Required Probability
Given, n = 5 and r = 3
Then, Success P = 3/4
Failure, q = 1 - 3/4 = 1/4
Man hit the target thrice
= ⁵c₃ (3/4)³ (1/4)² + ⁵c₄ (3/4)³ (1/4) + ⁵c₃(3/4)⁵
= (270/1024) + (405/1024) + (243/1024)
= 918/1024
= 459/512

Total number of favourable outcomes n(S) = 6³ = 216
Combinations of outcomes for getting sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4), (4, 6, 5), (6, 4, 5), (5, 5, 5),(6, 6, 3), (6, 3, 6), (3, 6, 6)
Now, outcomes on which first roll was a four, n(E) = (4, 5, 6),(4, 6, 5)
? P(E) = n(E)/n(S) = 2/216 =1/108

Number of 4- digit numbers which are formed with 1, 3, 5, 7, 9 = ⁵P₄
= 5 x 4 x 3 x 2 = 120 = n (s)

Number of 4 digit number which are formed with 1, 3, 5, 7, 9 and are divisible by 5 = ⁴P₃
= 4 x 3 x 2 = 24 = n (E)

? p(E) = n(E)/n( s) = 24/120
= 1/5

Total number of possible outcomes
= ¹²C₃ = 220
Number of events which do not contain blue marbles (3 marbles out of 7 marbles) = ⁷C₃ = 35
? Required probability = 1 - 35/220 = 37/44

 Vowels are A I A I O, 

 C    A   S    T   I    G   A    T   I    O   N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

n ( E ) = 5 P 5 2 ! * 2 ! ( A , I a r e 2 t i m e s ) * 6 P 6 2 ! ( T i s 2 t i m e s ) = 21600

n ( S ) = 11 ! 2 ! * 2 ! * 2 ! ( A , I a r e 2 t i m e s ) = 4989600

21600 4989600 = 0 . 043

Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2

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

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards
? The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

? required probability = 26/52 = 1/2.