P(winning prize at least on one ticket)
= 1 - P("Losing on all tickets")
= 1 - (0.8)⁴ = (1 + (0.8)²)(1 - (0.8)²)
= (1.64)(0.36)
= 0.5904

We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, probability of picking a perfect cube is
$$\eqalign{ & = \frac{4}{{100}} \cr & = \frac{1}{{25}} \cr} $$

Number of Black socks = 5
Total socks = 5 + 3 = 8
First we draw one sock
Probability of it being black = $$\frac{{5}}{{8}}$$
Now a second sock is picked
Probability of its being black = $$\frac{{4}}{{7}}$$
Total Probability $$ = \frac{5}{8} \times \frac{4}{7} = \frac{5}{{14}}$$

X = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
n(X) = 10, n(S) = 30
Hence required probability,
$$\eqalign{ & = \frac{{n(X)}}{{n(S)}} \cr & = \frac{{10}}{{30}} \cr} $$

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)}
$$\eqalign{ & \refore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{4}{{36}}\cr & = \frac{1}{9} \cr} $$

Possible event = 6 × 6 = 36
Favourable outcomes = {(3, 6), {4, 5}, {5, 4}, (6, 3)}
∴ Probability
$$\eqalign{ & = \frac{{{\text{Favourable}}}}{{{\text{total}}}} \cr & = \frac{4}{{36}} \cr & = \frac{1}{9} \cr} $$

The probability that first toy shows even number $$ = \frac{{10}}{{21}}$$
Since, toy is not replaced re are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows even number $$ = \frac{{9}}{{20}}$$
Required probability,
$$\eqalign{ & = \left( {\frac{{10}}{{21}}} \right) \times \left( {\frac{9}{{20}}} \right) \cr & = \frac{9}{{42}} \cr} $$

Given that re are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When four marbles are picked at random, n probability that none is blue is
$$\eqalign{ & = \frac{{{}^{12}{C_4}}}{{{}^{15}{C_4}}} \cr & = \frac{{12 \times 11 \times 10 \times 9}}{{15 \times 14 \times 13 \times 12}} \cr & = \frac{{11880}}{{32760}} \cr & = \frac{{33}}{{91}} \cr} $$

Given that re are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that both marbles are blue
$$\eqalign{ & = \frac{{{}^3{C_2}}}{{{}^{15}{C_2}}} \cr & = \frac{{3 \times 2}}{{15 \times 14}} \cr & = \frac{1}{{35}} \cr} $$
Probability that both are yellow
$$\eqalign{ & = \frac{{{}^2{C_2}}}{{{}^{15}{C_2}}} \cr & = \frac{{2 \times 1}}{{15 \times 14}} \cr & = \frac{1}{{105}} \cr} $$
Probability that one blue and or is yellow
$$\eqalign{ & = \frac{{{}^3{C_1} \times {}^2{C_1}}}{{{}^{15}{C_2}}} \cr & = \frac{{2 \times 3 \times 2}}{{15 \times 14}} \cr & = \frac{2}{{35}} \cr} $$
∴ Required probability
$$\eqalign{ & \frac{1}{{35}} + \frac{1}{{105}} + \frac{2}{{35}} \cr & = \frac{{3 + 1 + 6}}{{105}} \cr & = \frac{{10}}{{105}} \cr & = \frac{2}{{21}} \cr} $$

Total number of balls
= (8 + 7 + 6)
= 21
Let E = event that ball drawn is neir red nor green
= event that ball drawn is blue
$$\eqalign{ & \refore n(E) = 7 \cr & \refore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$