Total numbers in a die=6

P(mutliple of 3) = 2/6 = 1/3

P(multiple of 4) = 1/6

P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2

Total number of events=52

Number of aces =4

So,required probability = 4/52 =1/13

Let A be the event of A will tell truth. B be the event of B tell truth

  P ( A ) = 2 3 , P ( A c ) = 1 - P ( A ) = 1 3

P ( B ) = 4 5 , P ( B c ) = 1 - P ( B ) = 1 5 

When both agree then they say true or they say false together, that is 

  A ? B o r A c ? B c

Also these events will be mutually exclusive :

  P ( A ? B ) + P ( A c ? B c ) = P ( A ) P ( B ) P ( C ) = 2 3 * 4 5 + 1 3 * 1 5 = 3 5

Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18
Multiples of 5 below 20 are 5, 10, 15, 20
Required number of possibilities = 10
Total number of possibilities = 20
Required probability = 10/20 = 1/2.

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

=> n(S) = 8

E = { HHH, HHT, HTH, THH }

=> n(E) = 4

P(E) = 4/8 = 1/2

Total numbers = 25

 Numbers divisible by 4 or 7 are 4, 7, 8, 12, 14, 16, 20, 21, 24 = 9

 The probability (divisible by 4 or 7) = 9/25

Total number of persons = 9

Out of 9 persons 4 persons can be selected in 9 C 4 ways =126

1 man,1 woman and 2 children can be selected in 3 C 1 * 2 C 1 * 4 C 1 ways =36

So,required probability = 36/126 =2/7

Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability  =(12)×(16)
= 1/12

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299?199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40