Solution:

Clearly the coins are distinguishable in the sense that we can speak of the first coin and the second coin. Since either coin can turn up Head (H) or Tail(T), the possible outcomes may be
Heads on both coins = (H,H) = HH
Head on first coin and Tail on the other = (H,T) = HT
Tail on first coin and Head on the other = (T,H) = TH
Tail on both coins = (T,T) = TT
Thus, the sample space is S = {HH, HT, TH, TT}

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Solution:

Suppose 1 appears on blue die and 2 on the red die. We denote this outcome
by an ordered pair (1,2). Simlarly, if ‘3’ appears on blue die and ‘5’ on red, the outcome
is denoted by the ordered pair (3,5).
In general 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 number of elements of this sample space is `6 x× 6 = 36` and the sample space is
given below:
`{(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)}`

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Solution:

(i) Let Q denote a 1 rupee coin, H denotes a 2 rupee coin and R denotes a 5
rupee coin. The first coin he takes out of his pocket may be any one of the three coins
Q, H or R. Corresponding to Q, the second draw may be H or R. So the result of two
draws may be QH or QR. Similarly, corresponding to H, the second draw may be
Q or R.
Therefore, the outcomes may be HQ or HR. Lastly, corresponding to R, the second
draw may be H or Q.
So, the outcomes may be RH or RQ.
Thus, the sample space is S={QH, QR, HQ, HR, RH, RQ}
(ii) The number of accidents along a busy highway during the year of observation
can be either 0 (for no accident ) or 1 or 2, or some other positive integer.
Thus, a sample space associated with this experiment is `S= {0,1,2,...}`

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Solution:

Let us denote blue balls by `B_1, B_2, B_3` and the white balls by `W_1, W_2, W_3, W_4.`
Then a sample space of the experiment is
`S = { HB_1, HB_2, HB_3, HW_1, HW_2, HW_3, HW_4,
T1, T2, T3, T4, T5, T6}.`
Here HBi means head on the coin and ball Bi is drawn, `HW_i` means head on the coin
and ball Wi is drawn. Similarly, Ti means tail on the coin and the number i on the die.

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Solution:

In the experiment head may come up on the first toss, or the 2nd toss, or the 3rd toss and so on till head is obtained. Hence, the desired sample space is `S= {H, TH, T TH, T T T H, T T T T H,...}`

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