NCERT Solutions for Class 10 Maths Chapter 15: Probability - Exercise 15.2

Exercise: 15.2                            

1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?  

Ans: If Shyam and Ekta visit the same store on the same week, which begins on Tuesday and ends on Saturday, the total number of days in the week is $5$. 

Now, if they're both equally likely to go to the store, 

The outcome of the event of Shyam visiting the shop is $5$ 

The outcome of the event of Ekta visiting the shop is $5$ 

The total number of ways (combination):  (TT), (TW), (TTh),(TF), (TS), (WT), (WW), (WTh), (WF), (WS), (ThT), (ThW), (ThTh), (ThF), (ThS), (FT), (FW), (FTh), (FF), (FS), (ST), (SW), (STh), (SF), (SS). 

So, Total outcome events = $5\times 5$ 

= 25

(i) the same day? 

Ans: The Probability of them both arriving to the shop on the same day 

=the total number of ways they can go to the shop on the same day /the total number of ways.

= \[\frac{\left( T,T \right)\text{ },\text{ }\left( W,W \right)\text{ },\text{ }\left( Th,\text{ }Th \right),\text{ }\left( F,F \right),\text{ }\left( S,S \right)}{25}\] 

= $\frac{5}{25}$

=$\frac{1}{5}$

(ii) consecutive days?

Ans: The Probability of both of them arriving on the consecutive days  

= the total number of ways they can go to the shop on the consecutive days /the total number of ways

= $\frac{\left( \text{T,W} \right)\text{, }\left( \text{W,Th} \right)\text{, }\left( \text{Th, F} \right)\text{, }\left( \text{F,S} \right)\text{, }\left( \text{W,T} \right)\text{,}\left( \text{Th,W} \right)\text{, }\left( \text{F,Th} \right)\text{, }\left( \text{S,F} \right)}{\text{the total number of ways}}$

= $\frac{8}{25}$

(iii) different days?  

Ans: P(both of them arriving on same days) + P(both of them arriving not on same days)= 1 

$\frac{1}{5}$+ P(both of them reaching not on same days) = $1$ 

P(both of them reaching not on same days) = $1-\frac{1}{5}$

P(both of them reaching not on same days) = $\frac{4}{5}$

Hence, the probability of both of them not arriving on the same day = $\frac{4}{5}$

2. A die is numbered in such a way that its faces show the numbers $1$, $2$, $2$, $3$, $3$, $6$. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws: 

What is the probability that the total score is

(i) even?

(ii) $6$?

(iii) atleast $6$?

Ans: The complete table is shown as follows:

Here, total number of outcome=$6\times 6$ =36

\[\text{(i)The probability of achieving an even total score}=\frac{\text{The ways of getting even sum}}{\text{Total ways}}\]=\[\frac{18}{36}=\frac{1}{2}\]

\[\text{(ii)The probability of achieving}\;6=\frac{\text{The ways of getting even sum of }}{\text{Total ways}}=\frac{4}{36}\]=\[\frac{1}{9}\]

\[\text{ }~\text{(iii) The probability of getting atleast}\] 6 $\text{ }~\text{= }\frac{\text{The ways of getting the total sum as 6 or less than 6}}{\text{Total ways}}$ $=\frac{15}{36}$$=\frac{5}{12}$

3. A bag contains $5$ red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag. 

Ans: Given,

The total number of red balls is $5$

Assume, the total number of  blue balls is $x$

So, total balls present= $5+x$

Here, the probability of getting a blue ball is double the probability of getting a red ball so,

(probability of getting a blue ball) =\[2\] probability of getting a red ball -----(1)

(probability of getting a blue ball) = (Number of blue balls/ total balls) =\[\left( \frac{x}{x+5} \right)\]

probability of getting a red ball = (Number of red balls/ total balls)  =\[\left( \frac{5}{x+5} \right)\]

Putting in (1)

\[\left( \frac{x}{x+5} \right)\]

=\[2\times \left( \frac{5}{x+5} \right)\]

\[x=10\]

So, the total number of blue balls is 10

4: A box contains $12$ balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If  more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

Ans: Given, 

Total number of black balls=\[x\]

Total number of balls=12

The probability of getting a black ball = no of black balls/total balls = \[\frac{x}{12}\]

If 6 more balls are added 

The total number of black balls = \[x+6\]

The total number of balls = 18

 The probability of getting a black ball = no of black balls/ total balls = \[\frac{x+6}{18}\]

 The probability of getting a black ball is double as compared to before

So,

\[\frac{x+6}{18}=2\left( \frac{x}{12} \right)\]

\[x=3\]

5. A jar contains $24$ marbles, some are green, and others are blue. If a marble is drawn at random from the jar, the probability that it is green is $2/3$. Find the number of blue balls in the jar.

Ans: Given,

Total marble=24

$\text{ }\!\!~\!\!\text{ Assume green marble}=$x

$\text{The total number of blue marbles }=$24-x

$\text{The probability of getting a green marble}=\frac{\text{number of green marble}}{\text{total number of marbles}}=$$\frac{x}{24}$

Given, the probability of getting a green marble is $\frac{2}{3}$

Hence,

$\frac{x}{24}=\frac{2}{3}$

$x=16$

So, the total number of green marbles=16

$\text{The total number of blue marbles}=$24-168

NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

Opting for the NCERT solutions for Ex 15.2 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 15.2 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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