Chapter 16: Probability Exercise 16.1

Solution:-

From the question,

‘D’ denotes the event that the bulb is defective, and ‘N’ denotes the event of non-defective bulbs.

Then,

Total number of Sample space = 2 × 2 × 2 = 8

Thus, Sample space S = {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}.

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 outcomes when the die is thrown.

The problem can be solved by dividing it into 3 cases

Case 1: The outcome is Head, and the corresponding number on the die shows an odd number.

Total number of sample space = (1 × 3) = 3

Sample space S_HO = {(H,1), (H,3), (H,5)}

Case 2: The outcome is Head, and the corresponding number on the die shows an even number

Total number of sample space = (1 × 3 × 6) = 18

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

Case 3: The outcome is Tail

Total number of sample space=1

Sample space S_T = {(T)}

The overall sample spaces

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

(H,4,6), (H,6,1),(H,6,2),(H,6,3),(H,6,4),(H,6,5),(H,6,6),(T)}.

Solution:-

It is given that 1, 2, 3, and 4 are the numbers written on the four slips.

When two slips are drawn without replacement, the first event has 4 possible outcomes, and the second event has 3 possible outcomes because 1 slip is already picked.

Therefore, the total number of possible outcomes = (4 × 3) = 12

This sample space,

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)}.

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 outcomes when the die is thrown.

The following problem can be divided into two cases:

(i) The number on the die is even.

The sample space S_E={(2,H),(4,H),(6,H),(2,T),(4,T),(6,T)}

(ii) The number on the die is odd, and the coin is tossed twice.

The sample space

S_o = {(1,H,H), (3,H,H), (5,H,H), (1,H,T), (3,H,T), (5,H,T), (1,T,H), (3,T,H), (5,T,H), (1,T,T), (3,T,T), (5,T,T)}

Hence, the overall sample space for the problem = S_E + S_o

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

Solution:-

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

Let us assume R₁ and R₂ denote the event the red balls are drawn, and B₁, B₂, and B₃ denote the events black balls are drawn.

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

(i) Coin shows Tail.

So, the sample space S_T = {(TR₁), (TR₂), (TB₁), (TB₂), (TB₃)}

(ii) Coin shows Head.

So, the sample space S_H = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6)}.

Hence, the overall sample space for the problem = S_T + S_H

S = {( T,R₁), (T,R₂), (T,B₁), (T,B₂), (T,B₃), (H,1), (H,2), (H,3), (H,4), (H,5), (H,6)}.

Solution:-

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

As per the condition given in the question, a die is thrown repeatedly until a six comes up.

If six may come up for the first throw or six may come up on the second throw, this process will go continuously until six comes.

The sample space when 6 comes on the very first throw S₁ = {6}

The sample space when 6 comes on the second throw S₂ = {(1,6), (2,6), (3,6), (4,6), (5,6)}

This event can go on for infinite times.

So, the sample space is infinitely defined

S = {(6), (1,6), (2,6), (3,6), (4,6), (5,6), (1,1,6), (1,2,6)……}.