GSEB Class 12 Maths Solutions Chapter 13 સંભાવના Exercise 13.3

GSEB Solutions Class 12 Maths Chapter 13 संभावना Ex 13.3

 

Question 1. એક પાત્રમાં 5 લાલ રંગના અને 5 કાળા રંગના દડા છે. યાદચ્છિક રીતે એક દડો પસંદ કરવામાં આવે છે. તેનો રંગ નોંધીને તેને પાત્રમાં પાછો મૂકી દેવાય છે. તદુપરાંત, જે રંગ જે નોંધ્યો હતો તે રંગના 2 વધારાના ઇંડા પાત્રમાં મૂકવામાં આવે છે. અને ત્યાર બાદ એક દડો યાદચ્છિક રીતે પસંદ કરવામાં આવે છે, બીજો દડો લાલ રંગનો હોય તેની સંભાવના કેટલી?

Answer:Let event A be that the first ball chosen is red. Let event B be that the first ball chosen is black. The total number of balls is 10 (5 red + 5 black). So, the probability of choosing a red ball first, P(A), is \(\frac{5}{10} = \frac{1}{2}\). The probability of choosing a black ball first, P(B), is \(\frac{5}{10} = \frac{1}{2}\). When the first ball is red and put back, 2 more red balls are added. So, the number of balls becomes:

દડાઓની સંખ્યા	ઘટના A ઉદ્ભવ્યા બાદ	ઘટના B ઉદ્ભવ્યા બાદ
લાલ રંગના = 5	7	5
કાળા રંગનો = 5	5	7
કુલ = 10	12	12

Let event C be that the second ball chosen is red. If event A happened (first ball red), then there are 7 red balls out of 12 total. So, \(P(C | A) = \frac{7}{12}\). If event B happened (first ball black), then there are still 5 red balls out of 12 total. So, \(P(C | B) = \frac{5}{12}\). To find the probability that the second ball is red, P(C), we use the law of total probability: \(P(C) = P(A)P(C | A) + P(B)P(C | B)\) \(P(C) = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12}\) \(P(C) = \frac{7}{24} + \frac{5}{24}\) \(P(C) = \frac{12}{24}\) \(P(C) = \frac{1}{2}\) Therefore, the probability that the second ball chosen is red is \(\frac{1}{2}\).In simple words: First, we pick a ball and put it back, adding two more balls of the same color. Then, we pick another ball. The chance that this second ball is red is 1 out of 2.

🎯 Exam Tip: Remember to clearly define events and use the formula for total probability when a sequence of events affects the probabilities of subsequent events.

 

Question 2. એક થેલામાં 4 લાલ રંગના અને 4 કાળા રંગના દડા છે. બીજા થેલામાં 2 લાલ રંગના અને 6 કાળા રંગના દડા છે. બેમાંથી એક થેલો યાદચ્છિક રીતે પસંદ કરવામાં આવે છે અને એક દડો તે થેલામાંથી યાદચ્છિક રીતે પસંદ કરવામાં આવે છે. તે લાલ રંગનો માલૂમ પડે છે. દડો પહેલા થેલામાંથી પસંદ કર્યો હોય તેની સંભાવના શોધો.

Answer:Let event \(E_1\) be that the first bag is chosen. Let event \(E_2\) be that the second bag is chosen. Since a bag is chosen randomly from two bags, the probabilities are: \(P(E_1) = \frac{1}{2}\) \(P(E_2) = \frac{1}{2}\) The number of balls in each bag is:

	થેલો I	થેલો II
લાલ રંગના દડા	4	2
કાળા રંગના દડા	4	6

Total balls in Bag I = 4 + 4 = 8 Total balls in Bag II = 2 + 6 = 8 Let event A be that a red ball is drawn from the chosen bag. If Bag I is chosen, the probability of drawing a red ball is: \(P(A | E_1) = \frac{4}{8} = \frac{1}{2}\) If Bag II is chosen, the probability of drawing a red ball is: \(P(A | E_2) = \frac{2}{8} = \frac{1}{4}\) We want to find the probability that the ball came from Bag I, given that it is red, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(A)}\) Where \(P(A) = P(E_1)P(A | E_1) + P(E_2)P(A | E_2)\) So, \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2)}\) \(P(E_1 | A) = \frac{\frac{1}{2} \times \frac{4}{8}}{\frac{1}{2} \times \frac{4}{8} + \frac{1}{2} \times \frac{2}{8}}\) \(P(E_1 | A) = \frac{\frac{4}{16}}{\frac{4}{16} + \frac{2}{16}}\) \(P(E_1 | A) = \frac{\frac{4}{16}}{\frac{6}{16}}\) \(P(E_1 | A) = \frac{4}{6} = \frac{2}{3}\) The desired probability is \(\frac{2}{3}\).In simple words: We have two bags, and we pick one randomly, then draw a red ball. The chance that this red ball came from the first bag (which had more red balls) is 2 out of 3.

