Class 10, Maths, Chapter 15, Exercise 15.2, Solutions
Q.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: Total number of days (Tuesday to Saturday) = 5
Shyam can go to the shop in 5 ways
Ekta can go to the shop in 5 ways
Total number of outcomes = 5 x 5 = 25
(i) Possibility of visiting the shop on the same day
{T,T} {W, W} {Th , Th} {F,F} {S,S} = 5 Favourable outcomes
P (visiting the same day) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{5}{25}=\frac{1}{5}$
Hence, required probability = $\frac{1}{5}$
(ii) Consecutive days are (T, W), (W, T), (W, Th), (Th, W), (Th, F), (F, Th), (S, F), (F, S)
∴ Total Favourable outcomes = 8
P (visiting the consecutive days) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{8}{25}$
Hence, required probability = $\frac{8}{25}$
(iii) P (Visiting the different days) = 1 – P (Visiting the same day)
= 1 – $\frac{1}{5}=\frac{5-1}{5}=\frac{4}{5}$
Hence, required probability = $\frac{20}{25}=\frac{4}{5}$
Q.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:
(i) even? What is the probability that the total score is
(ii) 6?
(iii) at least 6?
Ans: Complete table is as under:
Clearly, total number of outcomes = 6 × 6 = 36
(i) Total score even when Favourable outcomes are 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8 and 12
∴ Favourable number of outcomes = 18 case
P (Total score is even number) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{18}{36}=\frac{1}{2}$
Hence, P(Total score is even number) = $\frac{1}{2}$
(ii) Total times when sum is = 4 case
∴ Favourable number of outcomes = 4
P (sum is 6) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{4}{36}=\frac{1}{9}$
Hence, P(sum is 6) = $\frac{4}{36}=\frac{1}{9}$
(iii) Let A be the event of the total score is at least 6. Then, elementary events Favourable to A are
7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12.
∴ Favourable number of elementary events = 15
Hence, P(A) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{15}{36}=\frac{5}{12}$
Q.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: let there be ‘x’ blue balls in the bag and number of red balls are 5.
∴ Total numbers of balls in the bag = 5 + x
Now, P1 = Probability of drawing a blue ball = $\frac{x}{5+x}$
P2 = Probability of drawing a red ball = $\frac{5}{5+x}$
According to question, P1 = 2P2
⇒ $\frac{x}{5+x}=2\times \frac{5}{5+x}$
⇒ x = 10
Hence, there are 10 blue balls in the bag.
Q.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 6 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: There are 12 balls in the box. Out of these 12 balls, one can be chosen in 12 ways
∴ Total number of outcomes = 12
Number of black ball = x
P (Getting a black ball) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{x}{12}$
Hence, p1 = P (Getting a black ball) = $\frac{x}{12}$
If 6 more black balls are put in the box, then total number of balls in the box = 12 + 6 = 18
Now, Number of black balls in the box = x + 6
∴ p2 = P (Getting a white ball) = $\frac{x+6}{18}$
According to question, p2 = 2p1
⇒ $\frac{x+6}{18}=2\times \frac{x}{12}$
⇒ $\frac{x+6}{18}=\frac{x}{6}$
⇒ x + 6 = 3x
⇒ 2x = 6
⇒ x = 3
Hence, the number of black balls were 3.
Q.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 $\frac{2}{3}$. Find the number of blue balls in the jar.
Ans: There are 24 marbles in the jar, some are green and others are blue.
∴ Total number of outcomes = 24
Let there be x green falls
∴ Favourable number of outcomes = x
∴ P (Green balls) = $\frac{x}{24}$ But, P (Green balls) = $\frac{2}{3}$ [Given]
⇒ $\frac{x}{24}=\frac{2}{3}$
⇒ x = $\frac{2}{3}$ x 24 = 16
⇒ Number of green marbles = 16
∴ Number of blue marbles = Total balls – Green balls = 24 – 16 = 8