Test 15 || Class 10th Maths || Ch. 15 प्रायिकता (Probability)

Test 15 || Class 10th Maths || Ch. 15 प्रायिकता (Probability)

EXERCISE 15.1

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 ___________.

(1). निम्नलिखित कथनों को पूरा

कीजिए :

(i)  घटना E  की प्रायिकता + घटना ‘E नहीं’ की प्रायिकता = __________ है।

(ii) उस घटना की प्रायिकता जो घटित नहीं हो सकती _______ है। ऐसी घटना _________ कहलाती है।

(iii) उस घटना की प्रायिकता जिसका  घटित होना निश्चित है ________ है।

ऐसी घटना _______ कहलाती है।

(iv) किसी प्रयोग की सभी प्रारंभिक घटनाओं की प्रायिकताओं का योग ________ है।

(v) किसी घटना की प्रायिकता _______ से बड़ी या उसके बराबर होती है तथा  _______ से छोटी या उसके बराबर होती है।

 Answere : 

(i) 1,  (ii) 0, Impossible , (iii) 1, Sure (iv) 1, (v) 0, 1

(i) 1,  (ii) 0, अनिश्चित , (iii) 1, निश्चित (iv) 1, (v) 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.

(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.

Answere : 

(i) not equally likely ,  (ii) not equally likely, (iii) equally likely  (iv) equally likely 

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?

फुटबॉल के खेल को प्रारंभ करते समय यह निर्णय लेने के लिए कि कौन-सी टीम पहले बॉल लेगी, इसके लिए सिक्का उछालना एक न्यायसंगत विधि क्यों माना जाता है?

Answer: Because of equal chance of both outcomes

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

4. निम्नलिखित में से कौन सी संख्या किसी घटना की प्रायिकता नहीं हो सकती?

(A)    2/3        (B) –1.5          (C) 15%       (D) 0.7

Answer: B cannot be a probability as it has negative value

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

5. यदि P(E) = 0.05 है, तो ‘E नहीं’ की प्रायिकता क्या है?

Solution: P(E) = 0.05

P(E) + P(not E) = 1

0.05 + P(not E) = 1

P(not E) = 1 – 0.05 = 0.95

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?

6. एक थैले में केवल नीबू की महक वाली मीठी गोलियाँ हैं। मालिनी बिना थैले में झाँके उसमें से एक गोली निकालती है। इसकी क्या प्रायिकता है कि वह निकाली गई गोली

(i) संतरे की महक वाली है?

(ii) नीबू की महक वाली है?

Solution:

(i) an orange flavoured candy?

Let the total Candies = x

P(E) for Orange Flavoured candy = Favourable outcome / Total outcomes

= 0 / x = 0  

Zero or impossible .

(ii) a lemon flavoured candy?

Solution: 

P(E) for a lemon flavoured candy candy = Favourable outcome / Total outcomes

= x / x = 1 

= 1 or Sure event

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?

7. यह दिया हुआ है कि 3 विद्यार्थियों के एक समूह में से 2 विद्यार्थियों के जन्मदिन एक ही दिन न होने की प्रायिकता 0.992 है। इसकी क्या प्रायिकता है कि इन 2 विद्यार्थियों का जन्मदिन एक ही दिन हो?

Solution: 

P(E) + P(not E) = 1

P(not E) = 1 – P(E)  = 1 – 0.992 = 1.000 – 0.992 = 0.008

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?

8. एक थैले में 3 लाल और 5 काली गेंदे हैं। इस थैले में से एक गेंद यादृच्छया निकाली जाती है। इसकी प्रायिकता क्या है कि गेंद

(i) लाल हो?

(ii) लाल नहीं हो?

Solution: 

 (i) is red?

Red Balls = 3

Black Ball = 5

Total number of outcomes = 3 + 5 = 8

Number of favourable outcomes = 3

Hence;

P(E) for a Red Ball = Favourable outcome / Total outcomes

P(Red) = 3 / 8

(ii) not red?

Solution:

P(E) for a not Red Ball = Favourable outcome for Not Red / Total outcomes

P(not Red) = 5 / 8

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?

