# Aptitude - Probability - Discussion

### Discussion :: Probability - General Questions (Q.No.1)

1.

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

[A].
 1 2
[B].
 2 5
[C].
 8 15
[D].
 9 20

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

 P(E) = n(E) = 9 . n(S) 20

 Sundar said: (Jun 18, 2010) Here, S = {1, 2, 3, 4, ...., 19, 20}. Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}. P(E) = 1/2. Is this correct? Please give your thoughts on this.

 Karthik said: (Jul 22, 2010) No. You are wrong Sundar. Because, 15 has already consider once so there is no need to consider it again. So the answer is 9/20.

 Veeru said: (Sep 26, 2010) Sundar & Manju thought wrong because 15 already be consider. So ans will be 9/20.

 Sneha said: (Sep 27, 2010) Yes, answer is 9/20. We should not consider a number more than one time (15 is a multiple of 3 and 5. So, it cannot be used more than once).

 Praveen Kumar Gautam said: (Oct 1, 2010) Answer will be 9/20, because 15 already be considered, so it can not be consider next time. So Sundar & Manju thought wrong.

 Souravbaidya said: (Nov 26, 2010) A circular dartboard of radius 1 foot is at a distance of 20 feet from you. You throw a dart at it and it hits the dartboard at some point Q in the circle. What is the probability that Q is closer to the center of the circle than the periphery? a) 0.75 b) 1 c) 0.5 d) 0.25

 Pawan said: (Dec 2, 2010) Hi Guys .. according to me the answer is 2/5 A = multiple of 3 B = multiple of 5 P(A U B) = P(A) + P(B) - P(A n B) n(A) = 6 ( 3,6,9,12,15,18) n(B) = 3 (5,10,15) n (A n B) = 1 (15 --> this comes in both) Hence, prob = 6/20 + 3/20 - 1/20 = 8/20 = 2/5

 Murugesh said: (Jan 1, 2011) Hi @pawan, according to your assumption, you had missed the 20 in n(B) so n(B)={5,10,15,20} then P(B)=4/20 Hence P(AUB)=6/20+4/20-1/20=9/20.

 Dipesh said: (Jan 26, 2011) Yes, answer is 9/20. We should not consider a number more than one time (15 is a multiple of 3 and 5. So, it cannot be used more than once)

 Satyendra Shukla said: (Feb 7, 2011) Answer will be 9/20, because 15 already be considered, so it can not be consider next time.

 Dharmesh Patel said: (Feb 16, 2011) Yes answer is Only and only 9/20.

 Sugnya said: (Feb 18, 2011) The answer is 1/2 because if there is same number we should include it. This not union or intersection model.

 Albert said: (Mar 26, 2011) Final answer is 9/20

 Kiruba said: (May 15, 2011) Thanks sneha. Ans is 9/20.

 Rach said: (May 26, 2011) Circular board of i feet has area Pi. The area of the region containing all points which are closed to center than to periphery will be Pi/4. Hence the proportion to hit near the center is 1/4=.25. So D is the correct answer.

 Rahul said: (May 31, 2011) @rach. Hai rach I didn't get you. Can you please explain clearly.

 Sreejith said: (Jun 17, 2011) Hello Rach, How did you get that as pi/4? Or anybody else knows?

 Pinki said: (Jun 24, 2011) How we calculate n(E) and n(S). ?

 Geeta Chauhan said: (Jul 26, 2011) I cant understand the ans please explain me.

 Rammu said: (Aug 19, 2011) n(E) = number of possible events n(S) = total number of samples Eg. Tickets which is numbered 1 to 20 is n(S) The possible tickets drawn which has number multiple of 3 and 5 is n(E).

 Arun said: (Aug 23, 2011) I can't understand. P (e) ={3, 6, 9, 12, 15, 18, 5, 10, 20}. We multiplying p (e) with s then we don't want to multiply s={7, 8 to 18}.

 Aruna said: (Sep 20, 2011) Hi murugesh what you explained is clear. Thank you.

 Sohail said: (Sep 21, 2011) 15 creates a big confusion. Because my answer is 1/2...:( p(3)={3,6,9,12,15,18} p(5)={5,10,15,20} P(3u5)=p(3)+p(5) =6/20+4/20 =6+4/20 =10/20 =1/2

 Aneel said: (Oct 3, 2011) Thank you murugesh for clarifying the doubt.