🎯 Exam Tip: Bayes' Theorem is crucial here. Ensure all prior and conditional probabilities are correctly calculated before applying the formula.

 

Question 3. કૉલેજના વિદ્યાર્થીઓ પૈકી 60 % વિદ્યાર્થીઓ છાત્રાલયમાં રહે છે. અને 40 % વિદ્યાર્થીઓ છાત્રાલયમાં રહેતા નથી તેમ જ્ઞાત છે. આગળના વર્ષના પરિણામ પરથી માહિતી મળે છે કે, છાત્રાલયમાં રહેતા વિદ્યાર્થીઓ પૈકી 30 % વિદ્યાર્થીઓએ વાર્ષિક પરીક્ષામાં A ગ્રેડ મેળવ્યો છે અને છાત્રાલયમાં નહિ રહેનારા વિદ્યાર્થીઓ પૈકીના 20 % વિદ્યાર્થીઓએ તેમની વાર્ષિક પરીક્ષામાં A ગ્રેડ મેળવ્યો છે. વર્ષાને કૉલેજમાંથી એક વિદ્યાર્થી યાદચ્છિક રીતે પસંદ કરવામાં આવ્યો અને તેણે A ગ્રેડ મેળવ્યો છે તેમ આપેલ હોય, તો આ વિદ્યાર્થી છાત્રાલયનો હોવાની સંભાવના કેટલી ?

Answer:Let event \(E_1\) be that a student lives in the hostel. Let event \(E_2\) be that a student does not live in the hostel. From the given data: \(P(E_1) = 60\% = \frac{60}{100} = \frac{6}{10}\) \(P(E_2) = 40\% = \frac{40}{100} = \frac{4}{10}\) Let event A be that a student gets an A grade in the annual examination. The probability that a student gets an A grade, given they live in the hostel: \(P(A | E_1) = 30\% = \frac{30}{100} = \frac{3}{10}\) The probability that a student gets an A grade, given they do not live in the hostel: \(P(A | E_2) = 20\% = \frac{20}{100} = \frac{2}{10}\) We need to find the probability that a student lives in the hostel, given that they got an A grade, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2)}\) \(P(E_1 | A) = \frac{\frac{6}{10} \times \frac{3}{10}}{\frac{6}{10} \times \frac{3}{10} + \frac{4}{10} \times \frac{2}{10}}\) \(P(E_1 | A) = \frac{\frac{18}{100}}{\frac{18}{100} + \frac{8}{100}}\) \(P(E_1 | A) = \frac{\frac{18}{100}}{\frac{26}{100}}\) \(P(E_1 | A) = \frac{18}{26} = \frac{9}{13}\) The desired probability is \(\frac{9}{13}\).In simple words: We know the chances of students getting an A grade whether they live in a hostel or not. If a student got an A grade, we want to know the chance they live in the hostel. This chance is 9 out of 13.

🎯 Exam Tip: Carefully identify the conditional probabilities and prior probabilities from the percentages given. Accuracy in calculations is vital for the final result.

 

Question 4. બહુવિકલ્પ કસોટીમાં પ્રશ્નનો જવાબ આપવામાં, વિદ્યાર્થી કાં તો જવાબ જાણે છે અથવા અનુમાન કરે છે. વિદ્યાર્થી જવાબ જાણે છે તેની સંભાવના \(\frac{3}{4}\) અને અનુમાન કરે છે તેની સંભાવના \(\frac{1}{4}\) છે. માની લો કે વિદ્યાર્થી જે જવાબનું અનુમાન કરે છે તે સાચો હોય તેની સંભાવના \(\frac{1}{4}\) છે. આપેલ હોય કે તેણે તે જવાબ સાચો આપ્યો છે ત્યારે વિદ્યાર્થીએ આપેલ જવાબ તે જાણતો હતો તેની સંભાવના કેટલી ?

Answer:Let event \(E_1\) be that the student knows the answer. Let event \(E_2\) be that the student guesses the answer. From the problem statement: \(P(E_1) = \frac{3}{4}\) \(P(E_2) = \frac{1}{4}\) Let event A be that the student gives the correct answer. If the student knows the answer, the probability of giving a correct answer is 1 (they are sure to be correct). \(P(A | E_1) = 1\) If the student guesses the answer, the probability of giving a correct answer is given as \(\frac{1}{4}\). \(P(A | E_2) = \frac{1}{4}\) We need to find the probability that the student knew the answer, given that they answered it correctly, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2)}\) \(P(E_1 | A) = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}}\) \(P(E_1 | A) = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}}\) \(P(E_1 | A) = \frac{\frac{3}{4}}{\frac{12}{16} + \frac{1}{16}}\) \(P(E_1 | A) = \frac{\frac{3}{4}}{\frac{13}{16}}\) \(P(E_1 | A) = \frac{3}{4} \times \frac{16}{13}\) \(P(E_1 | A) = \frac{12}{13}\) The desired probability is \(\frac{12}{13}\).In simple words: A student answers a question either by knowing it or by guessing. We want to find the chance that they knew the answer, given that their answer was correct. This chance is 12 out of 13.