9. एक डिब्बे में 5 लाल कंचे, 8 सफेद कंचे और 4 हरे कंचे हैं। इस डिब्बे में से एक कंचा यादृच्छया निकाला जाता है। इसकी क्या प्रायिकता है कि निकाला गया कंचा

(i) लाल है?

(Ii) सफेद है?

(iii) हरा नहीं है?

Solution: 

Red Balls = 5

White Ball = 8

Green Balls = 4

Total number of outcomes = 5 + 8 + 4 = 17

(i) Red?

Solution: Probability of red marbles;

P(E) for a not Red Ball = Favourable outcome / Total outcomes

P(E) = 5 / 17

(ii) White?

Solution: Probability of white marbles;

P(E) for a white Ball = Favourable outcome / Total outcomes

P (E) = 8 / 17

(iii) Not green?

Solution: Probability of green marbles;

Not green balls = 5 + 8 = 13

P(E) for a not green ball = Favourable outcome / Total outcomes

P(E) = (5 + 8) / 17 = 13 / 17

10. A piggy bank contains hundred 50p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 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 Rs 5 coin?

10. एक पिग्गी बैंक में, 50 पैसे के सौ सिक्के हैं, 1 रु के पचास सिक्के हैं, 2 रु के बीस सिक्के और 5 रु के दस सिक्के हैं। यदि पिग्गी बैंक को हिलाकर उल्टा करने पर कोई एक सिक्का गिरने के परिणाम समप्रायिक हैं, तो इसकी क्या प्रायिकता है कि वह गिरा हुआ सिक्का

(i) 50 पैसे का होगा?

(ii) 5 रु का नहीं होगा?

Solution: 

50p coins = 100

Rs 1 coin = 50

Rs 2 coin = 50

Rs 5 coin = 10

Total number of events = 100 + 50 + 20 + 10 = 180

(i) will be a 50 p coin?

P(E) for a 50 p coin = Favourable outcome for 50p cons / Total outcomes

P(E) for 50 p coins   =50 / 180 = 5 / 18

(ii) Will not be a Rs. 5 coin?

Solution: Number of Rs. 5 coins = 10

P(E) for not a Rs. 5 coin = Favourable outcome for not a Rs. 5 coin / Total outcomes

 P(E) Not Rs. 5  = (180 − 10)/180 = 170/180 = 17/18

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. What is the probability that the fish taken out is a male fish?

11. गोपी अपने जल-जीव कुंड के लिए एक दुकान से मछली खरीदती है। दुकानदार एक टंकी, जिसमें 5 नर मछली और 8 मादा मछली हैं, में से एक मछली यादृच्छया उसे देने के लिए निकालती है । इसकी क्या प्रायिकता है कि निकाली गई मछली नर मछली है?

Solution: Total number of events = 5 + 8 = 13

Number of male fish = 5

P(E) for not a male fish  = Favourable outcome for male fish  / Total outcomes

P(E) For male fish =5/13

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?

12. संयोग के एक खेल में, एक तीर को घुमाया जाता है, जो विश्राम में आने के बाद संख्याओं 1, 2, 3, 4, 5, 6, 7 और 8 में से किसी एक संख्या को इंगित करता है। यदि ये सभी परिणाम समप्रायिक हों तो इसकी क्या प्रायिकता है कि यह तीर इंगित

(i) 8 को करेगा?

(ii) एक विषम संख्या को करेगा?

(iii) 2 से बड़ी संख्या को करेगा?

(iv) 9 से छोटी संख्या को करेगा?

Solution:

(i) 8 ?

Total number of events = 8

Number of 8s = 1

P(E) for a Number of 8s = Favourable outcome for Number of 8s / Total outcomes

P(8)=1/8

(ii) an odd number?

Solution: Number of odd numbers = 4

P(E) for an odd number = Favourable outcome for an odd number / Total outcomes

P(E) for an odd number  = 4 / 8 = 1 / 2

(iii) a number greater than 2?

