NCERT Solutions Class 10 Maths Chapter 15

Q.1 Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ =­­­­­­­­­­­­­_____.
(ii) The probability of an event that cannot happen is_____. Such an event is called­­­­­­­­­­­­ ­_____.
(iii) The probability of an event that is certain to happen is_____. Such an event is called______.
(iv) The sum of the probabilities of all the elementary events of an experiment is ­­­­­­­­­­­­­_____.
(v) The probability of an event is greater than or equal to ­­­­­­­­­­­­­_____ and less than or equal to­­­­­­­­­­­­­_____.

Ans.

(i) Probability of an event E + Probability of the event ‘not E’ =­­­­­­­­­­­­­1.
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is ­­­­­­­­­­­­­1.
(v) The probability of an event is greater than or equal to ­­­­­­­­­­­­­0 and less than or equal to­­­­­­­­­­­­­ 1.

Q.2 Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

Ans.

(i) It does not have equally likely outcomes. The car may never start for some fault.
(ii) It does not have equally likely outcomes as player may take more attempts to shoot a basketball.
(iii) It has equally likely outcomes. The answer of a true-false question is either right or wrong.
(iv) It has equally likely outcomes. The baby is either a boy or a girl.

Q.3 Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Ans.

The outcomes of a coin toss are equally likely. So, the result of a coin toss is completely unpredictable.

Q.4 Which of the following cannot be the probability of an event?
(A) 2/3
(B) –1.5
(C) 15%
(D) 0.7

Ans.

The correct answer is (B).

Q.5 If P(E) = 0.05, what is the probability of ‘not E’?

Ans.

Probability of ‘not E’ = 1 – P(E) = 1 – 0.05 = 0.95

Q.6 A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Ans.

(i) There is no orange flavoured candy in the bag. So, the probability of taking out an orange flavoured candy is zero.
(ii) The probability of taking out lemon flavoured candy is 1 as there are only lemon flavoured candies in the bag.

Q.7 It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Ans.

The probability of 2 students not having the same birthday = P(E’) = 0.992
The probability of 2 students having the same birthday= 1 – P(E’) = 1 – 0.992 = 0.008

Q.8 A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Ans.

$\begin{array}{l}\text{Number of red balls}=3\\ \text{Number of black balls}=5\\ \text{Total number of outcomes}=3+5=8\\ \text{(i) Probability of getting a red ball}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{3}{8}\\ \text{(ii) Probability of not getting a red ball}\\ =1-\text{Probability of getting a red ball}\\ =1-\frac{3}{8}=\frac{5}{8}\end{array}$

Q.9 A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ? (ii) white ? (iii) not green?

Ans.

$\begin{array}{l}\text{Number of red marbles}=5\\ \text{Number of white marbles}=8\\ \text{Number of green marbles}=4\\ \text{Total number of outcomes}=5+8+4=17\\ \text{(i) Probability of getting a red marble}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{5}{17}\\ \text{(ii) Probability of getting a white marble}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{8}{17}\\ \text{(iii) Probability of getting a marble not green}\\ \text{=1}-\text{Probability of getting a green marble}\\ =1-\frac{4}{17}=\frac{13}{17}\end{array}$

Q.10 A piggy bank contains hundred 50p coins, fifty 1 coins, twenty 2 coins and ten 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a 5 coin?

Ans.

$\begin{array}{l}\text{Number of 50p coins}=100\\ \text{Number of \hspace{0.17em}1 coins}=50\\ \text{Number of \hspace{0.17em}2 coins}=20\\ \text{Number of \hspace{0.17em}5 coins}=10\\ \text{Total number of outcomes}=100+50+20+10=180\\ \text{(i) Probability of getting a 50p coin}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{100}{180}=\frac{5}{9}\\ \text{(ii)}\\ \text{Probability of not getting a\hspace{0.17em}5 coin}\\ =1-\text{Probability of getting a}\text{5 coin}\\ =1-\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =1-\frac{10}{180}=\frac{17}{18}\end{array}$

Q.11 Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see the following figure). What is the probability that the fish taken out is a male fish?