🎯 Exam Tip: Distinguish between knowing the answer (P(A|E1)=1) and guessing the answer (P(A|E2) is given) carefully. This distinction is vital for accurate application of Bayes' Theorem.

 

Question 5. એક પ્રયોગશાળા રક્ત પરીક્ષણમાં, જ્યારે તે ખરેખર રોગ હોય ત્યારે તે રોગને શોધી કાઢવામાં 99 % અસરકારક છે. તેમ છતાં, સ્વસ્થ વ્યક્તિનો પરીક્ષણ અહેવાલ ખોટો અને હકારાત્મક 0.5 % સુધી પણ આપે છે. (એટલે કે, જો સ્વસ્થ વ્યક્તિનું પરીક્ષણ કરાય, તો 0.005 સંભાવના સાથે પરીક્ષણ નિદાન કરશે કે તેને બીમારી છે.) જો વસતીના 0.1 % લોકોને ખરેખર બીમારી હોય, તો આપેલ હોય કે તેના પરીક્ષણનું પરિણામ હકારાત્મક છે તે પરિસ્થિતિમાં તેને બીમારી હોવાની સંભાવના કેટલી ?

Answer:Let event \(E_1\) be that a person has the disease. Let event \(E_2\) be that a person does not have the disease (is healthy). From the problem statement: The percentage of people with the disease in the population is 0.1%. \(P(E_1) = 0.1\% = \frac{0.1}{100} = 0.001\) The percentage of healthy people is: \(P(E_2) = 1 - P(E_1) = 1 - 0.001 = 0.999\) Let event A be that the test result is positive. The test is 99% effective at detecting the disease when it is present. So, the probability of a positive test given the person has the disease is: \(P(A | E_1) = 99\% = \frac{99}{100} = 0.99\) The test gives a false positive result for healthy people with a probability of 0.5%. So, the probability of a positive test given the person is healthy is: \(P(A | E_2) = 0.5\% = \frac{0.5}{100} = 0.005\) We want to find the probability that a person has the disease, given that their test result is positive, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2)}\) \(P(E_1 | A) = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005}\) \(P(E_1 | A) = \frac{0.00099}{0.00099 + 0.004995}\) \(P(E_1 | A) = \frac{0.00099}{0.005985}\) \(P(E_1 | A) = \frac{990}{5985}\) \(P(E_1 | A) = \frac{22}{133}\) (After dividing numerator and denominator by 45) The desired probability is \(\frac{22}{133}\).In simple words: A test is good at finding a disease but can sometimes give wrong positive results for healthy people. If someone gets a positive test result, we want to know the real chance they have the disease, considering how rare the disease is in the population. The chance is 22 out of 133.

🎯 Exam Tip: Pay close attention to distinguishing between true positive (P(A|E1)) and false positive (P(A|E2)) rates, as their values are crucial for accurate Bayesian inference.

 

Question 6. ત્રણ સિક્કા આપેલ છે. એક સિક્કાની બંને બાજુ છાપ છે. બીજો અસમતોલ સિક્કો છે. તેમાં છાપ મળવાની સંભાવના 75 % છે. અને ત્રીજો સમતોલ સિક્કો છે. ત્રણમાંથી એક સિક્કો યાદૈચ્છિક રીતે પસંદ કરીને ઉછાળ્યો. તે છાપ બતાવે છે, તો તે બે છાપ ધરાવતો સિક્કો હોય તેની સંભાવના કેટલી ?

Answer:Let event \(E_1\) be that the two-headed coin (both sides are heads) is chosen. Let event \(E_2\) be that the biased coin is chosen. Let event \(E_3\) be that the fair coin is chosen. Since one coin is chosen randomly from three coins, the probabilities are: \(P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}\) Let event A be that the coin shows a head. If the two-headed coin is chosen, the probability of getting a head is 1. \(P(A | E_1) = 1\) If the biased coin is chosen, the probability of getting a head is 75%. \(P(A | E_2) = 75\% = \frac{75}{100} = \frac{3}{4}\) If the fair coin is chosen, the probability of getting a head is \(\frac{1}{2}\). \(P(A | E_3) = \frac{1}{2}\) We want to find the probability that the two-headed coin was chosen, given that it showed a head, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2) + P(E_3)P(A | E_3)}\) \(P(E_1 | A) = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}}\) \(P(E_1 | A) = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{12} + \frac{1}{6}}\) \(P(E_1 | A) = \frac{\frac{1}{3}}{\frac{4}{12} + \frac{3}{12} + \frac{2}{12}}\) \(P(E_1 | A) = \frac{\frac{1}{3}}{\frac{9}{12}}\) \(P(E_1 | A) = \frac{1}{3} \times \frac{12}{9}\) \(P(E_1 | A) = \frac{4}{9}\) The desired probability is \(\frac{4}{9}\).In simple words: We pick one of three coins—a two-headed coin, a biased coin (75% heads), or a normal coin—and flip it. If it lands on heads, we want to know the chance it was the two-headed coin. This chance is 4 out of 9.

