A sequence a, ax, ax2, ......, axn, has odd number of terms. Then the median is

 

Let , S -  sample space        E - event of selecting 1 girl and 2 boys. 

Then, n(S) = Number ways of selecting 3 students out of 25 

                = 25C3 

                = 2300.

n(E) = 10C1×15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

                = 25C3  = 2300.

         n(E)= 10C1*15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

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

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7 =7C2 = 21

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =5C2 = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green 

            = event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

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.