Ans.

$\begin{array}{l}\text{Number of male fishes}=5\\ \text{Number of female fishes}=8\\ \text{Total number of outcomes}=5+8=13\\ \text{(i) Probability of getting a male fish}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{5}{13}\end{array}$

Q.12 A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see the following figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?


Ans.

\begin{array}{l}\text{(i) Probability of the arrow pointing at 8}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{8}\\ \text{(ii) Probability of the arrow pointing at an odd number}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{4}{8}=\frac{1}{2}\\ \text{(iii) Probability of the arrow pointing at a number greater than 2}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{6}{8}=\frac{3}{4}\\ \text{(iv) Probability of the arrow pointing at a number less than 9}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{8}{8}=1\end{array}

Q.13 A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Ans.

$\begin{array}{l}\text{Total number of outcomes}=6\\ \text{(i)}\\ \text{Prime numbers on the die are 2, 3 and 5.}\\ \therefore \text{Number of prime numbers on the die}=3\\ \text{Probability of getting a prime number}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{3}{6}=\frac{1}{2}\\ \text{(ii)}\\ \text{Numbers lying between 2 and 6 on the die are 3, 4 and 5.}\\ \therefore \text{Number of prime numbers on the die}=3\\ \text{Probability of getting a number lying between 2 and 6}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{3}{6}=\frac{1}{2}\\ \text{(iii)}\\ \text{Odd numbers on the die are 1, 3 and 5.}\\ \therefore \text{Number of odd numbers on the die}=3\\ \text{Probability of getting an odd number}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{3}{6}=\frac{1}{2}\end{array}$

Q.14 One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamond

Ans.

$\begin{array}{l}\text{Total number of outcomes}=52\\ \text{(i)}\\ \text{Number of kings of red colour}=2\\ \text{Probability of getting a king of red colour}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{2}{52}=\frac{1}{26}\\ \text{(ii)}\\ \text{Number of face cards}=12\\ \text{Probability of getting a face card}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{12}{52}=\frac{3}{13}\\ \text{(iii)}\\ \text{Number of red face cards}=6\\ \text{Probability of getting a red face card}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{6}{52}=\frac{3}{26}\\ \text{(iv)}\\ \text{Number of the jack of hearts}=1\\ \text{Probability of getting the jack of hearts}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{52}\\ \text{(v)}\\ \text{Number of spades}=13\\ \text{Probability of getting a spade}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{13}{52}=\frac{1}{4}\\ \text{(vi)}\\ \text{Number of the queen of diamond}=1\\ \text{Probability of getting the queen of diamond}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{52}\end{array}$

Q.15 Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Total number of outcomes}=5\\ \text{Probability of getting the queen}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{5}\\ \text{(ii)}\\ \text{Queen is put aside.}\\ \therefore \text{total number of outcomes}=4\\ \text{(a)}\\ \text{Probability of getting an ace}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{4}\\ \text{(b)}\\ \text{Probability of getting a queen}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{0}{4}=0\end{array}$

Q.16 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Ans.

$\begin{array}{l}\text{Number of defective pens}=12\\ \text{Number of good pens}=132\\ \text{Total number of outcomes}=132+12=144\\ \text{Probability of getting a good pen}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{132}{144}=\frac{11}{12}\end{array}$

Q.17 (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Number of defective balls}=4\\ \text{Total number of outcomes}=20\\ \text{Probability of getting a defective ball}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{4}{20}=\frac{1}{5}\\ \text{(ii)}\\ \text{Number of defective balls}=4\\ \text{Number of good balls}=16-1=15\\ \text{Total number of outcomes}=15+4=19\\ \text{Probability of not getting a defective ball}\\ =1-\text{Probability of getting a defective ball}\\ =1-\frac{4}{19}=\frac{15}{19}\end{array}$

Q.18 A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Ans.

