P(odd) = P (even) = 1 2 1(because there are 50 odd and 50 even numbers)

Sum or the three numbers can be odd only under the following 4 scenarios:

Odd + Odd + Odd =  1 2 * 1 2 * 1 2=  1 8

Odd + Even + Even =  1 2 * 1 2 * 1 2= 1 8

Even + Odd + Even =  1 2 * 1 2 * 1 2= 1 8

Even + Even + Odd =  1 2 * 1 2 * 1 2 =  1 8

Other combinations of odd and even will give even numbers. 

Adding up the 4 scenarios above:

=  1 8+  1 8+ 1 8+  1 8 =  4 8 =  1 2

Probability of occurrence of an event, 

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

? Probability of getting head in one coin = ½, 

? Probability of not getting head in one coin = 1- ½ = ½, 

Hence, 

All the 11 tosses are independent of each other.

? Required probability of getting only 2 times heads = 1 2 2 × 1 2 9 = 1 2 11

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus   n(E) = 13 + 3 = 16

Hence  n E ¯ = 52 - 16 = 36 

Therefore,  P E = 36 52 = 9 13

P ( neither A nor B) =  P A a n d B  

=   P A ? B=   P A ? B=  1 - P A ? B  

=  1 - 61 100 = 39 100

 Let S be the sample space

Then, n(S) =  52 C 2= 1326.

Let E = event of getting 2 kings out of 4.

n(E)  =  52 * 51 2 * 1= 6.

  4 C 2

=> 4 * 3 2 * 1

P ( E ) = n ( E ) n ( S ) = 6 1326 = 1 221

Let S be the sample space.
Here n(S)= 2 3 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events =  1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.