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## Karnataka State Syllabus Class 10 Maths Chapter 14 Probability Ex 14.1

1. Complete the following statements:

- Probability of an event E + Probability of the event ‘not E’ = ________
- The probability of an event that cannot happen is ________. Such an event is called ________.
- The probability of an event that is certain to happen is ________. Such an event is called ________.
- The sum of the probabilities of all the elementary events of an experiment is ________.
- The probability of an event is greater than or equal to ________. and less than or equal to ________.

Answer:

- P(E) + P\((\bar{E})\) = 1
- 0 (zero), impossible event.
- 1 (one), sure event (or) certain event.
- 1 (one)
- 0, 1

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.

Answer:

When driver attempts to start a car, then in normal case the car gets started but if the car has some fault, then the car may not start.

∴ The out comes are not equally likely.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

Answer:

The player may shoot the shot or miss the shot. It depends on several factors like training of player, and the amount of enthusiasm present in the player.

∴ The out comes are not equally likely.

(iii) A trial is made to answer a true – false question. The answer is right or wrong.

Answer:

When trail is made to answer a true – false Question, then either the answer is true or false, i.e., only one out come is possible.

∴Hence, this experiment has equally likely outcomes.

(iv) A baby is born. It is a boy or a girl.

Answer:

When a baby is born, then either the baby is to be a boy or a girl, i.e only one out come is possible.

∴ Hence, it is a equally likely out comes.

The Postive Factors of 144 are therefore all the numbers we used to divide (divisors) above to get an even number

Question 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?

Answer:

Tossing coin the out comes head and tail are equally likely, so, the result of an individual coin toss is completely unpredictable and so it is a fair way.

Question 4.

Which of the following cannot be the probability of an event?

A. \(\frac{2}{3}\)

B. – 1.5

C. 15%

D. 0.7

Answer:

B. -1.5.

– 1.5 is not a probability of an event because probability lies between 0 and 1.

Question 5.

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

Answer:

Question 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

- an orange flavoured candy?
- a lemon flavoured candy?

Answer:

- zero (0) it is impossible event.

Because bag contains only lemon flavoured candies only, i.e., bag not contains orange flavoured candies. - One (1) because bag contains lemon flavoured candies only.

∴ It is sure event.

Question 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?

Answer:

No. of students in the group is 3.

The probability of 2 students not having the same birthday is 0.992.

Then the possibility of having the same birthday = 1 – 0.992 = 0.008.

Question 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?

Answer:

Total number of balls in a bag = n(s) = 8.

(i) Let E be an event drawing a red ball n(E) = 3

Probability of drawing a red ball

Question 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?

Answer:

Total marbles in box n(S) = 5 + 8 + 4=17

∴ n(S) = 17

(i) Let E be event of getting a red marble. n(E) = 5.

(ii) Let F be an event of getting white marble n(F) = 8

(iii) Let G be an event of not getting green ball n(G) = 5 + 8 = 13

Question 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?

Answer:

Total number of possible out come = No% of 50 p coins + No% of ₹ 1 coins, + No% ₹ 2 coins + No% of ₹ 5 coins.

∴ n(S) = 100 + 50 + 20 + 10 = 180

(i) Let E be an a event of getting 50 P coin ∴ n(E) = 100

Probability of getting 50 p coin =

(ii) Let F be an a event of not getting ₹ 5 coin. P(F) = 180 – 10 = 170

Probability of not getting ₹ 5 coin =

Question 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 Fig. 14.4). What is the probability that the fish taken out is a male fish?

Answer:

Total number of fishes in aquarium = Number of female fishes + Number of male Fishes

n(s) = 5 + 8 = 13

Let E be an event of taken as male fish Probability of male fish =

Question 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. 14.5), 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?

Answer:

Number of all possible out comes is 8 ∴n(s) = 8

(i) Let A be a event of arrow will point ∴ 8 n(A) =1

Probability of arrow point to 8 =

(ii) Let B be a event of arrow point to odd number B = {1, 3, 5, 7} n(B) = 7

Probability of arrow point to odd number =

(iii) Let C be a event arrow point greater than 2 C = {3, 4, 5, 6, 7, 8} ∴ n(C) = 6

Probability of arrow pointing greater than 2 =

(iv) Let D be a event arrow point less than 9. D = {1,2, 3,4, 5, 6, 7,8}

n(D) = 8

Probability of arrow point less than 9

=

∴ It is a sure (or) certain event.