$\begin{array}{l}\text{Numbers of two-digit numbers among 1 to 90}=81\\ \text{Numbers of perfect square numbers among 1 to 90}=9\\ \text{Numbers of multiples of 5 which are less than or equal 90}=18\\ \text{Total number of outcomes}=90\\ \text{(i)}\\ \text{Probability of getting a two-digit number bearing disc}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{81}{90}=\frac{9}{10}\\ \text{(ii)}\\ \text{Probability of getting a perfect square number}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{9}{90}=\frac{1}{10}\\ \text{(iii)}\\ \text{Probability of getting a number divisible by 5}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{18}{90}=\frac{1}{5}\end{array}$

Q.19 A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Probability of getting A}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{2}{6}=\frac{1}{3}\\ \text{(ii)}\\ \text{Probability of getting D}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{6}\end{array}$

Q.20 Suppose you drop a die at random on the rectangular region shown in the following figure. What is the probability that it will land inside the circle with diameter 1m?


Ans.

$\begin{array}{l}\text{Favourable outcomes}=\text{Area of the circle}={\mathrm{\pi r}}^2=\mathrm{\pi }{(0.5)}^2\\ =0.25\mathrm{\pi }{\text{m}}^2\\ \text{Total outcomes}=\text{Area of the rectangle}=2m\times 3\mathrm{m}=6{\text{m}}^2\\ \text{Probability of landing inside the circle}\\ =\frac{\text{Area of the circle}}{\text{Area of the rectangle}}\\ =\frac{0.25\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{24}\end{array}$

Q.21 A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?

Ans.

$\begin{array}{l}\text{Number of defective pens}=20\\ \text{Number of good pens}=144-20=124\\ \text{Total number of outcomes}=144\\ \text{(i)}\\ \text{Probability of buying a pen}=\text{Probability of drawing a good pen}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{124}{144}=\frac{31}{36}\\ \text{(ii)}\\ \text{Probability of not buying pen}=1-\text{Probability of buying a pen}\\ =1-\frac{31}{36}=\frac{5}{36}\end{array}$

Q.22 Two dice, one blue and one grey, are thrown at the same time.
(i) Complete the following table:

(ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.

Ans.

(i) Total possible outcomes = 36
Possible outcomes for getting the sum 3 are (1, 2) and (2, 1).
Possible outcomes for getting the sum 4 are (1, 3); (3, 1) and (2, 2).
Possible outcomes for getting the sum 5 are (1, 4); (2, 3) (3, 2) and (4,1).
Possible outcomes for getting the sum 6 are (1, 5); (2, 4); (3, 3) (4, 2) and (5, 1).
Possible outcomes for getting the sum 7 are ((1, 6); (2, 5); (3, 4) (4, 3); (5, 2) and (6, 1).
Possible outcomes for getting the sum 9 are (3, 6); (4, 5) ; (5, 4) and (6, 3).
Possible outcomes for getting the sum 10 are (4, 6); (5, 5) and (6, 4).
Possible outcomes for getting the sum 11 are (5, 6) and (6, 5).
Now the complete table is as follows:

(ii) No. The eleven sums are not equally likely.

Q.23 A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Ans.

$\begin{array}{l}\text{Possible outcomes are}\\ \left\{\text{HHH, TTT, HTT, THH, HTH, THT, HHT, TTH}\right\}.\\ \text{Number of possible outcomes}=8\\ \text{Number of favourable outcomes}=2\\ \text{P(Hanif will win the game)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{2}{8}=\frac{1}{4}\\ \therefore \text{P(Hanif will lose the game)}\\ =1-\text{P(Hanif will win the game)}\\ =1-\frac{1}{4}=\frac{3}{4}\end{array}$

Q.24 A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Ans.

$\begin{array}{l}\text{Total number of outcomes}=6\times 6=36\\ \text{(i)}\\ \text{Possible outcomes when 5 comes up either time are}\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5).\\ \text{Number of favourable outcomes}=11\\ \text{P(5 will come up either time)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{11}{36}\\ \therefore \text{P(5 will not come up either time)}\\ =1-\text{P(5 will come up either time)}\\ =1-\frac{11}{36}=\frac{25}{36}\\ \text{(ii)}\\ \text{Number of cases when 5 will come up at least once}=11\\ \text{P(5 will come up at least once)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{11}{36}\end{array}$

Q.25 Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes — an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

Ans.