🎯 Exam Tip: When dealing with multiple mutually exclusive and exhaustive events, correctly applying the law of total probability in the denominator of Bayes' theorem is key.

 

Question 7. એક વીમાકંપનીએ 2000 સ્કૂટર ચાલકો, 4000 કાર-ચાલકો અને 6000 ટ્રક-ચાલકોનો વીમો ઉતાર્યો. તેમના દ્વારા થતા અકસ્માતોની સંભાવના અનુક્રમે 0.01, 0.03 અને 0.15 છે. વીમાધારકો પૈકીના એક વ્યક્તિને અકસ્માત થયો. તે સ્કૂટર- ચાલક હોવાની સંભાવના કેટલી ?

Answer:Let event \(E_1\) be that the insured person is a scooter driver. Let event \(E_2\) be that the insured person is a car driver. Let event \(E_3\) be that the insured person is a truck driver. Total number of insured people = 2000 + 4000 + 6000 = 12000. The probabilities of choosing each type of driver are: \(P(E_1) = \frac{2000}{12000} = \frac{2}{12} = \frac{1}{6}\) \(P(E_2) = \frac{4000}{12000} = \frac{4}{12} = \frac{1}{3}\) \(P(E_3) = \frac{6000}{12000} = \frac{6}{12} = \frac{1}{2}\) Let event A be that the person meets with an accident. The probabilities of accidents for each type of driver are given: \(P(A | E_1) = 0.01\) (for scooter drivers) \(P(A | E_2) = 0.03\) (for car drivers) \(P(A | E_3) = 0.15\) (for truck drivers) We want to find the probability that the person was a scooter driver, given that they had an accident, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2) + P(E_3)P(A | E_3)}\) \(P(E_1 | A) = \frac{\frac{1}{6} \times 0.01}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.03 + \frac{1}{2} \times 0.15}\) \(P(E_1 | A) = \frac{0.01/6}{0.01/6 + 0.03/3 + 0.15/2}\) \(P(E_1 | A) = \frac{0.01/6}{0.01/6 + 0.01 + 0.075}\) \(P(E_1 | A) = \frac{0.001666...}{0.001666... + 0.01 + 0.075}\) \(P(E_1 | A) = \frac{0.001666...}{0.086666...}\) To avoid rounding errors, let's keep as fractions: \(P(E_1 | A) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + \frac{0.03}{3} + \frac{0.15}{2}}\) Multiply numerator and denominator by 6 to clear denominators: \(P(E_1 | A) = \frac{0.01}{0.01 + 0.06 + 0.45}\) \(P(E_1 | A) = \frac{0.01}{0.52}\) \(P(E_1 | A) = \frac{1}{52}\) The desired probability is \(\frac{1}{52}\).In simple words: An insurance company covers different types of drivers, and each type has a different chance of having an accident. If someone gets into an accident, we want to know the chance they were a scooter driver. This chance is 1 out of 52.

🎯 Exam Tip: Clearly list the populations and their respective probabilities, along with the conditional probabilities of the event (accident) for each group. Converting decimals to fractions or finding common denominators can prevent calculation errors.

 

Question 8. એક ફેક્ટરી પાસે બે યંત્રો A અને B છે. ભૂતકાળની નોંધ બતાવે છે કે, યંત્ર A ઉત્પાદિત વસ્તુઓ પૈકી 60 % વસ્તુઓનું અને યંત્ર B 40 % વસ્તુઓનું ઉત્પાદન કરે છે. વધુમાં, યંત્ર A દ્વારા ઉત્પાદિત વસ્તુઓ પૈકી 2 % અને યંત્ર B દ્વારા ઉત્પાદિત વસ્તુઓ પૈકી 1 % વસ્તુઓ ખામીયુક્ત હતી. આ બધી વસ્તુઓ એક પુરવઠાગારમાં મૂકી દીધી અને ત્યાર બાદ આમાંથી એક વસ્તુ યાદૈચ્છિક રીતે પસંદ કરી અને તે ખામીયુક્ત માલૂમ પડી, તો તે યંત્ર B દ્વારા ઉત્પાદિત હોવાની સંભાવના કેટલી ?