 Saurav said: (Oct 12, 2011) Please give me logical solution of the following problem. A lady has fine gloves and hats in her closet- 18 blue, 32 red, and 25 yellow. The lights are out and it is totally dark. In spite of the darkness, she can make out the difference between a hat and a glove. She takes out an item out of the closet only if she is sure that if it is a glove. How many gloves must she take out to make sure she has a pair of each color? a) 50 b) 8 c) 60 d) 42 Thanks.

 Subrata said: (Nov 10, 2011) Hi The answer is 9/20 A={3,6,9,12,15,18} B={5,10,15,20} P(A)=6/20 P(B)=4/20 A intersection B = {15} P(A intersection B) = 1/20 P(A U B)= P(A)+P(B)-P(A intersection B) = 6/20 + 4/20 - 1/20 = 9/20

 Shekar said: (Feb 21, 2012) A(n)={3,6,9,12,15,18} B(n)={5,10,15,20} P(AuB)=P(A)+P(B)-P(A intersection B) P(AuB)=6/20+4/20-1/20 P(AuB)=9/20.

 Parthasarathy said: (Feb 26, 2012) A(n)={3,6,9,12,15,18} B(n)={5,10,15,20} P(AuB)=P(A)+P(B)-P(A intersection B) P(AuB)=6/20+4/20-1/20 P(AuB)=9/20.

 Naana said: (Apr 23, 2012) Please how do you come about the 1/2 (ie A intersection B) ? I am a bit confused. Thanks.

 Bangaru Babu said: (Sep 24, 2012) @Naana. I will solve your problem. ->In set A there are 6 sample points and in set of B there are 4 sample points ->Now we want the intersection of two sets ->So we select the common sample points in both sets i.e 15 ->Now in the set (A intersection B) we have only one sample point ->For finding probability of A intersection B i.e p(A intersection B) * So according to the definition of probability. P(A intersection B)== n(A intersection B)/no of events; So the result is P(A intersection B)=1/20.

 Sifuna Charles said: (Oct 10, 2012) Shekar your explanation is good but you should have included the intersection part of it for easy understanding.

 Anil Chauhan said: (Feb 9, 2013) A(n)={3,6,9,12,15,18} B(n)={5,10,15,20} P(AuB)=P(A)+P(B)-P(A intersection B) P(AuB)=6/20+4/20-1/20 P(AuB)=9/20.

 Tulasi said: (Mar 8, 2013) Hi, The answer is 9/20. A={3,6,9,12,15,18}. B={5,10,15,20}. P(A)=6/20. P(B)=4/20. A intersection B = {15}. P(A intersection B) = 1/20. P(A U B)= P(A)+P(B)-P(A intersection B). = 6/20 + 4/20 - 1/20 = 9/20.

 Ketan said: (May 30, 2013) p(3) = {3,6,9,12,15,18}. p(5) = {5,10,15,20}. P(3u5) = p(3)+p(5). I don't understand how it's come = 6/20+4/20 ?

 Chandan said: (Jun 3, 2013) As per Addition Rule of Probability, we can have two set of events - Mutually Exclusive & Non-Mutually Exclusive..where Mutually Exclusive events don't occur simultaneously or together and reverse logic with Non-Mutually Exclusive events. Having this theorem told in above para, now i come to the context: We have 2 events, Event A: Ticket picked is multiple of 3. Event B: Ticket picked is multiple of 5. Now P(A) = {3,6,9,12,15,18} = 6/20. And P(B) = {5,10,15,20} = 4/20. Since in P(A) & P(B) we could see 15 to be common, we could conclude that the events are Mutually Exclusive..so for such Mutually Exclusive events the Addition Rule of Probability goes like this: P(AUB) = P(A) + P(B) - P(A n B) Where, P(A n B) always looks for what is common in two events, which is 15 in this case, so P(A n B) = 1/20. Therefore, P(ticket picked is multiple of 3 or 5) = 6/20 + 4/20 - 1/20 = 9/20 is the answer.

