Since, A & B are independent events.

P(A)=0.45,

P(B)=0.35 (events are mutually exclusive)

P(A∩B)=0

8 couples are there out of which 4 persons are picked for a prize.

P (At least one couple in 4)=1−P ( No Couple in 4)

Favourable cases of getting 10 or greater than 10, if 5 appears on atleast one of dice.

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

Number of favourable cases =3

Total number of cases =6×6=36

∴ Required probability  = 3/36 = 1/12

Total outcomes =2 × 2 × 2 = 8

Unfavourable outcomes =2 = 2 , , i.e. either (a,b,c) or (b,c,a)

Hence, required probability = 1 - faces does not show same letter

Since the tickets are drawn with replacement, the same ticket may be repeated. Since the largest ticket is k, the tickets chosen are from 1 to k.

Total no. of ways of doing it = kⁿ

However, k may or may not have been drawn.

No. of ways of choosing numbers from 1 to (k - 1) = (k - 1)ⁿ

Let A be the event of obtaining an even sum and B be the event of obtaining a sum less 5

Then, we have to find P(A∪B).Since, A,B are not mutually exclusive, we have

P(A∪B)=P(A)+P(B)−P(A∩B)

[Since, there are 18 18 ways to get an even sum and 6 6 ways to get a sum less than 5 ie. (1,3), (3,1), (2,2), (1,2), (2,1),(1,1) and 4 ways to get an even sum less than 5, ie,(1,3), (3,1), (2,2), (1,1).]

The sum of two numbered on a dice is odd only, when one number is odd and second is even.

∴ Required probability

= 2× Probability of odd number × Probability of even number

[ ∵ Here, we multiply by 2 because either the even number is on first or on second dice.]