Answer:Let event \(E_A\) be that the item was produced by machine A. Let event \(E_B\) be that the item was produced by machine B. From the problem statement: Machine A produces 60% of items: \(P(E_A) = 60\% = \frac{60}{100} = \frac{6}{10}\) Machine B produces 40% of items: \(P(E_B) = 40\% = \frac{40}{100} = \frac{4}{10}\) Let event D be that an item is defective. Machine A produces 2% defective items: \(P(D | E_A) = 2\% = \frac{2}{100}\) Machine B produces 1% defective items: \(P(D | E_B) = 1\% = \frac{1}{100}\) We want to find the probability that the defective item was produced by machine B, which is \(P(E_B | D)\). Using Bayes' Theorem: \(P(E_B | D) = \frac{P(E_B)P(D | E_B)}{P(E_A)P(D | E_A) + P(E_B)P(D | E_B)}\) \(P(E_B | D) = \frac{\frac{4}{10} \times \frac{1}{100}}{\frac{6}{10} \times \frac{2}{100} + \frac{4}{10} \times \frac{1}{100}}\) \(P(E_B | D) = \frac{\frac{4}{1000}}{\frac{12}{1000} + \frac{4}{1000}}\) \(P(E_B | D) = \frac{\frac{4}{1000}}{\frac{16}{1000}}\) \(P(E_B | D) = \frac{4}{16} = \frac{1}{4}\) The desired probability is \(\frac{1}{4}\).In simple words: A factory uses two machines, A and B, to make things. We know how much each machine produces and how often they make faulty items. If we find a faulty item, we want to know the chance it came from machine B. This chance is 1 out of 4.

🎯 Exam Tip: Clearly identify the contribution of each machine to the total production and their individual defect rates. The ratio of the specific machine's defective output to the total defective output gives the posterior probability.

 

Question 9. એક નિગમમાં નિયામકોની સમિતિમાં હોદો મેળવવા માટે બે સમૂહો હરીફાઇ કરી રહ્યા છે. પ્રથમ અને દ્વિતીય સમૂહો જીતી તેની સંભાવનાઓ અનુક્રમે 0.6 અને 0.4 છે. વધુમાં, જો પ્રથમ સમૂહ જીતશે તો નવી ઉત્પાદિત વસ્તુ રજૂ કરવાની સંભાવના 0.7 છે અને દ્વિતીય સમૂહ માટે અનુરૂપ સંભાવના 0.3 છે. નવી ઉત્પાદિત વસ્તુ દ્વિતીય સમૂહ દ્વારા રજૂ થઈ હોય તેની સંભાવના કેટલી ?

Answer:Let event \(G_1\) be that the first group wins the competition. Let event \(G_2\) be that the second group wins the competition. From the problem statement: \(P(G_1) = 0.6\) \(P(G_2) = 0.4\) Let event P be that a new product is introduced. If the first group wins, the probability of introducing a new product is: \(P(P | G_1) = 0.7\) If the second group wins, the probability of introducing a new product is: \(P(P | G_2) = 0.3\) We want to find the probability that the second group introduced the new product, given that a new product was introduced, which is \(P(G_2 | P)\). Using Bayes' Theorem: \(P(G_2 | P) = \frac{P(G_2)P(P | G_2)}{P(G_1)P(P | G_1) + P(G_2)P(P | G_2)}\) \(P(G_2 | P) = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3}\) \(P(G_2 | P) = \frac{0.12}{0.42 + 0.12}\) \(P(G_2 | P) = \frac{0.12}{0.54}\) \(P(G_2 | P) = \frac{12}{54}\) \(P(G_2 | P) = \frac{2}{9}\) The desired probability is \(\frac{2}{9}\).In simple words: Two groups are trying to win a position, and we know their chances of winning. If a new product is launched, we want to know the chance that it was launched because the second group won. This chance is 2 out of 9.

🎯 Exam Tip: It is important to correctly identify the probabilities of each group winning and the conditional probabilities of product introduction under each group's victory. The Bayes' Theorem calculation then follows directly.

 

Question 10. ધારો કે એક છોકરી પાસો ઉછાળે છે. જો તેને 5 કે 6 મળે તો, તે સિક્કાને ત્રણ વખત ઉછાળે છે અને છાપની સંખ્યા નોંધે છે. જો તેને 1, 2, 3 અથવા 4 મળે તો તે સિક્કાને એક વખત ઉછાળે છે અને છાપ અથવા કાંટો મળ્યો તે નોંધે છે. જો બરાબર એક છાપ મળી હોય, તો તે પાસા પર 1, 2, 3 અથવા 4 મળ્યા હોવાની સંભાવના કેટલી ?