Question 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.

Answer:

Total number of possible out comes if die is thrown once. S = {1,2,3,4, 5, 6}

∴ n(S) = 6

(i) Let A be a event of getting prime number A = {2, 3, 5}

∴ n(A) = 3

Probability of getting prime number

(ii) Let B be a event of getting number between 2 and 6B = {3, 4, 5}

∴ n(B) = 3

Probability of getting number between 2 and 6

(iii) Let C is an a event of getting odd number C = {1,3,5} n(C) = 3

Probability of getting odd number

Question 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

Answer:

The total number of cards is 52

∴ n(S) = 52

(i) Let ’E_{1}’ be an event of getting king of red colour n(E_{1}) = 2

Probability of getting a red king

(ii) Let E_{2} be a event of getting a face card.

∴ n(E_{2}) = 4 kings + 4 queens + 4 Jacks = 12

Probability of getting a face card

(iii) Let E_{3} be an event of getting a red face card

∴ n(E_{3}) = 2 kings + 2 queens + 2 Jacks = 6

Probability of getting a red face card

(iv) Let E_{4} be a event of getting jack of hearts.

∴ n(E_{4}) = 1

Probability of getting jack of hearts

(v) Let E_{5} be a event of getting a spade

∴ n(E_{5}) = 13

∴ Probability of getting a spade

(vi) Let E_{6} be a event of getting queen of diamonds

∴ n(E_{6}) = 1

∴ Probability of getting queen of diamonds

Question 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?

Answer:

Total number of cards ∴ n(S) = 5.

(i) Let E_{1} be a event of getting Queen ∴ n(E_{1}) = 1

Probability of getting queen card

(ii) Now total number of cards

= 5 – 1 = 4 ∴ n(S) = 4

∴ n(S) = 4

(a) Let E_{2} be an event getting ace

∴ n(E_{2}) = 1

Probability of getting ace

(b) Let E_{3} be an event of getting second card Queen. ∴ n(E_{3}) = 0

Probability of getting Queen

Question 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 is taken out is a good one.

Answer:

Total number of all outcomes = number of defective pens + number of good pens

∴ n(S)= 12 + 132 = 144

Let E be an event of getting a good pen.

∴ n(E) = 132

Probability of getting good pen

Question 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 is 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?

Answer:

Total number of bulbs = 20

∴ n(S) = 20

(i) Let E_{1} be an event of defective bulbs

∴ Probability of defective bulb

(ii) Drawn bulb in (i) is not defective and is not replaced now total number of bulbs is = 20 – 1 = 19

n(S) = 19

Let E_{2} be a event of drawn bulb is not defective.

n(E_{2})= 16 – 1 = 15

Probability of getting not defective bulb

Question 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.

Answer:

Total number of all possible out comes from 1 to 90 discs ∴ n(S) = 90

(i) Let E_{1} be a event of drawn disc is two – digit number is E_{1} = {10, 11, ……. 90} ∴ n(E_{1}) = 81

Probability of drawn disc is two digit number =

(ii) Let E_{2} be a event of drawn disc is perfect square number E_{2} = {1, 4, 9, 16, 25, 36, 49, 64, 81} ∴ n(E_{2}) = 9

Probability of drawn disc is perfect square number =

(iii) Let E_{3} be a event of getting drawn number is divisible by 5

E_{3} = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}

n(E_{3}) = 18

Probability of drawing number is divisible by 5

Question 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?

Answer:

The total number of possible outcomes of letters of the die are S = {A, B, C, D, E, A}

∴ n(S) = 6

(i) Let E_{1} be an event of getting A is n(E_{1}) = 2

Probability of getting A’s

(ii) Let E_{2} be a event of getting D

∴ n(E_{2}) = 1

Probability of getting D

Question 20.

Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with a diameter 1m?