Solution: Number of numbers greater than 2 = 6

P(E) for a numbers greater than 2

= Favourable outcome for a numbers greater than 2  / Total outcomes

P(E) for a numbers greater than 2 = 6 / 8 = 3 / 4

(iv) a number less than 9?

Solution: since all 8 numbers are less than 9

P(E) for a number less than 9 = Favourable outcome for a number less than 9 / Total outcomes

P(E) for a numbers less than 9 = 8 / 8 = 1

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.

13. एक पासे को एक बार फैंका जाता है। निम्नलिखित को प्राप्त करने की प्रायिकता ज्ञात कीजिए :

(i) एक अभाज्य संख्या (ii) 2 और 6 के बीच स्थित कोई संख्या (iii) एक विषम संख्या

Solution: 

(i) a prime number

Total number of events = 6

Number of prime numbers (2, 3 , 5) = 3

P(E) for a prime numbers  = Favourable outcome for a prime numbers  / Total outcomes

P(E) for a prime numbers  = 3 / 6 = 1 / 2

(ii) a number lying between 2 and 6

Solution: numbers between 2 and 6  are (3, 4, 5) = 3

P(E) for a numbers between 2 and 6  

= Favourable outcome for a numbers between 2 and 6  / Total outcomes

P(E) for a numbers between 2 and 6  = = 3 / 6 = 1 / 2

(iii) an odd number.

Solution: Number of odd numbers (1, 3, 5) = 3

P(E) for an odd numbers = Favourable outcome for an odd numbers / Total outcomes

P(E) for an odd numbers = = 3 / 6 = 1 / 2

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

(14). 52 पत्तों की अच्छी प्रकार से फेटी गई एक गडडी में से एक पत्ता निकाला जाता है। निम्नलिखित को प्राप्त करने की प्रायिकता ज्ञात कीजिए :

(i) लाल रंग का बादशाह

(ii) एक फेस कार्ड अर्थात् तस्वीर वाला पत्ता

(iii) लाल रंग का तस्वीर वाला पत्ता

(iv) पान का गुलाम

(v) हुकुम का पत्ता

(vi) एक ईंट की बेगम

(i) a king of red colour

Solution: Total number of events = 52

Number of red king = 2

P(E) for a Red King  = Favourable outcome for a Red King  / Total outcomes

P(E) for a Red King =2 / 52 = 1 / 26

---

(ii) A face card

Solution: Number of face cards in a pack = 12

P(E) for a face cards in a pack  

= Favourable outcome for a face cards in a pack / Total outcomes

P(E) for a face cards in a pack = 12 / 52 = 3 / 13

(iii) A red face card

Solution: Number of red face cards in a pack = 6

P(E) for a red face cards = Favourable outcome for a red face cards / Total outcomes

P(E) for a red face cards in a pack = 6 / 52 = 3 / 26

(iv) Jack of hearts

Solution: Number of jack of hearts = 1

P(E) for a jack of hearts = Favourable outcome for a jack of hearts / Total outcomes

P(E) for a jack of hearts = 1 / 52

(v) A spade

Solution: Number of spade cards in a pack = 13

P(E) for a spade cards = Favourable outcome for a spade cards / Total outcomes

P(E) for a spade cards =13 / 52 = 1 / 4

(vi) Queen of diamonds

Solution: Number of queen of diamonds in a pack = 1

P(E) for a queen of diamonds = Favourable outcome for a queen of diamonds / Total outcomes

P(E) for a spade cards = 1 / 52

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?

15. ताश के पाँच पत्तों ईंट का दहला, गुलाम, बेगम, बादशाह और इक्का को पलट करके अच्छी प्रकार फैंटा जाता है। फिर इनमें से यादृच्छया एक पत्ता निकाला जाता है।

(i) इसकी क्या प्रायिकता है कि यह पत्ता एक बेगम है?

(ii) यदि बेगम निकल आती है, तो उसे अलग रख दिया जाता है और एक अन्य पत्ता निकाला जाता है। इसकी क्या प्रायिकता है कि दूसरा निकाला गया पत्ता

(a) एक इक्का है?