Answer:Let event \(E_1\) be that the girl gets 5 or 6 on the die. Let event \(E_2\) be that the girl gets 1, 2, 3, or 4 on the die. The total outcomes when rolling a die are {1, 2, 3, 4, 5, 6}. Number of outcomes for \(E_1\) = {5, 6} = 2 Number of outcomes for \(E_2\) = {1, 2, 3, 4} = 4 The probabilities for these events are: \(P(E_1) = \frac{2}{6} = \frac{1}{3}\) \(P(E_2) = \frac{4}{6} = \frac{2}{3}\) Let event A be that exactly one head is obtained. If \(E_1\) occurs (5 or 6 on the die), the coin is tossed three times. The sample space for three tosses is: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (8 outcomes) For exactly one head, the outcomes are {HTT, THT, TTH}. So there are 3 favorable outcomes. \(P(A | E_1) = \frac{3}{8}\) If \(E_2\) occurs (1, 2, 3, or 4 on the die), the coin is tossed once. The sample space is {H, T}. For exactly one head, the outcome is {H}. So there is 1 favorable outcome. \(P(A | E_2) = \frac{1}{2}\) We want to find the probability that the die showed 1, 2, 3, or 4 (event \(E_2\)), given that exactly one head was obtained, which is \(P(E_2 | A)\). Using Bayes' Theorem: \(P(E_2 | A) = \frac{P(E_2)P(A | E_2)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2)}\) \(P(E_2 | A) = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}}\) \(P(E_2 | A) = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}}\) \(P(E_2 | A) = \frac{\frac{1}{3}}{\frac{3}{24} + \frac{8}{24}}\) \(P(E_2 | A) = \frac{\frac{1}{3}}{\frac{11}{24}}\) \(P(E_2 | A) = \frac{1}{3} \times \frac{24}{11}\) \(P(E_2 | A) = \frac{8}{11}\) The desired probability is \(\frac{8}{11}\).In simple words: A girl rolls a die. If she gets 5 or 6, she flips a coin three times. If she gets 1, 2, 3, or 4, she flips it once. We are told she got exactly one head. We want to find the chance that she rolled 1, 2, 3, or 4 on the die. This chance is 8 out of 11.

🎯 Exam Tip: Clearly enumerate the sample spaces for coin tosses under different conditions (one toss vs. three tosses) to correctly determine the conditional probabilities of getting one head.

 

Question 11. એક કારખાનાદાર પાસે ત્રણ યંત્ર ચાલકો A, B અને C છે. પ્રથમ ચાલક A, 1 % ખામીયુક્ત વસ્તુઓનું ઉત્પાદન કરે છે. બીજા બે ચાલકો B અને C અનુક્રમે 5 % અને 7 % ખામીયુક્ત વસ્તુઓનું ઉત્પાદન કરે છે. A કામના નિશ્ચિત સમયનો, 50 % સમય કામ પર રહે છે. B 30 % સમય કામ પર રહે છે અને C 20 % સમય કાર્ય કરે છે. ખામીયુક્ત વસ્તુનું ઉત્પાદન થયું છે. તેનું ઉત્પાદન A દ્વારા થયું હોવાની સંભાવના કેટલી ?

Answer:Let event \(E_1\) be that the work is done by machine operator A. Let event \(E_2\) be that the work is done by machine operator B. Let event \(E_3\) be that the work is done by machine operator C. The time each operator works is given: \(P(E_1) = 50\% = \frac{50}{100} = \frac{5}{10}\) \(P(E_2) = 30\% = \frac{30}{100} = \frac{3}{10}\) \(P(E_3) = 20\% = \frac{20}{100} = \frac{2}{10}\) Let event A be that a produced item is defective. The defect rates for each operator are: \(P(A | E_1) = 1\% = \frac{1}{100}\) \(P(A | E_2) = 5\% = \frac{5}{100}\) \(P(A | E_3) = 7\% = \frac{7}{100}\) We want to find the probability that a defective item was produced by operator A, given that it is defective, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2) + P(E_3)P(A | E_3)}\) \(P(E_1 | A) = \frac{\frac{5}{10} \times \frac{1}{100}}{\frac{5}{10} \times \frac{1}{100} + \frac{3}{10} \times \frac{5}{100} + \frac{2}{10} \times \frac{7}{100}}\) \(P(E_1 | A) = \frac{\frac{5}{1000}}{\frac{5}{1000} + \frac{15}{1000} + \frac{14}{1000}}\) \(P(E_1 | A) = \frac{\frac{5}{1000}}{\frac{34}{1000}}\) \(P(E_1 | A) = \frac{5}{34}\) The desired probability is \(\frac{5}{34}\).In simple words: A factory has three workers, A, B, and C, who work for different amounts of time and have different rates of making faulty products. If a faulty product is found, we want to know the chance it was made by worker A. This chance is 5 out of 34.

🎯 Exam Tip: Accurately calculating the "prior" probabilities based on the percentage of time each operator works, and then combining them with their respective defect rates, is essential for a correct application of Bayes' Theorem.

 

Question 12. 52 પત્તાંની થોકડીમાંથી એક પત્તું ખોવાઇ ગયું છે. બાકી રહેલાં પત્તાંની થોકડીમાંથી બે પત્તાં યાદૈચ્છિક રીતે પસંદ કરવામાં આવ્યાં અને માલૂમ પડ્યું કે તે બંને ચોકઠનાં પત્તાં છે. ખોવાયેલ પત્તું ચોકટનું હોય તેની સંભાવના શોધો.

