Chapter 16: Probability Exercise 16.1

Solution:-

Since either coin can turn up the Head (H) or Tail (T), are the possible outcomes,

When 1 coin is tossed once the sample space = 2

Then,

The coin is tossed 3 times the sample space = 2³ = 8

Thus, the sample space is S = {HHH, THH, HTH, HHT, TTT, HTT, THT, TTH}

Solution:-

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

Then, the total number of sample spaces = (6 × 6)

= 36

Thus, the sample space is

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)}

Solution:-

Since either coin can turn up the Head (H) or Tail (T), are the possible outcomes,

When 1 coin is tossed once the sample space = 2

Then,

The coin is tossed 3 times the sample space = 2⁴ = 16

Thus, the sample space is S = {HHHH, THHH, HTHH, HHTH, HHHT, TTTT, HTTT, THTT, TTHT, TTTH, TTHH, HHTT, THTH, HTHT, THHT, HTTH}

Solution:-

Since either coin can turn up Head (H) or Tail (T), are the possible outcomes,

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible numbers that come when the die is thrown.

Then, total number of space = (2 × 6) = 12

Thus, the sample space is,

S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

Solution:-

Since either coin can turn up Head (H) or Tail (T), are the possible outcomes,

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible numbers that come when the die is thrown.

When the head is encountered,

Then, number of space = (1 × 6) = 6

Sample Space S_H= {H1, H2, H3, H4, H5, H6}

Now, the tail is encountered, so sample space S_T = {T}

Therefore, the total sample space S = {H1, H2, H3, H4, H5, H6, T}.

Solution:-

It is given that,

2 boys and 2 girls are in Room X

1 boy and 3 girls in Room Y

Let us assume b1, b2 and, g1, g2 be 2 boys and 2 girls in Room X.

And also, assume b3 and g3, g4, and g5 be 1 boy and 3 girls in Room Y.

The problem is solved by dividing it into two cases:

Case 1: Room X is selected

Sample Space S_x = {(X,b1),(X,b2),(X,g1),(X,g2)}

Case 2: Room Y is selected

Sample Space S_y ={(Y,b3),(Y,g3),(Y,g4),(Y,g5)}

The overall sample space

S = {(X,b1),(X,b2),(X,g1),(X,g2),(Y,b3),(Y,g3),(Y,g4),(Y,g5)}.

Solution:-

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible numbers that come when the die is thrown.

And also, assume die of red colour be ‘R’, die of white colour be ‘W’, die of blue colour be ‘B’.

So, the total number of sample space = (6 × 3) = 18

The sample space of the event is

S={(R,1),(R,2),(R,3),(R,4),(R,5),(R,6),(W,1),(W,2),(W,3),(W,4),(W,5),(W,6) (B,1),(B,2),(B,3),(B,4),(B,5),(B,6)}.

Solution:-

Let us assume the boy be ‘B’, and the girl be ‘G’.

(i) The sample space if we are interested in knowing whether it is a boy or girl in the order of their births, S = {GG, BB, GB, BG}.

(ii) The sample space if we are interested in the number of girls in the family when there are two children in the family, then

Sample space S = {2, 1, 0}.

Solution:-

It is given that a box contains 1 red and 3 identical white balls.

Let us assume ‘R’ be the event of the red ball being drawn and ‘W’ be the event of the white ball being drawn.

Given in the question that white balls are identical; therefore the event of drawing any one of the three white balls is the same.

Then, total number of sample space = (2² – 1) = 3

∴ Sample space S = {WW, WR, RW}.

Solution:-

Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

Since either coin can turn up Head (H) or Tail (T), are the possible outcomes,

Let us take,

Case 1: Head is encountered

Sample space S₁ = {HT, HH}

Case 2: Tail is encountered

Sample Space S₂ = {(T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

Then,

The Overall Sample space

S = {(HT), (HH), (T1), (T2), (T3), (T4), (T5), (T6)}.