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

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{9}{20}\)

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 

= \(\frac{(7\times6)}{(2\times1)}\)

=21.

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

So, n(E) = = Number of ways of drawing 2 balls out of (2 + 3) balls.

= ⁵C₂

_{= \(\frac{(5\times4)}{(2\times1)}\)}

_=10.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{10}{21}\)

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.

So, n(E) = 7.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{7}{21}=\frac{1}{3}.\)

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

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

So,  \(P(E) = \frac{n(E)}{n(S)}=\frac{4}{36}=\frac{1}{9}.\)

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}.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{7}{8}\)

n 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)}

So, n(E) = 27.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{27}{36}=\frac{3}{4}.\)

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 `

= \(\frac{(25\times24\times23)}{(3\times2\times1)}\)

= 2300.

n(E)  = (¹⁰C₁ x ¹⁵C₂)

=  \(\left[10\times\frac{(15\times14)}{(2\times1)}\right]\)

=1050.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{1050}{2300}=\frac{21}{46}.\)

P (getting a prize) =  \(\frac{10}{(10+25)}=\frac{10}{35}=\frac{2}{7}.\)

Let S be the sample space.

Then, n(S) = ⁵²C₂ =  \(\frac{(52\times51)}{(2\times1)}\)   = 1326.

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

So, n(E) = ⁴C₂ = \(\frac{(4\times3)}{(2\times1)}\)  = 6.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{6}{1326}=\frac{1}{221}.\)

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) }

So, n(E) = 15.

So, \(P(E) = \frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}.\)