(b) एक बेगम है?

(i) What is the probability that the card is the queen?

Solution: Total number of events = 5

Number of queen = 1

P(E) for a queen card = Favourable outcome for a queencard / Total outcomes

P(E) for a queen card = 1 / 5

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is

(a) an ace?

Solution: Total number of events = 4

P(E) for an ace = Favourable outcome for an ace / Total outcomes

P(E) for an ace = 1 / 4

(b) a queen?

Solution:Now, there is no queen in the pack

P(E) for a queen = Favourable outcome for a queen / Total outcomes

P(E) for a queen = 0 / 4 = 0

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.

16. किसी कारण 12 खराब पेन 132 अच्छे पेनों में मिल गए हैं। केवल देखकर यह नहीं बताया जा सकता है कि कोई पेन खराब है या अच्छा है। इस मिश्रण में से, एक पेन यादृच्छया निकाला जाता है। निकाले गए पेन की अच्छा होने की प्रायिकता ज्ञात कीजिए।

Solution: Total number of events = 132 + 12 = 144

Number of favourable events = 132

P(E) for a good one pen = Favourable outcome for a good one pen / Total outcomes

P(E) for a good one pen =132 / 144 = 11 / 12

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 ?

(17). (i) 20 बल्बों के एक समूह में 4 बल्ब खराब हैं। इस समूह में से एक बल्ब यादृच्छया निकाला जाता है। इसकी क्या प्रायिकता है कि यह बल्ब खराब होगा?

(ii) मान लीजिए (i) में निकाला गया बल्ब खराब नहीं है और न ही इसे दुबारा बल्बों के साथ मिलाया जाता है। अब शेष बल्बों में से एक बल्ब यादृच्छया निकाला जाता है। इसकी क्या प्रायिकता है कि यह बल्ब खराब नहीं होगा?

Solution: (i)Total Number of events = 20

Number of a defective bulb = 4

P(E) for a defective bulb = Favourable outcome for a defective bulb / Total outcomes

P(E) for a defective bulb = 4 / 20 = 1 / 5

Solution: (ii) Total number of events = 19

Number of non-defective bulb = 15

P(E) for a non-defective bulb = Favourable outcome for a non-defective bulb / Total outcomes

P(E) for a non-defective bulb =15 / 19

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.

(18). एक पेटी में 90 डिस्क हैं, जिन पर 1 से 90 तक संख्याएँ अंकित हैं। यदि इस पेटी में से एक डिस्क यादृच्छया निकाली जाती है तो इसकी प्रायिकता ज्ञात कीजिए कि इस डिस्क पर अंकित होगी :

(i) दो अंकों की एक संख्या

(ii) एक पूर्ण वर्ग संख्या

(iii) 5 से विभाज्य एक संख्या।

(i) a two-digit number

Solution: Total number of events = 90

Number of two-digit number = 90 – 9 = 81

(single digit numbers are from 1 to 9)

P(E) for a two-digit number = Favourable outcome for a two-digit number / Total outcomes

P(E) for a two-digit number =81 / 90 = 9 / 10

(ii) a perfect square number

Solution: Number of favourable events can be counted by making following list of square numbers:

Number of a perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81) = 9

P(E) for a perfect squares = Favourable outcome for a perfect squares / Total outcomes

P(E) for a perfect squares = 9 / 90 = 1 / 10

(iii) a number divisible by 5.

Solution: The largest number divisible by 5 is 90

Number divisible by 5 can be counted as 5, 10 , 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90) = 18

P(E) for a number divisible by 5

= Favourable outcome for a number divisible by 5/ Total outcomes

P(E) for a number divisible by 5 = 18 / 90= 1 / 5

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?

(19). एक बच्चे के पास ऐसा पासा है जिसके फलकों पर निम्नलिखित अक्षर अंकित हैं :

[A], [B], [C], [D], [E], [A]

इस पासे को एक बार फेंका जाता है। इसकी क्या प्रायिकता है कि

(i) A प्राप्त हो?

(ii) D प्राप्त हो?

(i) A?

