MATHS (Each question 2 marks)

The four balls in the box are R, W, W, W.
When two balls are drawn at random without replacement.
then the sample space is given by S = {RW, WR, WW}

Let H denote the event of a head and T denote the event of a tail.
Total number of elements in sample space =2 x 2 x 2 x 2 = 16 (because in single toss of coin sample space will be 2 and the coin is tossed 4 times)
We define the possible outcomes by an ordered set (w,x, y, z)
w denotes the event the coin is tossed for the first time
x denotes the event the coin is tossed for the second time
y denotes the event the coin is tossed for the third time
z denotes the event the coin is tossed for the fourth time
Sample space
S = {(H,H,H,H),(H,H,H,T),(H,H,T,H),(H,T,H,H),(T,H,H,H),(H,H,T,T),(H,T,H,T),(T,H,H,T),(T,H,T,H),(T,T,H,H),(H,T,T,H),(H,T,T,T),(T,H,T,T),(T,T,H,T),(T,T,T,H),(T,T,T,T)}

Let 1,2,3,4,5,6 denote the event the respective numbers comes when the die is thrown
The total number of sample space = (6 x 6) = 36
We define the possible outcomes by an ordered set (x, y)
where,
x denotes the digit occurring on the first die
y denotes the digit occurring on the second die
The sample spaces
S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3)(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

In the given experiment,on throwing a die, six may come up on the first throw, the second throw,
the third throw, the fourth throw and so on till six comes up.
Hence, the sample space (S) of this experiment is given by,
S= {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (2, 1, 6), (2, 2, 6), (2, 3, 6), (2, 4, 6), (2, 5, 6), (3, 1, 6), (3, 2, 6), (3, 3, 6), (3, 4, 6), (3, 5, 6), (4, 1, 6), ... (4, 5, 6), (5, 1, 6), ... (5, 5, 6), ... }.

Here numbers 1, 2,3 and 4 are written separately on four slips of paper, put in a box and mixed thoroughly.
When two slips are drawn from the box without replacement then sample space is given by S = {(1, 2), (1, 3), (1, 4), (2,1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

When a coin is tossed then outcomes are H, T.
When the coin shows T, then there is no action.
When the coin shows H, a die is thrown then outcomes will be 1, 2, 3, 4, 5, 6.
When the die shows 1, 3, 5, then there is no action.
When the die shows 2, 4, 6, a die is thrown again and then outcomes are 1, 2, 3, 4, 5, 6.
Hence the required sample space (S) is given by,
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

Given that, defective bulbs are denoted by D and non-defective bulbs are denoted by N.
So the sample space when 3 bulbs are selected at random from a lot is given by
S - {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}

A coin is tossed then outcomes are H, T
If H comes again thown, then outcomes are H, T.
When a die is rolled then outcomes are 1, 2, 3, 4, 5, 6
Hence in this experiment the required sample space (S),
is given by S= {HH, HT, T1, T2, T3, T4, T5, T6}.

Since in each die the number of outcomes are 6.
So, the number of elements of this sample space is 6 × 6 = 36
Each outcome can be denoted by the ordered pair (x, y), where x is the number appeared on the blue die and y is the number appeared on the red die.
Therefore, this sample space is given by
S = {(x, y): x is the number on the blue die and y is the number on the red die}.
The sample space is given below:
S= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}