UP Board Solutions Class 8 Maths Chapter 16 Sambhavna,Prayikta

संभावना (प्रायिकता)

Exercise 16(A)

 

Question 1. एक सिक्का कई बार उछालकर उसके शीर्ष (चित्र) तथा पूँछ (पट) आने की संख्या निम्नांकित सारणी में लिखी गई है। अपनी अभ्यास पुस्तिका में सारणी में रिक्त स्थानों की पूर्ति कीजिए ।
Answer:

In simple words: This table asks you to find the missing counts for heads (चित) and tails (पट) based on the total number of coin tosses. If you know the total and one outcome (e.g., heads), you can find the other (tails) by subtracting the known outcome from the total.

🎯 Exam Tip: Pay close attention to subtraction for each row. Ensure that the sum of the 'चित आने की संख्या' and 'पट आने की संख्या' for any given row equals the 'सिक्का के उछालों की संख्या' for that same row.

 

Question 2. एक पाँसे को कई बार फेंककर उसके ऊपर आने वाली संख्याएँ आगे अंकित सारणी में लिखी गई है। अपनी अभ्यास पुस्तिका में सारणी में रिक्त स्थान की पूर्ति कीजिए
Answer:

In simple words: This table shows how many times each number (1 to 6) appeared when a die was thrown multiple times. You need to fill in the blanks, which implies continuing the sequence or completing rows based on the existing patterns. In this solution, the blank cells are filled in as a completed reference for students.

🎯 Exam Tip: Check for consistency in rows where totals are implied or missing values need to be determined by summing up individual outcomes. Ensure all numbers within a row (for 1-6) sum up to the total number of throws mentioned for that experiment, if available.

 

Question 3. एक समांगी पाँसे के 48 बार फेंकने पर प्रत्येक फलक के ऊपर आने की संभावनाओं को समान मान लेने पर ज्ञात कीजिए कि अंक. 1, 2, 3, 4, 5, 6 में से प्रत्येक कितनी बार ऊपर आएगा?
Answer: अंकों की संख्या 6 पाँसा फेंका गया = 48 बार प्रत्येक अंक दिए पाँसों की संख्या = 48 ÷ 6 = 8 बार
In simple words: If a fair die is thrown many times, each face (1, 2, 3, 4, 5, 6) is expected to appear an equal number of times. To find this, divide the total number of throws (48) by the number of faces on the die (6).

🎯 Exam Tip: For a fair die, the probability of each outcome is equal. The expected frequency of each outcome is calculated by dividing the total number of trials by the number of possible outcomes.

 

Question 4. एक समांगी पाँसे के 54 बार फेंकने पर यह पाया गया कि सम अंकों के ऊपर आने की संख्या 25 है, तो ज्ञात कीजिए कि विषम अंकों के ऊपर अपने की कुल संख्या कितनी होगी?
Answer: कुल फेंके गए पाँसे = 54 बार सम अंकों के लिए फेंके गए पाँसे = 25 बार विषम अंकों के लिए फेंके गए पाँसों की संख्या = 54-25 = 29 बार ।
In simple words: The total number of times a die was thrown (54) is the sum of times even numbers appeared and times odd numbers appeared. Subtract the count of even numbers (25) from the total throws (54) to find the count of odd numbers.

🎯 Exam Tip: Understand that when a die is thrown, an outcome is either an even number or an odd number (mutually exclusive and exhaustive events). The total number of trials is always the sum of the frequencies of these complementary events.

Exercise 16(B)

 

Question 1. दो सिक्के एक साथ 40 बार उछाले गए। यदि HH, HT, TH क्रमशः : 9,8, 12 बार आए हों, । तो ज्ञात कीजिए कि TT कितनी बार आया होगा?
Answer: सिक्के उछाले गए = 40 बार तीन परिणामों (HH, HT, TH) के लिए उछाले गए सिक्के (9+8+12) = 29 बार TT के लिए उछाले गए सिक्के = 40-29 = 11 बार
In simple words: The total number of coin tosses (40) is the sum of the frequencies of all possible outcomes (HH, HT, TH, TT). To find the frequency of TT, sum the frequencies of HH, HT, and TH, and then subtract this total from the grand total of tosses.

🎯 Exam Tip: Identify all possible outcomes when two coins are tossed (HH, HT, TH, TT). The sum of their individual frequencies must always equal the total number of trials performed.

