Examples - Chapter 14 Probability class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 15 - Probability - Ncert Solutions class 10 - Maths

Example 1 :

Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
Solution : In the experiment of tossing a coin once, the number of possible outcomes is two - Head (H) and Tail (T). Let $\mathrm{E}$ be the event 'getting a head'. The number of outcomes favourable to E, (i.e., of getting a head) is 1 . Therefore,
$
P(E)=P(\text { head })=\frac{\text { Number of outcomes favourable to } E}{\text { Number of all possible outcomes }}=\frac{1}{2}
$

Similarly, if $\mathrm{F}$ is the event 'getting a tail', then
$
P(F)=P(\text { tail })=\frac{1}{2} \quad \text { (Why ?) }
$

Example 2 :

A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the
(i) yellow ball?
(ii) red ball?
(iii) blue ball?

Solution :

Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.

Let $\mathrm{Y}$ be the event 'the ball taken out is yellow', B be the event 'the ball taken out is blue', and R be the event 'the ball taken out is red'.
Now, the number of possible outcomes $=3$.
(i) The number of outcomes favourable to the event $\mathrm{Y}=1$.

So,
$
\mathrm{P}(\mathrm{Y})=\frac{1}{3}
$

Similarly,
(ii) $\mathrm{P}(\mathrm{R})=\frac{1}{3}$ and
(iii) $\mathrm{P}(\mathrm{B})=\frac{1}{3}$.

Remarks :
1. An event having only one outcome of the experiment is called an elementary event. In Example 1, both the events E and F are elementary events. Similarly, in Example 2, all the three events, Y, B and R are elementary events.
2. In Example 1, we note that : $P(E)+P(F)=1$

In Example 2, we note that : $P(Y)+P(R)+P(B)=1$
Observe that the sum of the probabilities of all the elementary events of an experiment is 1 . This is true in general also.

Example 3 :

Suppose we throw a die once. (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or equal to 4 ?

Solution :

(i) Here, let E be the event 'getting a number greater than 4'. The number of possible outcomes is six : $1,2,3,4,5$ and 6 , and the outcomes favourable to $E$ are 5 and 6. Therefore, the number of outcomes favourable to $\mathrm{E}$ is 2 . So,
$
P(E)=P(\text { number greater than } 4)=\frac{2}{6}=\frac{1}{3}
$
(ii) Let $\mathrm{F}$ be the event 'getting a number less than or equal to 4 '.

Number of possible outcomes $=6$
Outcomes favourable to the event $\mathrm{F}$ are 1, 2, 3, 4 .
So, the number of outcomes favourable to $\mathrm{F}$ is 4 .
Therefore,
$
\mathrm{P}(\mathrm{F})=\frac{4}{6}=\frac{2}{3}
$

Are the events $\mathrm{E}$ and $\mathrm{F}$ in the example above elementary events? No, they are not because the event $\mathrm{E}$ has 2 outcomes and the event $\mathrm{F}$ has 4 outcomes.

Example 4 :

One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will
(i) be an ace,
(ii) not be an ace.

Solution : Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck. Let $\mathrm{E}$ be the event 'the card is an ace'.
The number of outcomes favourable to $E=4$
The number of possible outcomes $=52$ (Why?)
Therefore,
$
\mathrm{P}(\mathrm{E})=\frac{4}{52}=\frac{1}{13}
$
(ii) Let $\mathrm{F}$ be the event 'card drawn is not an ace'.
The number of outcomes favourable to the event $F=52-4=48$ (Why?)

The number of possible outcomes $=52$
Therefore,
$
\mathrm{P}(\mathrm{F})=\frac{48}{52}=\frac{12}{13}
$

Remark: Note that $\mathrm{F}$ is nothing but $\overline{\mathrm{E}}$. Therefore, we can also calculate $\mathrm{P}(\mathrm{F})$ as follows: $\mathrm{P}(\mathrm{F})=\mathrm{P}(\overline{\mathrm{E}})=1-\mathrm{P}(\mathrm{E})=1-\frac{1}{13}=\frac{12}{13}$.