$\begin{array}{l}\text{(i)}\\ \text{The given statement is incorrect. There are 4 possible}\\ \text{outcomes which are}(\text{H},\text{H}),(\text{T},\text{T}),(\text{H},\text{T}),(\text{T},\text{H}).\\ \text{P(getting two heads)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P(getting two tails)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{4}\\ \text{P(getting one head and one tail)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{2}{4}=\frac{1}{2}\\ \text{(ii) Correct because the two outcomes are equally likely.}\end{array}$

Q.26 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.

$\begin{array}{l}\text{There are 5 days namely Tuesday, Wednesday, Thursday,}\\ \text{Friday, Saturday.}\\ \text{Shyam can go to the shop on any of the five days.}\\ \text{Also, Ekta can go to the shop on any of the five days.}\\ \text{So, total number of outcomes}=5\times 5=25\\ \text{(i)}\\ \text{Outcomes of visiting the shop on same day are}\\ \text{(T, T), (W, W), (Th, Th), (F, F), (S, S).}\\ \text{P(both will visit the shop on the same day)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{5}{25}=\frac{1}{5}\\ \text{(ii)}\\ \text{Outcomes of visiting the shop on consecutive days are}\\ \text{(T, W), (W, Th), (Th, F), (F, S), (W, T), (Th, W), (F, Th), (S, F).}\\ \text{Number of odd numbers}=3\\ \therefore \text{P(both will visit on consecutive days)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{8}{25}\\ \text{(iii)}\\ \text{P(both will visit on different days)}\\ =1-\text{both will visit on the same day}\\ =1-\frac{1}{5}=\frac{4}{5}\end{array}$

Q.27 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) at least 6?

Ans.

The given table can be completed as below.

$\begin{array}{l}\text{Total number of possible outcomes}=6\times 6=36\\ \text{(i)}\\ \text{Number of times when the total score is even}=18\\ \text{P(total score being even)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{18}{36}=\frac{1}{2}\\ \text{(ii)}\\ \text{Number of times when the total score is 6}=4\\ \text{P(total score being 6)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{4}{36}=\frac{1}{9}\\ \text{(iii)}\\ \text{Number of times when the total score is at least 6}=15\\ \text{P(total score being at least 6)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{15}{36}=\frac{5}{12}\end{array}$

Q.28 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.

$\begin{array}{l}\text{Let number of blue balls be}\mathrm{n}.\\ \text{Number of red balls}=5\\ \text{Total number of balls in the bag}=5+\mathrm{n}\\ \text{P(drawing a red ball)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{5}{5+\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P(drawing a blue ball)}=2\times \text{P(drawing a red ball)}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{n}}{5+\mathrm{n}}=2\times \frac{5}{5+\mathrm{n}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=10\\ \therefore \text{Number of blue balls}=10\end{array}$

Q.29 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.

$\begin{array}{l}\text{Number of black balls}=\mathrm{x}\\ \text{Total number of balls in the bag}=12\\ \text{P(drawing a black ball)}\\ =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ =\frac{\mathrm{x}}{12}\\ \text{Now, 6 more black balls are putted in the box.}\\ \therefore \text{Number of black balls}=\mathrm{x}+6\\ \text{Total number of balls in the bag}=12+6=18\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}P(drawing a black ball)}=2\times \frac{\mathrm{x}}{12}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}+6}{18}=2\times \frac{\mathrm{x}}{12}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=3\end{array}$

Q.30 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 marbles in the jar.

Ans.

$\begin{array}{l}\text{Let the number of green marbles}=\mathrm{x}\\ \text{Total number of marbles in the jar}=24\\ \text{P(drawing a green marble)}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{3}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{3}=\frac{\mathrm{x}}{24}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=16\\ \text{Number of blue marbles}=24-\mathrm{x}=24-16=8\end{array}$