 

Question 2. एक सिक्का 1000 बार उछाला गया और पाया गया कि चित 455 बार आया। ज्ञात कीजिए पट आने का प्रतिशत कितना है?
Answer:सिक्के उछाले गए = 1000 बार सिक्के के चित आने की संख्या = 455 पट आने की संख्या = 1000 – 455 = 545
1000 बार उछालने पर पट आने की संख्या = 545 तो 100 बार उछालने पर पट आने का प्रतिशत =x100 = 54.5%
In simple words: First, determine the number of times 'पट' (tails) appeared by subtracting the number of 'चित' (heads) from the total number of tosses. Then, calculate the percentage of 'पट' by dividing its frequency by the total tosses and multiplying by 100.

🎯 Exam Tip: Remember the percentage calculation formula: (Favorable Outcomes / Total Outcomes) × 100. Be careful to calculate the number of 'पट' correctly before finding its percentage.

 

Question 3. दो सिक्कों को एक साथ 400 बार उछालने पर देखा गया कि
दो चित 90 बार
एक चित 210 बार
कोई भी चितं नहीं 100 बार
इनसे प्रत्येक घटना के घटित होने का प्रतिशत ज्ञात कीजिए।
Answer:दो चितों का प्रतिशत = \( \frac{90}{400} \times 100 = 22.5\% \)
एक चित होने का प्रतिशत = \( \frac{210}{400} \times 100 = 52.5\% \)
कोई भी चित नहीं = \( \frac{100}{400} \times 100 = 25\% \)
In simple words: For each given event (two heads, one head, no heads), calculate its percentage by dividing the number of times it occurred (its frequency) by the total number of tosses (400) and then multiplying the result by 100.

🎯 Exam Tip: Apply the percentage formula (Frequency / Total Trials) × 100 systematically to each event. Ensure your calculations are accurate and that the sum of all percentages is close to 100% (allowing for rounding).

 

Question 4. एक पाँसे को 1000 बार फेंकने पर प्राप्त परिणामों 1,2,3,4,5,6 बारम्बारताएँ निम्नांकित सारणी में दी हुई हैं। 1, 2, 3, 4, 5, 6 में प्रत्येक के आने का प्रतिशत ज्ञात कीजिए ।

Answer:दो चितों का प्रतिशत = \( \frac{90}{400} \times 100 = 22.5\% \)
एक चित होने का प्रतिशत = \( \frac{210}{400} \times 100 = 52.5\% \)
कोई भी चित नहीं = \( \frac{100}{400} \times 100 = 25\% \)
In simple words: The provided answer illustrates how to calculate the percentage of specific outcomes (two heads, one head, no heads) by dividing their observed frequency by the total number of trials (400, in this example) and then multiplying by 100. Although the question asks for die roll percentages, the answer provided uses coin toss data.

🎯 Exam Tip: When calculating percentages, it's crucial to correctly identify the observed frequency for the specific event and the total number of trials. The formula is (Observed Frequency / Total Trials) × 100.

 

Question 5. दो पाँसे एक साथ फेंके जाते हैं और पाँसों पर ऊपर आने वाले अंकों का योगफल लिया जाता है। निम्नांकित घटनाओं को समुच्चय के रूप में लिखिए -
(i) प्राप्त योग सम संख्या हो,
(ii) प्राप्त योग 3 का अपवर्त्य हो,
(iii) प्राप्त योग 4 से न्यून हो,
(iv) प्राप्त योग 10 से अधिक हो,
Answer: शिक्षक की सहायता से शिक्षार्थी स्वयं करें ।।
In simple words: This question requires you to list all possible pairs of outcomes from throwing two dice (like (1,1), (1,2), etc.) and then identify which of these pairs satisfy each given condition based on the sum of their numbers. For example, for condition (i), list all pairs whose numbers add up to an even sum.

🎯 Exam Tip: First, list all 36 possible outcomes when throwing two dice (from (1,1) to (6,6)). Then, systematically go through each condition (sum is even, sum is a multiple of 3, sum is less than 4, sum is greater than 10) and filter the 36 outcomes to find the specific pairs that meet each condition.

 

Question 6. तीन सिक्के एक साथ उछाले जाते हैं तो निम्नांकित घटनाओं को समुच्चय के रूप में लिखिए।
(i) कोई चित प्रकट नहीं होता,
(ii) केवल एक चित होता है,
(iii) कम से कम दो चित प्रकट होते हैं,
(iv) तीनों चित आते हैं।
Answer:
1. कोई चित प्रकट नहीं होता से आशय है कि तीनों पूँछ है = TTT
2. दो पूँछ और दर्शाए = HTT, THT, TTH
3. एक पूँछ और दर्शाए = HHT, HTH, HHT
4. तीनों सिर = HHH
In simple words: When three coins are tossed, there are 8 possible outcomes (e.g., HHH, HHT, etc.). This question asks you to list the specific outcomes (expressed as combinations of Heads 'H' and Tails 'T') that correspond to each given condition, such as getting no heads, exactly one head, at least two heads, or all three heads.