Example 5 :

Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62 . What is the probability of Reshma winning the match?
Solution :

Let $\mathrm{S}$ and $\mathrm{R}$ denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
The probability of Sangeeta's winning $=P(S)=0.62$ (given)
The probability of Reshma's winning $=\mathrm{P}(\mathrm{R})=1-\mathrm{P}(\mathrm{S})$
[As the events $R$ and $S$ are complementary]
$
=1-0.62=0.38
$

Example 6 :

Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution :

Out of the two friends, one girl, say, Savita's birthday can be any day of the year. Now, Hamida's birthday can also be any day of 365 days in the year.
We assume that these 365 outcomes are equally likely.
(i) If Hamida's birthday is different from Savita's, the number of favourable outcomes for her birthday is $365-1=364$
So, $\mathrm{P}$ (Hamida's birthday is different from Savita's birthday) $=\frac{364}{365}$
(ii) $\mathrm{P}$ (Savita and Hamida have the same birthday)
$
\begin{aligned}
& =1-\mathrm{P} \text { (both have different birthdays) } \\
& =1-\frac{364}{365} \quad[\text { Using } \mathrm{P}(\overline{\mathrm{E}})=1-\mathrm{P}(\mathrm{E})] \\
& =\frac{1}{365}
\end{aligned}
$

Example 7:

There are 40 students in Class $\mathrm{X}$ of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?

Solution :

There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl $=25$ (Why?) Therefore, $\mathrm{P}($ card with name of a girl $)=\mathrm{P}(\mathrm{Girl})=\frac{25}{40}=\frac{5}{8}$
(ii) The number of outcomes favourable for a card with the name of a boy $=15$ (Why?) Therefore, $\mathrm{P}($ card with name of a boy $)=\mathrm{P}(\mathrm{Boy})=\frac{15}{40}=\frac{3}{8}$

Note: We can also determine $\mathrm{P}$ (Boy), by taking
$
P(\text { Boy })=1-P(\text { not Boy })=1-P(\text { Girl })=1-\frac{5}{8}=\frac{3}{8}
$

Example 8 :

A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) white?
(ii) blue?
(iii) red?

Solution :

Saying that a marble is drawn at random is a short way of saying that all the marbles are equally likely to be drawn. Therefore, the
$
\text { number of possible outcomes }=3+2+4=9
$
(Why?)
Let $\mathrm{W}$ denote the event 'the marble is white', B denote the event 'the marble is blue' and $R$ denote the event 'marble is red'.
(i) The number of outcomes favourable to the event $\mathrm{W}=2$

So,
$
\mathrm{P}(\mathrm{W})=\frac{2}{9}
$

Similarly,
(ii) $\mathrm{P}(\mathrm{B})=\frac{3}{9}=\frac{1}{3}$ and
(iii) $\mathrm{P}(\mathrm{R})=\frac{4}{9}$

Note that $\mathrm{P}(\mathrm{W})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{R})=1$.

Example 9 :

Harpreet tosses two different coins simultaneously (say, one is of ₹ 1 and other of ₹ 2). What is the probability that she gets at least one head?
Solution :

We write $\mathrm{H}$ for 'head' and $\mathrm{T}$ for 'tail'. When two coins are tossed simultaneously, the possible outcomes are $(\mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{T})$, which are all equally likely. Here $(\mathrm{H}, \mathrm{H})$ means head up on the first coin (say on $₹ 1$ ) and head up on the second coin (₹ 2). Similarly (H, T) means head up on the first coin and tail up on the second coin and so on.

The outcomes favourable to the event $\mathrm{E}$, 'at least one head' are $(\mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T})$ and ( $\mathrm{T}, \mathrm{H}$ ). (Why?)
So, the number of outcomes favourable to $\mathrm{E}$ is 3 .
Therefore,
$
\mathrm{P}(\mathrm{E})=\frac{3}{4}
$
i.e., the probability that Harpreet gets at least one head is $\frac{3}{4}$.

