S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

  E 3 = E 1 ? E 2 = { 6 , 12 , 18 } 

=> n(E3) = 3

? E = E 1 ? E 2 = E 1 + E 2 - E 3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

? P ( E ) = n ( E ) n ( S ) = 12 18 = 2 3

n(S) =  C 2 52 = 1326

 Let  A = event of getting both red cards

and B = event of getting both queens

then  A ? B = event of getting two red queens

n(A) =  C 2 26 = 325,   n(B) =  C 2 4 = 6

  n ( A ? B ) = C 2 2 = 1

  ? P A = 325 1326 , P B = 6 1326

  P A ? B = 1 1326

?P ( both red or both queens) =  P A ? B

=  P A + P B - P A ? B = 325 1326 + 1 221 - 1 1326= 55 221

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14=4/7.

Total number of cases=p+q

Favourable cases=p

Probability of that event is=p/(p+q)

Let A be the event of driver A drive safely after consuming liquor. 

Let B be the event of driver B drive safely after consuming liquor. 

Let C be the event of driver C drive safely after consuming liquor. 

P(A)=2/5,P(B)=3/7,P(C)=3/4

The events A, B and C are independent . Therefore,

P A ? B ? C = P ( A ) P ( B ) P ( C ) = 2 5 * 3 7 * 3 4 = 9 70

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

Probability of atleast one failure
= 1 - no failure > 31/32
= 1 - P 5 > 31/32
=   P 5 <1/32
= p < 1/2
Also p > 0
Hence p lies in [0,1/2].

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address....so the remaining one letter automatically reach to their correct address
P(E) = favorable outcomes /total outcomes
Here favorable outcomes are '0'.
So probability is '0'.

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

P(E) =n(E)/n(S)=2/52=1/26.