Answer:Let \(E_1\) be the event that the lost card is a diamond. Let \(E_2\) be the event that the lost card is a heart. Let \(E_3\) be the event that the lost card is a club. Let \(E_4\) be the event that the lost card is a spade. There are 13 cards of each suit in a standard deck of 52 cards. So, the probability of the lost card being from any specific suit is: \(P(E_1) = \frac{13}{52} = \frac{1}{4}\) \(P(E_2) = \frac{13}{52} = \frac{1}{4}\) \(P(E_3) = \frac{13}{52} = \frac{1}{4}\) \(P(E_4) = \frac{13}{52} = \frac{1}{4}\) Let A be the event that two cards drawn from the remaining 51 cards are both diamonds. If \(E_1\) occurs (a diamond is lost), then in the remaining 51 cards, there are 12 diamonds and 39 non-diamonds. The probability of drawing two diamonds from these 51 cards is: \(P(A | E_1) = \frac{\text{Number of ways to choose 2 diamonds from 12}}{\text{Number of ways to choose 2 cards from 51}} = \frac{^{12}C_2}{^{51}C_2}\) If \(E_2\) occurs (a heart is lost), then in the remaining 51 cards, there are 13 diamonds and 38 non-diamonds (12 hearts, 13 clubs, 13 spades). The probability of drawing two diamonds from these 51 cards is: \(P(A | E_2) = \frac{^{13}C_2}{^{51}C_2}\) Similarly, if \(E_3\) occurs (a club is lost), then in the remaining 51 cards, there are 13 diamonds. \(P(A | E_3) = \frac{^{13}C_2}{^{51}C_2}\) And if \(E_4\) occurs (a spade is lost), then in the remaining 51 cards, there are 13 diamonds. \(P(A | E_4) = \frac{^{13}C_2}{^{51}C_2}\) We want to find the probability that the lost card was a diamond, given that two drawn cards are diamonds, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2) + P(E_3)P(A | E_3) + P(E_4)P(A | E_4)}\) \(P(E_1 | A) = \frac{\frac{1}{4} \times \frac{^{12}C_2}{^{51}C_2}}{\frac{1}{4} \times \frac{^{12}C_2}{^{51}C_2} + \frac{1}{4} \times \frac{^{13}C_2}{^{51}C_2} + \frac{1}{4} \times \frac{^{13}C_2}{^{51}C_2} + \frac{1}{4} \times \frac{^{13}C_2}{^{51}C_2}}\) Cancel out \(\frac{1}{4}\) and \(\frac{1}{^{51}C_2}\) from all terms: \(P(E_1 | A) = \frac{^{12}C_2}{^{12}C_2 + ^{13}C_2 + ^{13}C_2 + ^{13}C_2}\) \(P(E_1 | A) = \frac{^{12}C_2}{^{12}C_2 + 3 \times ^{13}C_2}\) Now, calculate the combinations: \(^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66\) \(^{13}C_2 = \frac{13 \times 12}{2 \times 1} = 78\) Substitute these values: \(P(E_1 | A) = \frac{66}{66 + 3 \times 78}\) \(P(E_1 | A) = \frac{66}{66 + 234}\) \(P(E_1 | A) = \frac{66}{300}\) Divide by 6: \(P(E_1 | A) = \frac{11}{50}\) The desired probability is \(\frac{11}{50}\).In simple words: A card is lost from a deck. We then pick two cards from the remaining ones and find that both are diamonds. We want to know the chance that the *lost* card was also a diamond. This chance is 11 out of 50.

🎯 Exam Tip: When a card is lost, the composition of the remaining deck changes depending on which card was lost. Accurately calculate conditional probabilities (using combinations) for each scenario of the lost card, then apply Bayes' theorem.

 

પ્રશ્નો 13 તથા 14માં વિધાન સાચું બને તે રીતે આપેલ વિકલ્પોમાંથી યોગ્ય વિકલ્પ પસંદ કરો :

 

Question 13. A સત્ય બોલે છે તેની સંભાવના \(\frac{4}{5}\) છે. એક સિક્કો ઉછાળ્યો છે. A માહિતી આપે છે કે છાપ મળી છે. ખરેખર છાપ હતી તેની સંભાવના ............... હોય.