🎯 Exam Tip: First, systematically list all 8 possible outcomes when tossing three coins (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT). Then, carefully match these outcomes to each specified condition (no heads, exactly one head, at least two heads, three heads) to form the correct sets.

 

Question 7. दो पाँसों को एक साथ फेंकने पर दोनों पर सम अंकों के ऊपर आने की घटना का समुच्चय ज्ञात कीजिए।
Answer: दूसरे, चौथे व छठे बार फेंके जाने वाले पाँसों पर सम अंक 2, 4, 6= (2, 2), (2,4), (2,6), (4, 2), (4,4), (4, 6), (6, 2), (6, 4), (6, 6)
In simple words: When two dice are thrown, you need to find all possible pairs of outcomes where both the number on the first die and the number on the second die are even numbers. The even numbers on a standard die are 2, 4, and 6.

🎯 Exam Tip: Identify the even numbers on a single die (2, 4, 6). Then, systematically combine these even numbers from both dice to form all possible ordered pairs, ensuring that both elements in each pair are even.

 

Question 8. दो पाँसों को एक साथ फेंकने पर दोनों पर विषम अंकों के ऊपर आने की घटना का समुच्चय लिखिए ।
Answer: प्रथम, तीसरी व पाँचवी बार फेंके जाने वाले पाँसों पर विषम अंक 1, 3, 5= (1,1), (1,3), (1,5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)
In simple words: This question asks you to list all the possible pairs of outcomes when two dice are thrown, such that both the number appearing on the first die and the number appearing on the second die are odd. The odd numbers on a standard die are 1, 3, and 5.

🎯 Exam Tip: Identify the odd numbers on a single die (1, 3, 5). Then, systematically list all combinations of these odd numbers from both dice to form all possible ordered pairs where both elements are odd.

 

Question 9. दो पाँसों को एक साथ फेंकने पर दोनों पर अंकों का योग विषम संख्या आने का समुच्चय लिखिए ।
Answer: पहले पाँसे पर सम अंक, दूसरे पर विषम अंक, तीसरे पर सम अंक, चौथे पर विषम अंक, पाँचवें पर सम और छठे पर विषम अंक रखने पर = (1,2), (1,4), (1,6), (2, 1), (2, 3), (2,5), (3, 2), (3,4), (3,6), (4,1), (4, 3), (4, 5), (5, 2), (5,4), (5, 6), (6, 1), (6, 3), (6, 5)
In simple words: To achieve an odd sum when rolling two dice, one die must show an odd number and the other must show an even number. This answer lists all possible pairs where one number is odd and the other is even, resulting in an odd sum.

🎯 Exam Tip: An odd sum from two dice always results from one odd number and one even number. You can systematically list all pairs (Odd, Even) and (Even, Odd) to ensure all possibilities are covered, or list all 36 outcomes and filter for odd sums.

 

Question 10. दो पाँसों को एक साथ फेंकने पर दोनों पर अंकों का योग अभाज्य संख्या होने का समुच्चय लिखिए ।
Answer: शिक्षार्थी उपरोक्त प्रश्न की तरह हल करें।
In simple words: This question asks you to list all the pairs of outcomes when two dice are thrown such that the sum of the numbers on the dice is a prime number. Prime numbers that can be formed by summing two dice rolls range from 2 (1+1) to 11 (5+6 or 6+5).

🎯 Exam Tip: First, list all possible sums from throwing two dice (ranging from 2 to 12). Identify which of these sums are prime numbers (2, 3, 5, 7, 11). Then, for each prime sum, list all the pairs of die rolls that result in that specific sum.

 

Question 11. एक लाटरी में 100 इनाम हैं जबकि उसके 100000 टिकट बिके हैं। इस लाटरी का एक टिकट खरीदने वाले व्यक्ति की इनाम जीतने की संभावना कितनी है?
Answer: इनाम प्राप्त करने की संभावना = \( \frac{100}{100000} = \frac{1}{1000} \)
In simple words: The probability of winning in a lottery is calculated by dividing the total number of favorable outcomes (the number of prizes) by the total number of possible outcomes (the total number of tickets sold).

🎯 Exam Tip: Remember the basic probability formula: Probability = (Number of favorable outcomes) / (Total number of possible outcomes). Always simplify the resulting fraction to its lowest terms for the final answer.