Class 10, Maths, Chapter 15, Exercise 15.1, Solutions
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) 1
(ii) 0, impossible event
(iii) 1, sure or certain event
(iv) 1
(v) 0, 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) A driver attempts to start a car. The car starts or does not start”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot”, we are not justified to assume that each outcome is as likely to occur as the other. Thus, the experiment has no equally likely outcomes.
(iii) A trial is made to answer a true- false question. The answer is right or wrong. We know, in advance, that the result can lead in one of the two possible ways — either right or wrong. We can reasonably assume that each outcome, right or wrong, is likely to occur as the other. Thus, the outcomes right or wrong, are equally likely.
(iv) A baby is born. It is a boy or a girl”. We know, in advance, that the outcome can load in one of two possible outcomes — either a boy or a girl. We are justified to assume that each outcome, boy or girl,, is likely to occur as the other. Thus, the outcomes boy or girl, are equally likely.
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 tossing of a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game as we know that the tossing of the coin only land in one of two possible ways — either head up or tail up. the outcomes head and tail are equally likely. So, the result of the tossing of a coin is completely unpredictable.
Q.4. Which of the following cannot be the probability of an event?
(A) $\frac{2}{3}$
(B) –1.5
(C) 15%
(D) 0.7
Ans: Since the probability of an event E is a number P(E) such that
0 ≤ P(E) ≤ 1
∴ -1.5 cannot be the probability of an event.
∴ (B) is the correct answer.
Q.5. If P(E) = 0.05, what is the probability of ‘not E’?
Ans: Since P(E) + P (Ē) = 1
∴ P (Ē) = 1 – P (E) = 1 – 0.05 = 0.95
Q.6. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavored candy?
(ii) a lemon flavored candy?
Ans: (i) As the bag contain only lemon-flavored candy. So, P (orange flavor candy) = 0
(ii) As the bag contain only lemon flavored candy. This event is a certain event so its probability is 1.
∴ P (lemon flavor candy) = 1
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: Let E be the event of not having the same birthday., P(Ē) = 0.992
probability of having same birthday,
P(E) + P (Ē) = 1
⇒ P(E) + 0.992 = 1
⇒ P(E) = 1 – 0.992 = 0.008
Hence, the probability that the 2 students have the same birthday is 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: Total number of balls = 3 + 5 = 8
(i) Probability of a red ball
⇒ P (Red ball) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}$
⇒ P (Red ball) = $\frac{3}{8}$
Hence, P (getting a red ball) = $\frac{3}{8}$
(ii) Probability of not a red ball
⇒ P (Red ball) + P (Not a red ball) = 1
⇒ P (Not a red ball) = 1
⇒ P (Not a red ball) = 1 + $\frac{3}{8}=\frac{8-3}{8}=\frac{5}{8}$
Hence, P (not getting a red ball) = $\frac{5}{8}$
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: Total number of marbles in the box = 5 (Red) + 8 (white) + 4 (Green) = 17 marbles
∴ Total number of elementary events = 17
(i) There are 5 red marbles in the box
P (Red marble) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}$
P (Red marble) = $\frac{5}{17}$
Hence, P (getting a red marble) = $\frac{5}{17}$
(ii) There are 8 white marbles in the box.
P (White marble) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}$
P (white marble) = $\frac{8}{17}$
Hence, P (getting a white marble) = $\frac{8}{17}$
(iii) Number of green marbles = 4
⇒ P (Green marble) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{4}{17}$
∵ P (Green marble) + P (Not a green marble) = 1
⇒ $\frac{4}{17}$+ P (Not a green marble) = 1
⇒ P (Not a green marble) = 1- $\frac{4}{17}=\frac{17-4}{17}=\frac{13}{17}$
Hence, P (not getting a green marble) = $\frac{13}{17}$
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: Total number of coins in a piggy bank
Number of 50 p coins = 100
Number of ₹1 coins = 50
Number of ₹2 coins = 20
Number of ₹5 coins = 10
Total number of coins in a piggy bank = 100 + 50 + 20 + 10 = 180
(i) Number of 50 p coins = 100
P (50 p coins) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{100}{180}=\frac{5}{9}$
Hence, P (falling out of a 50 p coin) = $\frac{5}{9}$
(ii) Number of ₹5 coins = 10
P (₹ 5 coins) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{10}{180}=\frac{1}{18}$
P (falling out not a ₹ 5 coin) = 1 – P (₹ 5 coin) =1- $\frac{1}{18}=\frac{18-1}{18}=\frac{17}{18}$
Hence, P (falling out a coin other than Rs 5 coin) = $\frac{17}{18}$
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 Figure). What is the probability that the fish taken out is a male fish?
