** According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 14.*

**NCERT Solutions for Class 11 Maths Chapter 16 Probability** contains answers for Exercise 16.1 questions in the textbook. Chapter 16, Probability of Class 11 Maths, is categorised under the CBSE Syllabus for the academic year 2023-24. The students will be able to boost their confidence levels by solving the textbook questions using the solutions PDF. These solutions are prepared by subject-matter experts at BYJUâ€™S to help students gain conceptual knowledge more effectively. They can download the NCERT Solutions of Class 11 Maths in PDF for free and practise offline to score well in the annual exam.

## NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.1

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### Access Other Exercise Solutions of Class 11 Maths Chapter 16 Probability

Exercise 16.2 Solutions: 7 Questions

Exercise 16.3 Solutions: 21 Questions

Miscellaneous Exercise Solutions: 10 Questions

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NCERT Solutions for Class 11 Maths Chapter 16

#### Access Solutions for Class 11 Maths Chapter 16 Exercise 16.1

**In each of the following questions, 1 to 7, describe the sample space for the indicated experiment. **

**1. A coin is tossed three times.**

**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 = 2^{3} = 8

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

**2. A die is thrown two times.**

**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 Ã— 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)}

**3. A coin is tossed four times.**

**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 = 2^{4} = 16

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

**4. A coin is tossed, and a die is thrown. **

**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 Ã— 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)}

**5. A coin is tossed, and then a die is rolled only in case a head is shown on the coin.**

**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 Ã— 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}.

**6. 2 boys and 2 girls are in Room X, and 1 boy and 3 girls are in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.**

**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)}.

**7. One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected randomly and rolled, and its colour and the number on its uppermost face are noted. Describe the sample space.**

**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 â€˜Râ€™, die of white colour be â€˜Wâ€™, die of blue colour be â€˜Bâ€™.

So, the total number of sample space = (6 Ã— 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)}.

**8. An experiment consists of recording the boy-girl composition of families with 2 children. **

**(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?**

**(ii) What is the sample space if we are interested in the number of girls in the family?**

**Solution:-**

Let us assume the boy be â€˜Bâ€™, and the girl be â€˜Gâ€™.

(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}.

**9. A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.**

**Solution:-**

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

Let us assume â€˜Râ€™ be the event of the red ball being drawn and â€˜Wâ€™ 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 = (2^{2} â€“ 1) = 3

âˆ´ Sample space S = {WW, WR, RW}.

**10. An experiment consists of tossing a coin and then throwing it a second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.**

**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 S_{1 }= {HT, HH}

Case 2: Tail is encountered

Sample Space S_{2 }= {(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)}.

**11. Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.**

**Solution:-**

From the question,

â€˜Dâ€™ denotes the event that the bulb is defective, and â€˜Nâ€™ denotes the event of non-defective bulbs.

Then,

Total number of Sample space = 2 Ã— 2 Ã— 2 = 8

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

**12. A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?**

**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 Ã— 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 Ã— 3 Ã— 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)}.

**13. The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.**

**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 Ã— 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)}.

**14. An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.**

**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)}.

**15. A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows a head, we throw a die. Find the sample space for this experiment.**

**Solution:-**

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

Let us assume R_{1}Â and R_{2} denote the event the red balls are drawn, and B_{1}, B_{2}, and B_{3} 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 }= {(TR_{1}), (TR_{2}), (TB_{1}), (TB_{2}), (TB_{3})}

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

**16. A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?**

**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 S_{1} = {6}

The sample space when 6 comes on the second throw S_{2} = {(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)â€¦â€¦}.

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