Answer:

Area of the rectangular region = length × breadth = 3 × 2 = 6 m_{2}

Diameter of circle = 1m

∴ radius = \(\frac{d}{2}=\frac{1}{2}\)m

Area of circle =

Probability of die will lie on inside circle

Question 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?

Answer:

Total number of ball pens = 144

n(S)=144

(i) Let E_{1} be a event of nuri will buy pen

∴ n(E_{1}) = 144 – 20 = 124

Probability of she will buy

(ii) Let E_{2} be a event of nuri will not buy pen. ∴ n(E_{2}) = 20

Probability of she will not buy

Question 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.

Answer:

Number of all possible out comes are tossing two dice simultaneously are

S = (1,1) 0,2) (1,3) (1,4) (1,5) (1,6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6).

∴ n(S) = 36

(i) Let E_{1} be a event of getting sum is equal to 3 is E_{1} = (1, 2) (2, 1)

n(E_{1}) = 2

Probability of getting 2 dice sum is 3

(ii) Let E_{2} be a event of getting sum of two dice is equal to 4 E_{2} = (1, 3) (3, 1) (2,2)

∴ n(E_{2}) = 3

Probability of getting sum of 2 dice is 4

(iii) Let E_{3} be a event of getting sum of two dice is equal to 5, E_{3} = (1, 4) (4, 1) (2, 3) (3, 2)

∴ n(E_{3}) = 4

Probability of getting sum of 2-dices is equal to 5 =

(iv) Let E_{4} be a event of getting sum of two dice is equal to 6, E_{4} = (1, 5) (5, 1) (3, 3) (4, 2) (2, 4)

∴ n(E_{4}) = 5.

Probability of getting sum of 2-dices is equal to 6 =

(v) Let E_{5} be a event of getting sum of two dice is equal to 7, E_{5} = (1, 6) (6, 1) (2, 5) (5, 2) (4, 3) (3, 4)

∴n(E_{5}) = 6

Probability of getting sum of 2-dices is equal to 7

(vi) Let E_{6} be a event of getting sum of 2-dice is equal to 9. E_{6} = (3, 6) (4, 5) (5, 4) (6, 3)

∴ n(E_{6}) = 4

Probability of getting sum of 2-dice is equal =

(vii) Let E_{7} be a event of getting sum of two dice is equal to 10. E_{7} = (5, 5) (6, 4) (4, 6)

∴ n(E_{7}) = 3

Probability of getting sum of 2 dice is equal to

(viii) Let E_{8} be a event of getting sum of 2-dice is equal to 11. E_{8} = (5, 6) (6, 5)

∴ n(E_{8}) = 2

Probability of getting sum of 2-dice is equal to

(ii) Probability of Number of eleven sums are not equally likely from the above table.

Question 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.

Answer:

Let ‘H’ denote head and T denote the tail. Then possible out comes are S = {HHH, HTH, HTT, LIHT, THH, THT, TTH, TTT} ∴ n(S) = 8

Let E be a event lose the game i.e he does not get all three heads (or) all three tails. ∴ E = {HTH, HTT, HHT, THH, THT, TTH} n(E) = 6

Probability of lose

Question 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]

Answer:

Throwing a die twice simultaneously

S = (1, 1) (1, 2) (1,3) (1,4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5,4) (5, 5) (5,6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6).

∴ n(S) = 36

(i) Let E_{1} be a event 5 will not come up either time.

Then favourable out comes E_{1} = (1, 1) (1, 2) (1, 3) (1,4) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 6).

∴ n (E_{1}) = 25

Probability of 5 will not come up

(ii) Let E be a event of 5 will come up either once E_{2} = (1, 5) (2, 5) (3, 5) (4, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6,5)

∴ n(E_{2}) = 11

Probability of 5 will come up

Question 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 \(\frac{1}{3}\)

(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 \(\frac{1}{2}\)

Answer:

(i) Not correct

Because by tossing two coins simultaneously the total number of possible out comes are 4 (Four)

∴ P (one tail (or) one head)

= P(TH,HT) = \(\frac{2}{4}\) or \(\frac{1}{2}\)

(ii) Correct

The two out comes considered in the question are equally likely.