Ans: Number of male fishes = 5
Number of female fishes = 8
Total number of fishes in the tank = 5 + 8 = 13
P (Taking out male fish) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{5}{13}$
Hence, P (taking out a male fish) = $\frac{5}{13}$
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 Fig. 15.5), 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: Out of 8 number, an arrow can point any of the numbers in 8 ways.
∴ Total number of outcomes = 8
(i) There is only ‘8’on the spinning plant
P (Arrow points at 8) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{1}{18}$.
Hence, P (arrow points at 8) = $\frac{1}{8}$
(ii) There are 4 odd numbers (viz. 1, 3, 5 and 7)
P (Arrow points at an odd number) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{4}{8}=\frac{1}{2}$
Hence, P(Arrow points at an odd number) = $\frac{1}{2}$
(iii) There are 6 numbers greater than 2 (viz. 3, 4, 5, 6, 7, and 8)
P (Arrow points at a number > 2) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{6}{8}=\frac{3}{4}$
Hence, P(Arrow points at a number > 2) = $\frac{6}{8}=\frac{3}{4}$
(iv) There are 8 numbers less than 9 (viz. 1,2, 3, …, 8)
P (Arrow points at a number < 9) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{8}{8}=1$
Hence, P(arrow points at a number < 9)
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: In a single throw of a die we can get any one of the six numbers
⟹ {1,2,3,4,5,6}
∴ Number of possible outcomes = 6
(i) Prime no. on a dice = {2,3,5}
Total prime number = 3
P (Prime number) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{3}{6}=\frac{1}{2}$
Hence, P (Prime number) = $\frac{3}{6}=\frac{1}{2}$
(ii) Number lying between 2 and 6 = {3,45}
Total number = 3
P (Number lying between 2 and 6) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{3}{6}=\frac{1}{2}$
Hence, P (Number lying between 2 and 6) = $\frac{1}{2}$
(iii) Odd number on a dice = {1,3,5}
Total number = 3
P (Odd number on a dice) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{3}{6}=\frac{1}{2}$
Hence, P (Odd number on a dice) = $\frac{1}{2}$
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 diamonds
Ans: Total number of cards = 52
Here, out of 52 cards, one card can be drawn in 52 ways. So, total number of outcomes = 52.
(i) There are two suits of red cards viz. diamond and heart. Each suit contains one king.
Total number of red king = 2
P (A red king) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{2}{52}=\frac{1}{26}$
Hence, P(a king of red colour) = $\frac{2}{52}=\frac{1}{26}$
(ii) In a deck of 52 cards: kings, queens and jacks are called face cards. Thus, there are 12 face cards. So, one face card can be chosen in 12 ways.
Total face cards = 12
P (A face card) =
Hence, P (a face card) = $\frac{3}{13}$
(iii) There are two suits of red cards viz. diamond and heart. Each suit contains 3 face cards.
Total number =
P (A red face card) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{6}{52}=\frac{3}{26}$
Hence, P (a red face card) = $\frac{3}{26}$
(iv) There is only one jack of hearts
Total number =1
P (Jack of heart) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{1}{52}$
Hence, P (Jack of heart) = $\frac{1}{52}$
(v) There are 13 cards of spade:
Total number of spade = 13
P (a spade card) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{13}{52}=\frac{1}{4}$
Hence, P (a spade) = $\frac{1}{4}$
(vi) There is only one queen of diamonds.