(A) \(\frac{4}{5}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{5}\)
Answer: (A) \(\frac{4}{5}\)Let \(E_1\) be the event that a head appears on the coin. Let \(E_2\) be the event that a tail appears on the coin. Since it's a fair coin: \(P(E_1) = \frac{1}{2}\) \(P(E_2) = \frac{1}{2}\) Let A be the event that person A states that a head appeared. The probability that A speaks the truth is \(\frac{4}{5}\). If a head actually appeared (\(E_1\)), and A says it's a head, then A is speaking the truth. \(P(A | E_1) = \frac{4}{5}\) If a tail actually appeared (\(E_2\)), and A says it's a head, then A is not speaking the truth. The probability of A not speaking the truth is \(1 - \frac{4}{5} = \frac{1}{5}\). \(P(A | E_2) = \frac{1}{5}\) We want to find the probability that there was actually a head, given that A stated it was a head, which is \(P(E_1 | A)\). Using Bayes' Theorem: \(P(E_1 | A) = \frac{P(E_1)P(A | E_1)}{P(E_1)P(A | E_1) + P(E_2)P(A | E_2)}\) \(P(E_1 | A) = \frac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2} \times \frac{4}{5} + \frac{1}{2} \times \frac{1}{5}}\) \(P(E_1 | A) = \frac{\frac{4}{10}}{\frac{4}{10} + \frac{1}{10}}\) \(P(E_1 | A) = \frac{\frac{4}{10}}{\frac{5}{10}}\) \(P(E_1 | A) = \frac{4}{5}\) The desired probability is \(\frac{4}{5}\).In simple words: A person, who usually tells the truth, says a coin flip resulted in heads. We want to find the actual chance that it was indeed heads. The chance is 4 out of 5.

🎯 Exam Tip: Correctly identifying the conditional probabilities P(A|E1) (truthful statement) and P(A|E2) (false statement) is the key to solving this problem using Bayes' Theorem.

 

Question 14. જો P(B) ≠ 0 અને A C B હોય તેવી બે ઘટનાઓ A અને B માટે નીચેનામાંથી કયું સત્ય છે ?

(A) P(A | B) = \(\frac{P(B)}{P(A)}\)
(B) P(A | B) < P(A)
(C) P(A | B) ≥ P(A)
(D) આમાંથી એક પણ હિ
Answer: (C) P(A | B) ≥ P(A)We are given that \(P(B) \neq 0\) and event A is a subset of event B (A C B). This means that if event A occurs, event B must also occur. When A C B, it implies that the intersection of A and B is A itself, i.e., \(A \cap B = A\). The formula for conditional probability \(P(A | B)\) is: \(P(A | B) = \frac{P(A \cap B)}{P(B)}\) Since \(A \cap B = A\), we can substitute this into the formula: \(P(A | B) = \frac{P(A)}{P(B)}\) Now, let's analyze the relationship between \(P(A | B)\) and \(P(A)\). Since B is an event and \(P(B) \neq 0\), we know that \(0 < P(B) \leq 1\). If \(0 < P(B) \leq 1\), then \(\frac{1}{P(B)} \geq 1\). Therefore, \(P(A | B) = P(A) \times \frac{1}{P(B)}\). Since \(\frac{1}{P(B)} \geq 1\), it means \(P(A | B) \geq P(A)\). The correct option is (C) \(P(A | B) \geq P(A)\).In simple words: If event A is part of event B, and B has a chance of happening, then the chance of A happening *given that B has already happened* is either equal to or greater than the chance of A happening alone. This is because if A happens, B must happen too, making A relatively more likely if we know B occurred.

🎯 Exam Tip: Understand the implications of "A is a subset of B" for probabilities: \(P(A \cap B) = P(A)\) and \(P(A) \leq P(B)\). This foundational understanding is crucial for correctly interpreting conditional probabilities in such scenarios.

Question 14. જો \( P(B) \neq 0 \) અને \( A \subset B \) હોય તેવી બે ઘટનાઓ A અને B માટે નીચેનામાંથી કર્યું સત્ય છે ?
(A) \( P(A | B) = \frac{P(B)}{P(A)} \)
(B) \( P(A | B) < P(A) \)
(C) \( P(A | B) \ge P(A) \)
(D) આમાંથી એક પણ હિ
Answer: (B) P(A | B) < P(A)
Answer: ઘટના B એ ઘટના A ને સમાવી લે છે અને તેની સંભાવના શૂન્ય નથી. આથી, જો A એ B નો યોગ્ય ઉપગણ હોય, તો A ની ઘટનાની સંભાવના B ની ઘટનાની સંભાવના કરતા ઓછી હશે. શરતી સંભાવનાના સૂત્ર મુજબ, \( P(A | B) = \frac{P(A \cap B)}{P(B)} \). કારણ કે \( A \subset B \), \( A \cap B = A \) થાય છે.
તેથી, \( P(A | B) = \frac{P(A)}{P(B)} \). જ્યારે \( A \) એ \( B \) નો યોગ્ય ઉપગણ હોય, ત્યારે \( P(A) < P(B) \) થાય છે, તેથી \( \frac{P(A)}{P(B)} < P(A) \). આમ, વિકલ્પ (B) સાચું છે કે \( P(A | B) < P(A) \).
In simple words: When event A is a smaller part of event B, and B has a probability greater than zero, the chance of A happening when B has already happened is found by dividing the probability of A by the probability of B. Because B is a larger event than A (or at least equal), this fraction is usually less than the probability of A alone.

🎯 Exam Tip: Understanding the relationship between subsets in probability and the conditional probability formula is key to solving such problems accurately.