Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

Total number of elementary events = 50 C 5
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = 29 C 2 ways and remaining two in 20 C 2 ways.

Therfore,favourable number of events =  29 C 2 * 20 C 2

Hence,required probability =  29 C 2 * 20 C 2 / 50 C 5 =551 / 15134

Let   A = Event that A speaks the truth

B = Event that B speaks the truth 

Then P(A) = 75/100 = 3/4

P(B) = 80/100 = 4/5

P(A-lie) =  1 - 3 4= 1/4 

P(B-lie) = 1 - 4 5= 1/5

Now, A and B contradict each other =[A lies and B true] or [B true and B lies]

 = P(A).P(B-lie) + P(A-lie).P(B) 

 =  3 5 * 1 5 + 1 4 * 4 5 = 7 20  

 =  7 20 * 100= 35%

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now,  P A E 1 = Probability of drawing a red ball when the first bag has been selected = 4/7

   P A E 2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) =  P E 1 × P A E 1 + P E 2 × P A E 2 

                          =  1 2 × 4 7 + 1 2 × 2 6 = 19 42

We know that,

Total number of balls n(S) = 26

Number of vowels n(E) = 5

Hence, required probability = n(E)/n(S) = 5/26.

3 letters can be choosen out of 9 letters in 9 C 3 ways.

More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in  ( 4 C 2 * 5 C 1 ) + 4 C 3 ways

Hence,required probability = 4 C 2 * 5 C 1 + 4 C 3 9 C 3 = 17/42

One person can select one house out of 3= 3 C 1 ways =3.

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

Therefore, probability that all thre apply for the same house is 1/9

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

Total number of elementary events =  10 C 5

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is  3 C 1 * 7 C 4

So,required probability = 3 C 1 * 7 C 4/ 10 C 5 = 5/12.