Note : You can also find $\mathrm{P}(\mathrm{E})$ as follows:

$
P(E)=1-P(\overline{\bar{E}})=1-\frac{1}{4}=\frac{3}{4} \quad\left(\text { Since } P(\overline{\bar{E}})=P(\text { no head })=\frac{1}{4}\right)
$

Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite? If not, check it now.

There are many experiments in which the outcome is any number between two given numbers, or in which the outcome is every point within a circle or rectangle, etc. Can you now count the number of all possible outcomes? As you know, this is not possible since there are infinitely many numbers between two given numbers, or there are infinitely many points within a circle. So, the definition of (theoretical) probability which you have learnt so far cannot be applied in the present form. What is the way out? To answer this, let us consider the following example :

Example $10^*$ :

In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?
Solution :

Here the possible outcomes are all the numbers between 0 and 2. This is the portion of the number line from 0 to 2 (see Fig. 14.1).

Let $\mathrm{E}$ be the event that 'the music is stopped within the first half-minute'.
The outcomes favourable to $\mathrm{E}$ are points on the number line from 0 to $\frac{1}{2}$. The distance from 0 to 2 is 2 , while the distance from 0 to $\frac{1}{2}$ is $\frac{1}{2}$.

Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the distance favourable to the event $E$ is $\frac{1}{2}$.

So, $\quad \mathrm{P}(\mathrm{E})=\frac{\text { Distance favourable to the event } \mathrm{E}}{\text { Total distance in which outcomes can lie }}=\frac{\frac{1}{2}}{2}=\frac{1}{4}$
Can we now extend the idea of Example 10 for finding the probability as the ratio of the favourable area to the total area?

Example 11* :

A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig. 14.2. What is the probability that it crashed inside the lake shown in the figure?

Solution:

The helicopter is equally likely to crash anywhere in the region. Area of the entire region where the helicopter can crash
$
=(4.5 \times 9) \mathrm{km}^2=40.5 \mathrm{~km}^2
$

Area of the lake $=(2.5 \times 3) \mathrm{km}^2=7.5 \mathrm{~km}^2$
Therefore, $\mathrm{P}$ (helicopter crashed in the lake) $=\frac{7.5}{40.5}=\frac{75}{405}=\frac{5}{27}$
Example 12 :

A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?

Solution :

One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes.
(i) The number of outcomes favourable (i.e., acceptable) to Jimmy $=88$ (Why?) Therefore, P (shirt is acceptable to Jimmy) $=\frac{88}{100}=0.88$
(ii) The number of outcomes favourable to Sujatha $=88+8=96$ (Why?) So, $P$ (shirt is acceptable to Sujatha) $=\frac{96}{100}=0.96$

Example 13 :

Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) 8 ?
(ii) 13 ?
(iii) less than or equal to 12 ?

Solution:

When the blue die shows ' 1 ', the grey die could show any one of the numbers $1,2,3,4,5,6$. The same is true when the blue die shows ' 2 ', ' 3 ', ' 4 ', ' 5 ' or ' 6 '. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.

Note that the pair $(1,4)$ is different from $(4,1)$. (Why?)
So, the number of possible outcomes $=6 \times 6=36$.
(i) The outcomes favourable to the event 'the sum of the two numbers is 8 ' denoted by $\mathrm{E}$, are: $(2,6),(3,5),(4,4),(5,3),(6,2)$ (see Fig. 14.3)
i.e., the number of outcomes favourable to $\mathrm{E}=5$.
Hence,
$
\mathrm{P}(\mathrm{E})=\frac{5}{36}
$
(ii) As you can see from Fig. 14.3, there is no outcome favourable to the event F, 'the sum of two numbers is 13 '.
So,
$
\mathrm{P}(\mathrm{F})=\frac{0}{36}=0
$
(iii) As you can see from Fig. 14.3, all the outcomes are favourable to the event G, 'sum of two numbers $\leq 12$ '.
So,
$
\mathrm{P}(\mathrm{G})=\frac{36}{36}=1
$