Solution: Total number of events = 6

Number of A = 2

P(E) for A = Favourable outcome for A / Total outcomes

P(E) for A = 2 / 6 = 1 /3

(ii) D?

Solution: Number of D = 1

P(E) for D = Favourable outcome for D / Total outcomes

P(E) for D = 1 / 6

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?

(20)* मान लीजिए आप एक पासे को आकृति 15.6 में दर्शाए आयताकार क्षेत्र में यादृच्छया रूप से गिराते हैं। इसकी क्या प्रायिकता है कि वह पासा 1m व्यास वाले वृत्त के अंदर गिरेगा?

Solution: Area of rectangle = 6 sq m

Area of circle = π r ²  = π × 0.5 ² = 0.25 π = π / 4

Area of rectangle gives the total number of events and area of circle gives the number of favourable events.

P(E) for area of circle = Favourable outcome for area of circle / Total Area of Rectangle

P(E) for area of circle =  (π/4) / 6  = π / 24

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 ?

(21). 144 बॉल पेनों के एक समूह में 20 बॉल पेन खराब हैं और शेष अच्छे हैं। आप वही पेन खरीदना चाहेंगे जो अच्छा हो, परंतु खराब पेन आप खरीदना नहीं चाहेंगे। दुकानदार इन पेनों में से, यादृच्छया एक पेन निकालकर आपको देता है। इसकी क्या प्रायिकता है कि

(i) आप वह पेन खरीदेंगे?

(ii) आप वह पेन नहीं खरीदेंगे?

(i) She will buy it?

Solution: Total number of events = 144

She will buy it if it is a good pen = 124

P(E) for a good pen = Favourable outcome for a good pen / Total outcomes

P(E) for a good pen =  124 / 144= 31 / 36

(ii) She will not buy it?

Solution: Number of defective pen = 20

She will buy it if it is a defective pen = 20

P(E) for a defective pen = Favourable outcome for a defective pen / Total outcomes

P(E) for a defective pen = 20 / 144 = 5 / 36

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

Event 'Sum of 2 dice'	3	3	4	5	6	7	8	9	10	11	12
Probability	1/36						5/36				1/36

 (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.

उदाहरण 13 को देखिए। (i) निम्नलिखित सारणी को पूरा कीजिए :

(ii) एक विद्यार्थी यह तर्क देता है कि ‘यहाँ कुल 11 परिणाम 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 और 12 हैं। अतः, प्रत्येक की प्रायिकता 1

11 है।’ क्या आप इस तर्क से सहमत हैं? सकारण उत्तर दीजिए।

Solutions:

A dice is thrown twice means total no. of possible outcomes are 36

The probability that sum will be 3,4,5,6,7,9,10,11?

Steps:

No of possible outcomes to get the sum as 2 = (1,1) 

No of possible outcomes to get the sum as 3 = (2,1),(1,2)

No of possible outcomes to get the sum as 4 = (2,2),(1,3),(3,1)

No of possible outcomes to get the sum as 5 = (3,2),(2,3),(4,1),(1,4)

No of possible outcomes to get the sum as 6 = (5,1),(1,5),(3,3),(4,2),(2,4)

No of possible outcomes to get the sum as 7 = (4,3),(3,4),(6,1),(1,6),(5,2),(2,5)

No of possible outcomes to get the sum as 8 = (4,4),(6,2),(2,6),(5,3),(3,5)

No of possible outcomes to get the sum as 9 = (5,4), (4,5),(6,3),(3,6)

No of possible outcomes to get the sum as 10 = (5,5),(6,4),(4,6)

No of possible outcomes to get the sum as 11 = (6,5),(5,6)

 No of possible outcomes to get the sum as 12 = (6,6)

Event 'Sum of 2 dice'	3	3	4	5	6	7	8	9	10	11	12
Probability	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36
	1/36	1/18	1/12	1/9	5/36	1/6	5/36	1/9	1/12	1/18	1/36

(ii)Probability of each of them is not 1/11 as these are not equally likely.