 Rajini said: (Jun 29, 2013) Please anyone help me for solving this problem: Condition 1: Whenever a white ball + black ball is taken out a black ball is placed in. Condition 2: Whenever 2 black balls are taken out one white ball is placed in. Condition 3: Whenever 2 white balls are taken out one white ball is placed inside. Suppose user gives this input: Black balls: 5. White balls: 4. Output should be any of these: White ball/black ball/undetermined.

 Ravi said: (Jul 2, 2013) ANS IS 9/20. PROBABILITY = NUMBER OF FAVORABLE EVENTS/NUMBER OF TOTAL EVENTS. Events of greeting multiple 3 of 5= 3, 6, 9, 12, 15, 18, 5, 10, 20. P(E) = P(F)/P(E). P(E) = 9/20.

 Adhamki Dileep Kumar said: (Sep 26, 2013) Sample space p(s) = 20c1 = 20. Events occurred p(e) = (3,6,9,12,15,18,5,10,15,20) = 10. = (3,6,9,12,15,18,5,10,20)=9 Probability = p(e)/p(s). = 9/20.

 Sharada said: (Nov 13, 2013) 15 also should come in the list right?

 Niha said: (Dec 2, 2013) Yes, 9/20 is the correct answer.

 Goutam said: (Dec 28, 2013) Yes answer is 9/20 because if we consider two event A & B which are multiple of 3 and 5 respectively then the required event is P(AuB)= P(A)+ P(B)-p(AB) = 6/20+4/20-1/20 = 9/20.

 Rupali said: (Mar 13, 2014) Hi all. Can anyone explain how 1/20 come?

 Abhisek Mukherjee said: (Apr 1, 2014) Use just simple probability formula, P(A+B) = P(A)+P(B)-P(AB).

 Saiful Islam said: (May 23, 2014) Actually, I don't understand your explanation boss. Please explain elaborately. How can it will be 9/20?

 Nikhil Merwade said: (Jun 19, 2014) Why can't we use the formula of nCr...??... 1 ticket can be selected from 20 tickets in 20C1 ways. Ticket removed is a multiple of 3 or 5 will be 20C6 * 20C4 Answer will be (20C6 * 20C4)/20C1... Why can't this be proper method?

 Aarti Gosavi said: (Aug 13, 2014) There are three drawers in a table. One drawer contains two gold coins another drawer contains two silver coins and third drawer contains one silver and one gold coin. One of the drawers is pulled out and a coin is taken out. It turns out to be a silver. What is probability of drawing a gold coin if one of the other two drawers is pulled out next and one of the coin in it is drawn at random?

 Santosh said: (Sep 2, 2014) Yah by using P(AUB)formula we can calculate this, So no doubt persists-P(A)=6/20, P(B)=4/20,and P(A^B)=1/20(since 15 common for 3,5), Finally answer is 6/20+4/20-1/20 = 9/20.

 Keerthi said: (Sep 10, 2014) Here he said 3 or 5 but not 3&5 so answer will be 1\2. Since or means we should add &and means multiply.

 Mosca said: (Nov 6, 2014) Everyone is wrong. The answer is D: 9/20. I am the best at maths see. This is because it is out of 20 and the multiplier of 3 and 5 is 9. You see? So, it has to be 9/20.

 Shakthi said: (May 31, 2015) Where this 5 gone in the event?

 Sagar Valecha Soni said: (Jul 29, 2015) Ya 9/20 is correct I am agree from this answer.

 Garima said: (Sep 13, 2015) According to me the answer should be 1/2. Multiples of 3 and 5. (1, 3, 5, 6, 9, 10, 12, 15, 18, 20). Please let me know where am I wrong?

 Divya said: (Sep 22, 2015) How to cal multiple of larger number? Can any one give a answer for this? Ex: Same question number is 1-300.

 Samuel said: (Jan 20, 2016) @Divya. Factorize or divide it.

 Rutu said: (Mar 18, 2016) Why s=20 selected? please give me answer.

 Mahesh said: (Mar 19, 2016) What's, and what is e please explain?

 Shailendra Rathore said: (Apr 19, 2016) 9/20 is the right answer, and thanks for explaining the answer in different ways.

 Saurav said: (Jul 3, 2016) As both the events A and B are independent and non-mutually exclusive we can use the following formula:. P (A or B) = P (A) + P (B) - P (A and B). = 6/20 + 4/20 - (6/20 * 4/20) [probability intersection rule for dependent events]. = 10/20 - 3/50. = 50 - 6/100. = 44/100 = 11/25. But this answer is not in the option.

