# Probability Chapter 16 Class 11 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 11 Maths Chapter 16‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Maths Chapter 16 Exercise Solution’ using the download button.

### Chapter 16: Probability

#### 16.1 Introduction

In earlier classes, we studied the concept of probability as a measure of uncertainty of various
phenomena.

We have obtained the probability of getting an even number in throwing a die as 3/6 i.e., 1/2 Here the total possible outcomes are 1,2,3,4,5 and 6 (six in number).

The outcomes in favor of the event of ‘getting an even number’ are 2,4,6 (i.e., three in number).

In general, to obtain the probability of an event, we find the ratio of the number of outcomes favorable to the event, to the total number of equally likely outcomes. This theory of probability is known as the classical theory of probability.

In Class IX, we learned to find the probability on the basis of observations and collected data. This is called the statistical approach of probability.

Both theories have some serious difficulties. For instance, these theories can not be applied to the activities/experiments which have an infinite number of outcomes.

In classical theory we assume all the outcomes to be equally likely. Recall that the outcomes are called equally likely when we have no reason to believe that one is more likely to occur than the other.

In other words, we assume that all outcomes have an equal chance (probability) to occur.

Thus, to define probability, we used equally likely or equally probable outcomes.

This is logically not a correct definition. Thus, another theory of probability was developed by A.N. Kolmogorov, a Russian mathematician, in 1933.

### NCERT Solutions Class 11 Maths Chapter 16 Probability

1. A coin is tossed three times.

Solution:-

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

So, when 1 coin is tossed once the sample space = 2

Then,

Coin is tossed 3 times the sample space = 23 = 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 × 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 Head (H) or Tail (T), these are the possible outcomes.

So, when 1 coin is tossed once the sample space = 2

Then,

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}

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

Solution:-

Since either coin can turn up Head (H) or Tail (T), these 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)}

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

Solution:-

Since either coin can turn up Head (H) or Tail (T), these 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, number of space = (1 × 6) = 6

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

Now, tail is encountered, Sample space ST= {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:-

From the question, 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 are 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,b1),(X,b2),(X,g1),(X,g2)}

Case 2: Room Y is selected

Sample Space Sy ={(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 color, one of white color and one of blue colour are placed in a bag. One die is selected at random and rolled, 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 color 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)}