This is demonstrated in the solution of (i)

We do not agree with this argument because there are different number of possible outcomes for each sum. According to the above table, we can see that each sum has different probability.

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.

(23). एक खेल में एक रुपए के सिक्के को तीन बार उछाला जाता है और प्रत्येक बार का परिणाम लिख लिया जाता है। तीनों परिणाम समान होने पर, अर्थात् तीन चित या तीन पट प्राप्त होने पर, हनीफ खेल में जीत जाएगा, अन्यथा वह हार जाएगा। हनीफ के खेल में हार जाने की प्रायिकता परिकलित कीजिए।

Solution: Possible outcomes of 3 tosses of coin

1st toss = HHH
2nd toss = HHT or HTH or THH
3rd toss = TTH or THT or HTT
4th toss = TTT

Total number of events (HHH, HHT, HTT, TTT, TTH, THH, HTH, THT) = 8

Hanif will loss if he did not find HHH or TTT = 6 times

Number of favourable events for losing (HHT, HTT, TTH, THH, HTH, THT) = 6

P(E) for losing the game = Favourable outcome for losing the game / Total outcomes

P(E) for losing the game = 6 / 8 = 3 / 4

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]

(24). एक पासे को दो बार फेकां जाता है। इसकी क्या प्रायिकता है कि

(i) 5 किसी भी बार में नहीं आएगा?

(ii) 5 कम से कम एक बार आएगा?

[¹संकेत : एक पासे को दो बार फेंकना और दो पासों को एक साथ फेंकना एक ही प्रयोग माना जाता है।]

(i) 5 will not come up either time?

Solution: Possible outcomes of 2 throws of a die can be shown by following table:

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

Total number of events = 36

Number of times when 5 does not come up either of times = 25

P(E) for 5 does not come up either of times

= Favourable outcome for 5 does not come up either of times / Total outcomes

P(E) for 5 does not come up either of times =  (36 – 11) / 36 = 25 / 36

(ii) 5 will come up at least once?

Solution: Number of 5 comes at least once = 11

P(E) for Number 5 comes at least once

= Favourable outcome for Number 5 comes at least once / Total outcomes

P(E) for Number 5 comes at least once = 11 / 36

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.

(25). निम्नलिखित में से कौन से र्तक सत्य हैं और कौन से र्तक असत्य हैं? सकारण उत्तर दीजिए।

(i) यदि दो सिक्कों को एक साथ उछाला जाता है, तो इसके तीन संभावित परिणाम- दो चित, दो पट या प्रत्येक एक बार हैं। अतः, इनमें से प्रत्येक परिणाम की प्रायिकता 1/3 है।

(ii) यदि एक पासे को फेंका जाता है, तो इसके दो संभावित परिणाम एक विषम संख्या या एक सम संख्या हैं। अतः एक विषम संख्या ज्ञात करने की प्रायिकता 1/2 है।

Solution: Possible outcomes: HH, HT, TH, TT

P(2 heads) = 1 / 4
P(2 tails) = 1 / 4
P(both) = 2 / 4 = 1 / 2

So, given answer is incorrect.

(i) Incorrect

If two coins are tossed simultaneously then ,

Total possible outcomes are

(H,H), (T,T), (H,T), (T,H) = 4

No of outcomes to get two heads = (H,H) = 1

No of outcomes to get two tails = (T,T) =1

No of outcomes to one of each = (H,T), (T,H) = 2

Probability of getting two head = 1 / 4

Probability of getting two tails = 1 / 4

Probability of getting one of each = 2 /4 = 1 / 2

We can observe that, the probability of each of the outcome is not 1/3.

(ii) Correct

Solution: This is correct. Number of both outcomes is equal.

Total no of possible outcomes when a dice is thrown =(1,2,3,4,5,6)

 No of possible outcomes to get odd number (1, 3, 5) = 3

No of possible outcomes to get even number (2, 4, 6) = 3

Probability of getting odd number = 3 / 6 = 1 / 2

Probability of getting even number = 3 / 6 = 1 / 2

Thus, We can say that the probability of getting an odd or even number is 1/2.