**2.(i) **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?

+
1
2
2
3
3
6
1
2
3
3
4
4
7
2
3
4
4
5
5
8
2
3
4
4
5
5
8
3
4
5
5
6
6
9
3
4
5
5
6
6
9
6
7
8
8
9
9
12
Total possible outcomes when two dice are thrown =
(1) Number of times when sum is even = 18

**1.(iii) **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 (iii) different days?

Total possible ways Shyam and Ekta can visit the shop =
(1) Case that both will visit the same day.
Shyam can go on any day between Tuesday to saturday in 5 ways.
For any day that Shyam goes, Ekta will go on a different day in ways.
Total ways that they both go in the same day =

**1.(i) ** 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?

Total possible ways Shyam and Ekta can visit the shop =
(1) Case that both will visit the same day.
Shyam can go on any day between Tuesday to saturday in 5 ways.
For any day that Shyam goes, Ekta will go on the same day in 1 way.
Total ways that they both go in the same day =

**25.(ii) **Which of the following arguments are correct and which are not correct? Give reasons for your answer. (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

The possible outcomes when a die is thrown= {1,2,3,4,5,6}
Total number of possible outcomes = 6
Number of odd number, {1,3,5} = 3
And, number of even numbers {2,4,6} = 3
Hence, both these events are equally likely

**25.(i)** 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

The possible outcomes when two coins are tossed = {HH, HT, TH, TT}
Total number of possible outcomes = 4
Hence, the given statement is not correct. This is because one of each can occur in two different ways. Hence the mentioned events are not equally likely.

**24.(ii)** A die is thrown twice. What is the probability that 5 will come up at least once?

When a die is thrown twice, the possible outcomes =
Total number of possible outcomes =
The outcomes when 5 comes up at least once =
{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}
Number of such favourable outcomes = 11
Therefore, the probability that 5 comes at least once is

**24.(i)** A die is thrown twice. What is the probability that 5 will not come up either time?

When a die is thrown twice, the possible outcomes =
Total number of possible outcomes =
The outcomes when 5 comes up either on them =
{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}
Number of such favourable outcomes = 11
Therefore, the probability that 5 will not come either time is

**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.

The possible outcomes when a coin is tossed 3 times: (Same as 3 coins tossed at once!)
{HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
Number of total possible outcomes = 8
For Hanif to win, there are only two favourable outcomes: {HHH, TTT}
Number of favourable outcomes = 2
Therefore, the probability that Hanif will lose is

**22.(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.

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 of 1/11. We do not agree with this argument because there are a different number of possible outcomes for each sum. we can see that each sum has a different probability.

**22.(i) ** Refer to Example 13. (i) Complete the following table:

The table becomes:
Sum on two dice
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
1/18
1/12
1/9
5/36
1/6
5/36
1/9
1/12
1/18
1/36

**21.(ii) ** 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 (ii) She will not buy it ?

Total number of pens = 144
Total number of defective pens = 20
She will buy if the pen is good.
Therefore, probability that she will not buy = probability that the pen is defective =

**21.(i) ** 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 ?

Total number of pens = 144
Total number of defective pens = 20
Number of good pens = 144-20 = 124
She will buy if the pen is good.
Therefore, the probability that she buys = probability that the pen is good =

**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?

Here, Total outcome is the area of the rectangle and favourable outcome is area of the circle.
Area of the rectangle =
Area of the circle =

**19.(ii)** 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 (ii) D?

The six faces of the die contains : {A,B,C,D,E,A}
Total number of letters = 6
(i) Since there is only one D,
number of favourable outcomes = 1
Therefore, the probability of getting D is

**19.(i)** 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?

The six faces of the die contains : {A,B,C,D,E,A}
Total number of letters = 6
(i) Since there are two A's,
number of favourable outcomes = 2
Therefore, the probability of getting A is

**18.(iii) **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 a number divisible by 5.

Total number of discs = 90
Numbers between 1 and 90 that are divisible by 5 are {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90}
Therefore, total number of discs having numbers that are divisible by 5 = 18.

**18.(ii) ** 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 a perfect square number

Total number of discs = 90
Perfect square numbers between 1 and 90 are {1, 4, 9, 16, 25, 36, 49, 64, 81}
Therefore, total number of discs having perfect squares = 9.

**18.(i) ** 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 a two-digit number

Total number of discs = 90
Number of discs having a two-digit number between 1 and 90 = 81

**17.(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 ?

Total number of bulbs = 20
Hence, total possible outcomes = 20
Number of defective bulbs = 4
Hence, the number of favourable outcomes = 4

**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?

Total number of bulbs = 20
Hence, total possible outcomes = 20
Number of defective bulbs = 4
Hence,number of favourable outcomes = 4

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