(\frac{Total\,prime\,number}{Total\,number\,of\,outcomes}\) = \(\frac{4}{6}=\frac{2}{3}\)

Therefore probability of getting a multiple of 2 or 3 = \(\frac{2}{3}\)

Given: A die is thrown.

The total number of outcomes is six, n (S) = 6

By using the formula,

P (E) = favourable outcomes / total possible outcomes

(i) Let E be the event of getting a prime number

E = {2, 3, 5}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 6

= 1/2

(ii) Let E be the event of getting 2 or 4

E = {2, 4}

n (E) = 2

P (E) = n (E) / n (S)

= 2 / 6

= 1/3

(iii) Let E be the event of getting a multiple of 2 or 3

E = {2, 3, 4, 6}

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 6

= 2/3

(i) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6

Total number of outcome = 6

Prime numbers are: 1, 3 and 5

Total number of prime numbers = 3

Probability of getting a prime number = Total prime numbers/Total number of outcomes

= 3/6

= 1/2

∴ Probability of getting a prime number = 1/2

(ii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6

Total number of outcome = 6

Probability of getting 2 and 4 is Total numbers/Total number of outcomes

= 2/6

= 1/3

∴ Probability of getting 2 and 4 is 1/3

(iii) Outcomes of a die are: 1, 2, 3, 4, 5, 5 and 6

Multiples of 2 and 3 are = 2, 3, 4 and 6

Total number of multiples are 4

Probability of getting a multiple of 2 or 3 is Total numbers/Total number of outcomes

= 4/6

= 2/3

∴ Probability of getting a multiple of 2 or 3 is 2/3

Correct option is (B) 1/4

If two coins are tossed then possible outcomes are S = {HH, HT, TH, TT}

Let event E = Getting two heads = {HH}

\(\therefore\) n(E) = 1 & n(S) = 4

\(\therefore\) P(E) \(=\frac{n(E)}{n(S)}=\frac14\)

Hence, the probability of getting 2 heads is \(\frac{1}{4}.\)

Correct option is  B) 1/4

Correct option is: C) 1

Given: A dice is thrown once

Required to find:

(i) Probability of getting a prime number

(ii) Probability of getting 2 or 4

(iii) Probability of getting a multiple of 2 or 3.

(iv) Probability of getting an even number

(v) Probability of getting a number greater than five.

(vi) Probability of lying between 2 and 6

Total number on a dice is 6 i.e., 1, 2, 3, 4, 5 and 6.

(i) Prime numbers on a dice are 2, 3, and 5. So, the total number of prime numbers is 3.

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

Thus, probability of getting a prime number = \(\frac{3}{6}\) = \(\frac{1}{2}\)

(ii) For getting 2 and 4, clearly the number of favourable outcomes is 2.

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

Thus, the probability of getting 2 or 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\)

(iii) Multiple of 2 are 3 are 2, 3, 4 and 6.

So, the number of favourable outcomes is 4

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

Thus, the probability of getting an multiple of 2 or 3 = \(\frac{4}{6}\) = \(\frac{2}{3}\)

(iv) An even prime number is 2 only.

So, the number of favourable outcomes is 1.

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

Thus, the probability of getting an even prime number = \(\frac{1}{6}\)

(v) A number greater than 5 is 6 only.

So, the number of favourable outcomes is 1.

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

Thus, the probability of getting a number greater than 5 = \(\frac{1}{6}\)

(vi) Total number on a dice is 6.

Numbers lying between 2 and 6 are 3, 4 and 5

So, the total number of numbers lying between 2 and 6 is 3.

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

Thus, the probability of getting a number lying between 2 and 6 = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Given A pair of dice is thrown

Required to find: Probability that the total of numbers on the dice is greater than 9

First let’s write the all possible events that can occur

(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),

It’s seen that the total number of events is 6² = 36

Favourable events i.e. getting the total of numbers on the dice greater than 9 are

(5,5), (5,6), (6,4), (4,6), (6,5) and (6,6).

So, the total number of favourable events i.e. getting the total of numbers on the dice greater than 9 is 6

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

Thus, the probability of getting the total of numbers on the dice greater than 9 = 6/36 = 1/6

Given: Three coins are tossed together.

By using the formula,

P (E) = favourable outcomes / total possible outcomes

Total number of possible outcomes is 2³ = 8

(i) Let E be the event of getting exactly two heads

E = {(H, H, T) (H, T, H) (T, H, H)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 8

(ii) Let E be the event of getting at least two heads

E= {(H, H, T) (H, T, H) (T, H, H) (H, H, H)}

n (E)=4

P (E) = n (E) / n (S)

= 4 / 8

= 1/2

(iii) Let E be the event of getting at least one head and one tail

E = {(H, T, T) (T, H, T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)}

n (E) = 6

P (E) = n (E) / n (S)

= 6 / 8

= 3/4

Correct option is: B) 1/4

We have, \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

From the given data, P(A ∩ B) = \(\cfrac4{13}\)

And P(B) = \(\cfrac9{13}\)

Hence, P(\(\cfrac{A}{B}=\cfrac{\frac{4}{13}}{\frac{9}{13}}\))

= \(\cfrac49\)

Correct option is: B) 1/4

Given: Three coins are tossed simultaneously.

