Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) = 10 C 2 10 = 10 * 9 2 * 1 =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

                = 6 C 2 + 4 C 2 =  6 * 5 2 * 1 + 4 * 3 2 * 1= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

Let A, B, C be the respective events of solving the problem and  A , B , C be the respective events of not solving the problem. Then A, B, C are independent event

? A , B , C are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

  P A = 1 2 , P B = 2 3 , P C = 3 4

? P( none  solves the problem) = P(not A) and (not B) and (not C)  

                  =  P A ? B ? C 

                  =  P A P B P C          ? A , B , C a r e I n d e p e n d e n t                       

                  =   1 2 × 2 3 × 3 4  

                  =  1 4  

Hence, P(the problem will be solved) = 1 - P(none solves the problem) 

                =  1 - 1 4= 3/4

We have n(s) = 52 C 2 52 = 52*51/2*1= 1326. 

Let A = event of getting both black cards 

     B = event of getting both queens 

A?B = event of getting queen of black cards 

n(A) = 52 * 51 2 * 1 =  26 C 2 = 325, n(B)= 26 * 25 2 * 1= 4*3/2*1= 6  and  n(A?B) =  4 C 2 = 1 

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and 

P(A?B) = n(A?B)/n(S) = 1/1326 

P(A?B) = P(A) + P(B) - P(A?B) = (325+6-1) / 1326 = 330/1326 = 55/221

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

  P ( E ) = n ( E ) n ( S ) = 10 35 = 2 7

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

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

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

1 year = 365 days . A leap year has 366 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364days

366 ? 364 = 2 days

In a leap year there will be 52 Sundays and 2 days will be left.

These 2 days can be:

1. Sunday, Monday

2. Monday, Tuesday

3. Tuesday, Wednesday

4. Wednesday, Thursday

5. Thursday, Friday

6. Friday, Saturday

7. Saturday, Sunday

Of these total 7 outcomes, the favourable outcomes are 2.

Hence the probability of getting 53 days = 2/7

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 =  15 C 3  = 15 * 14 * 13 3 * 2 * 1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5 C 3 = 5 * 4 2 * 1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.

Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18