Total number of queen of diamond = 1
P (the queen of diamonds) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{1}{52}$
Hence, P (the queen of diamonds) $=\frac{1}{52}$
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: Out of five cards the ten, jack, queen, king and ace of diamonds, one card can be drawn in 5 ways. So, total number of outcomes = 5
(i) There is only one queen
Total number of queens = 1
P (Getting queen) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{1}{5}$
Hence, P (Getting queen) = $\frac{1}{5}$
(ii) After keeping the queen card aside, we are left with 4 cards. So, total number outcomes = 4
(a) There is only one ace
Total number of an ace card = 1
P (an ace) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{1}{4}$
Hence, P (an ace) = $\frac{1}{4}$
(b) There is no card as queen.
Total number of queen card = 0
P (the queen) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{0}{4}=0$
Hence, P (the queen = 0.
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: Total number of defective pens = 12
Total number of good pens = 132
Total number of pens = 12 + 132 = 144
Out of 144 pens, one pen can be chosen in 144 ways.
∴ Total number of outcomes = 144
Probability of getting a good pen,
Total number of good pens = 132
P (Getting a good pen) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{132}{144}=\frac{11}{12}$
Hence, P (Getting a good pen) = $\frac{11}{12}$
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: (i) Out of 20 bulbs one bulb can be chosen in 20 ways.
∴ Total number of outcomes = 20
Probability of getting a defective bulb,
Total number of defective bulbs = 4
P (Getting a defective bulb) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{4}{20}=\frac{1}{5}$
Hence, P (Getting a defective bulb) = $\frac{4}{20}=\frac{1}{5}$
(ii) On drawing a non-defective bulb of 20 bulbs we are left with 19 bulbs including 4 defective bulbs.
∴ Total number of outcomes = 19
There are 19 – 4 = 15 non-defective bulbs out of which one bulb can be drawn in 15 ways.
P (Getting a non-defective bulb) =
Hence, P (Getting a non-defective bulb) = $\frac{15}{19}$
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: There are 90 discs bearing numbers 1 to 90 in the box of which one disc can be drawn in 90 ways
∴ Total number of outcomes = 90
(i) There are 90 – 9, i.e., 81 discs bearing two-digit number in the box of which one disc can be drawn in 81 ways
∴ Favourable number of outcomes = 81
P (Getting a disc bearing a two-digit number) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{81}{90}=\frac{9}{10}$
Hence, P (Getting a disc bearing a two-digit number) = $\frac{9}{10}$
(ii) Those numbers from 1 to 90 which are perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, i.e., squares of l, 2, 3, 4, 5, 6, 7, 8 and 9 respectively.
Therefore, there are 9 discs marked with numbers which are perfect squares
∴ Favourable number of outcomes = 9
P (Getting a disc marked perfect square) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{9}{90}=\frac{1}{10}$
Hence, P (Getting a disc marked with a number which is a perfect square) = $\frac{1}{10}$
(iii) Those numbers from 1 to 90 which are divisible by 5 are 5, 10, 15, 20, 25, 30, 36, 40, 45, 50, 55, 60, 65, 70, 75. 80, 85 and 90. They are 18 in number
Therefore. there are 18 discs marked with the numbers which are divisible by 5
∴ Favourable number of outcomes = 18
P (Getting a number divisible by 5) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{18}{90}=\frac{1}{5}$
Hence, P (Getting a disc marked with a number which is divisible by 5) = $\frac{1}{5}$
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: In a single throw of a die we can got any one of the six letters A, B, C, D, E, A marked on its faces. Therefore,
Total numbers of outcomes = 6
(i) Total number of face having ‘A’ = 2
P (Getting A) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{2}{6}=\frac{1}{3}$ .
Hence, P (Getting A) = $\frac{1}{3}$
(ii) total number of face having ‘D’ = 1
P (Getting D) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{1}{6}$
Hence, P (Getting D) = $\frac{1}{6}$
Q.20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?
Ans: Length of rectangle = 3 m
Width of rectangle = 2 m
Total area of the rectangle =
Diameter of circle = 1 m
∴ radius = ½ m
Area of the circle =
Now, P (die to land inside the circle) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{{}^{\pi }/{}_{4}}{6}=\frac{\pi }{24}$
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: There is a lot of 144 hall pons. Out of these 144 ball pens 20 are defective ball pens.