When three coins are tossed then the outcome will be anyone of these combinations.

TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.

So, the total number of outcomes is 8.

(i) For exactly two heads, the favourable outcome are THH, HHT, HTH

So, the total number of favourable outcomes is 3.

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

Thus, the probability of getting exactly two heads is 3/8

(ii) For getting at least two heads the favourable outcomes are HHT, HTH, HHH, and THH

So, the total number of favourable outcomes is 4.

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

Thus, the probability of getting at least two heads when three coins are tossed simultaneously = 4/8 = 1/2

(iii) For getting at least one head and one tail the cases are THT, TTH, THH, HTT, HHT, and HTH.

So, the total number of favourable outcomes i.e. at least one tail and one head is 6

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

Thus, the probability of getting at least one head and one tail = 6/8 = 3/4

(iv) For getting an outcome of no tail, the only possibility is HHH.

So, the total number of favourable outcomes is 1.

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

Thus, the probability of getting no tails is 1/8.

Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively. 

Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) = P (C) = P (D). 

Since out of 16 balls, 11 are not white, therefore, P (A) = \(\frac{11}{16}\).

∴ Required probability = P (A) . P (B) . P (C) . P (D)

= \(\frac{11}{16}\) x \(\frac{11}{16}\) x \(\frac{11}{16}\) x \(\frac{11}{16}\) = \(\big(\frac{11}{16}\big)^4.\)

Correct option is: C) 7/8

When two coins are tossed the sample space will be 

S = {(H, H), (H, T), (T, H), (T, T)} 

n(S) = 4

(i) probability of getting 1 head and one tail = 2/4 = 1/2

(ii) Probability of getting almost two tails = 4/4 = 1

Given total number of times a coin is tossed = 1000

Number of times a head comes up = 445

Number of times a tail comes up = 555

(i) Probability of getting head = number of heads/total number of trails

= (445/1000)

= 0.445

(ii) Probability of getting tail = number of tail/total number of trails

= (555/1000)

= 0.555

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} n(S)=8 

Let A: At least one head & at least one tail. 

A = {HTH, HHT, HTT, THH, THT, TTH}, n(A) = 6 

P(A) = \(\frac{n(A)}{n(S)} = \frac{6}{8} \)

= \(\frac{3}{4}\)

Given that the probability of showing head by a coin when tossed = p 

∴ Prob. of coin showing a tail = 1 – p 

Now pn = prob. that no two or more consecutive heads occur when tossed n times. 

∴ p₁ = prob. of getting one or more on no head = prob. of H or T = 1 

Also p₂ = prob. of getting one H or no H 

= P (HT) + P (TH) + P (TT) 

= p(1 – p) + p(1 – p)p + (1 – p) (1 – p) 

= 1 – p2, For n ≥ 3 

P_n = prob. that no two or more consecutive heads occur when tossed n times.

= p (last outcome is T) P (no two or more consecutive heads in (n – 1) throw) + P (last outcome H) P((n – 1)th throw results in a T) P (no two or more consecutive heads in (n – 2) n throws) 

= (1 – p)P_{n -1} + p(1 – p)p_{n – 2} 

The total no. of outcomes = 6ⁿ 

We can choose three numbers out of 6 in ⁶C₃ ways. By using three numbers out of 6 we can get 3ⁿ sequences of length n. But these sequences of length n which use exactly two numbers and exactly one number. 

The number of n – sequences which use exactly two numbers 

= ³C₂ [2ⁿ – 1ⁿ – 1ⁿ] = 3(2ⁿ – 2) and the number of n sequence which are exactly one number 

= (³C₁) (In) = 3 

Thus, the number of sequences, which use exactly three numbers 

= ⁶C₃ [3ⁿ – 3(2ⁿ – 2) – 3] = ⁶C₃ [3ⁿ – 3(2ⁿ) + 3]

∴ Probability of the required event,

= ⁶C₃[3ⁿ – 3(2ⁿ) + 3]/6ⁿ

Let W₁ (B₁) be the event that a white (a black) ball is drawn in the first draw and let W be the event that a white ball is drawn in the second draw. Then 

P(W) = P(B₁). P(W| B₁) + P(W₁). P (W| W₁) 

= n/m + n. m/m + n + k + m/m + n. m + k/m + n + k 

= m(n + m + k)/(m + n) (m + n + k) = m/m + n 

Solution: Total number of outcomes = 24

P (Green) = 2/3

If number of green balls is G then;

G/24 = 2/3

or, G = (24*2)/3 = 16

Hence, number of blue balls = 24 – 16 = 8

(d) \(\frac{31}{66}\)

Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.

Then

n(S) = ¹²C₂ = \(\frac{12\times11}{2}\) = 66

∴ Let A : Event of drawing two red balls

⇒ n(A) = ⁷C₂ = \(\frac{7\times6}{2}\) = 21

B : Event of drawing two green balls

⇒ n(B) = ⁵C₂ = \(\frac{5\times4}{2}\) = 10

∴ P(Event of drawing 2 balls of same colour) = P(Drawing two red balls) or P(drawing two green balls)

= P(A) + P(B) = \(\frac{n(A)}{n(S)}\) + \(\frac{n(B)}{n(S)}\) = \(\frac{21}{66}\) + \(\frac{10}{66}\) = \(\frac{31}{66}\).