Number of non-defective ball pens = 144 – 20 = 124
(i) Probability of getting good pen by Nuri
P (she will buy) = P (a non-defective pen) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{124}{144}=\frac{31}{36}$
Hence, P (She will but) = $\frac{31}{36}$
(ii) P (she will not buy) = P (a defective pen) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{20}{144}=\frac{5}{36}$
Hence, P (she will not buy) = $\frac{5}{36}$
Q.22. Refer to Example 13. (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 $\frac{1}{11}$. Do you agree with this argument? Justify your answer.
Ans: Total possible outcomes = 6 x 6 = 36
To get the sum of 2, possible outcomes = {1,1} = 1 outcome
Hence, required probability = $\frac{1}{36}$
To get the sum of 3, possible outcomes = {1,2} {2,1} = 2 outcomes
Hence, required probability = $\frac{2}{36}$
To get the sum of 4, possible outcomes = {2,2} {1,3} {3,1} = 3 outcomes
Hence, required probability = $\frac{3}{36}$
To get the sum of 5, possible outcomes = {4,1} {1,4} {3,2} {2,3} = 4 outcomes
Hence, required probability = $\frac{4}{36}$
To get the sum of 6, possible outcomes = {5,1} {1,5} {3,3} {2,4} {4,2} = 5 outcomes
Hence, required probability = $\frac{5}{36}$
To get the sum of 7, possible outcomes = {6,1} {1,6} {2,5} {5,2} {3,4} {4,3} = 6 outcomes
Hence, required probability = $\frac{6}{36}$
To get the sum of 8, possible outcomes = {6,2} {2,6} {5,3} {3,5} {4,4} = 5 outcomes
Hence, required probability = $\frac{5}{36}$
To get the sum of 9, possible outcomes = {6,3} {3,6} {5,4} {4,5} = 4 outcomes
Hence, required probability = $\frac{4}{36}$
To get the sum of 10, possible outcomes = {5,5} {4,6} {6,4} = 3 outcomes
Hence, required probability = $\frac{3}{36}$
To get the sum of 11, possible outcomes = {6,5} {5,6} = 2 outcomes
Hence, required probability = $\frac{2}{36}$
To get the sum of 12, possible outcomes = {6,6}= 1 outcomes
Hence, required probability = $\frac{1}{36}$
Thus, the complete table is as under:
(ii) I do not agree with the argument given here because the sum are not equally likely as Justification has already been given in part (i).
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: Consider the experiment in which a coin is tossed thrice. The outcomes associated with this experiment are given by HHH, HHT, HTH, THH, TTH, HTT, THT, TTT
∴ Total number of outcomes = 8
Hanif will lose the game if he gets HHT, HTH, THH, TTH, HTT, THT.
∴ Favourable number of outcomes = 6
P (lose the game) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{6}{8}=\frac{3}{4}$
Hence, required probability .
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?
Ans: as the die is thrown twice. Therefore, total outcomes = 6 x 6 = 36
(i) Total number of outcomes when 5 come either time.
= {5,1} {5,2} {5,3} {5,4} {5,5} {5,6}
{1,5} {2,5} {3,5} {4,5} { } {6,5}
Total number of cases = 11
Now, P (5 come either time) =
And, P (5 note come either time) = 1- $\frac{11}{36}=\frac{36-11}{36}=\frac{25}{36}$
Hence, the required probability = $\frac{25}{36}$
(ii) 5 come up at least once = 11 case
P (5 come at least once) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{11}{36}$
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
(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
Ans: (i) Incorrect,
Justification:
If two coins are tossed simultaneously there are four possible outcomes
= {H, H} {T,T} {H,T} {T,H}
∴ Total number of outcomes = 4
Now, for each, let say Probability of getting {H,H}
P (H,H) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{1}{4}$
(ii) Correct, the two outcomes considered in the question are equally likely.
Justification: If a die is thrown, then possible outcomes = {1,2,3,4,5,6}
∴ Total number of outcomes = 6
Probability of getting an odd number (1,3,5) = 3
P (Getting odd number) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{3}{6}=\frac{1}{2}$
Similarly,
Probability of getting an even number (2,4,6) = 3
P (Getting even number) = $\frac{Number\,\,of\,\,Favorable\,\,outcomes}{Total\,\,no.\,\,of\,\,outcomes}=\frac{3}{6}=\frac{1}{2}$