 Rani said: (Jul 14, 2016) Where this 5 gone in the event?

 Mani said: (Aug 3, 2016) Where this 6 gone in the event?

 Nicole said: (Aug 3, 2016) I'm confused, on how thy got D as an answer?

 Gad said: (Aug 6, 2016) A = multiple of 3 (3, 6, 9, 12, 15, 18). Possible outcome = 6. B = multiple of 5 (5, 10, 15, 20). Possible outcome = 4. Therefore, P (A or B) = P (A) + P (B). 6/20 + 4/20 = 10/20 = 1/2.

 Dabir Masood said: (Sep 1, 2016) Yes I too agree that 1/2 is the correct answer for this question.

 Anonymous said: (Sep 18, 2016) The correct answer is 9/20.

 Deva Bala said: (Sep 26, 2016) The option d) 9/20 is the correct answer.

 Prawesh Pradhan said: (Sep 27, 2016) P(A) + P(B) - P(A and B). 6 + 4 - 1 = 9. P(E) = n(E)/n(S). P(E) = 9/20.

 Prawesh Pradhan said: (Sep 27, 2016) From 1 to 20, numbers which are divisible by 3 are 3, 6, 9, 12, 15 & 18 -> total 6. Numbers divisible by 5 are 5, 10, 15, 20 total 4. So, 6 + 4 = 10. But 15 is counted twice. Therefore 10 - 1= 9. Result is 9/20.

 Kavana said: (Oct 18, 2016) From 1 to 340, the probability of getting multiples of 8 and 5 is? Please answer this question.

 Ujjwal Shukla said: (Oct 18, 2016) Find the probability that a number from 1 to 300 is divisible by 3 or 7. Can anyone answer this question.

 Ravi said: (Dec 13, 2016) According to me, From 1 to 20, numbers its 1, 2, 3, 4....., 19, 20 then number which is a multiple of 3 or 5. Lets take 3 (3, 6, 9, 12, 15, 18 and 5 = (5, 10, 15, 20). So in 3 we have 6 and 5 we have 4 so it's 6/20 and 4/20 and 1/20. Solution = 6 + 4 - 1/20 == 9/20 answer.

 Ravi said: (Dec 13, 2016) According to me, From 1 to 20, numbers its 1, 2, 3, 4....., 19, 20 then number which is a multiple of 3 or 5. Lets take 3 (3, 6, 9, 12, 15, 18 and 5 = (5, 10, 15, 20). So in 3 we have 6 and 5 we have 4 so it's 6/20 and 4/20 and 1/20. Solution = 6 + 4 - 1/20 == 9/20 answer.

 Arjun Patidar said: (Feb 7, 2017) Ans is 9/20 because, P(A) + P(B) - P(A andB) = P(A OR B ) For p (a) outcome are divide to 20 for getting multiple of 3 we get in rational left after point so 6. Then also of 5 we get 4 numbers can multiple of five. After it puts in probability case such as 6/20 + 4/20 - 1/20. 1/20 comes to get no. of both that is only one number 15 and also use short cut as take the LCM of 3&5 we get 15, divide 20 by it to get some thing 1. Left fraction we get 1/ 20 or 1 out of 20 put all in formula. 6/20 + 4/20 - 1/20 = 9/20 answer.

 Isha said: (Mar 2, 2017) Please help me to solve this- A number is selected at random from first 50 number. Find the probability that it is a multiple of 3 and 4.

 Vikram Reddy said: (May 3, 2017) No, here 15 is repeated in 3&5 so the answer is 2/5.

 Nithya Arun said: (Jun 17, 2017) How can we calculate the same problem with big numbers? i.e if tickets numbered 1-200.

 Shashi said: (Jul 19, 2017) Since the number of events are 20. So possible events are {3,6,9,12,15,18,5,10,20}. So answer is 9/20.

 Naina said: (Jul 27, 2017) A coin is tossed live times. What is the probability that there is at the least one tail? Can anyone help me with this question?

 Thrideep said: (Aug 22, 2017) I think The answer is 3/4.

 Goutam said: (Nov 3, 2017) 9/20 is correct answer. 5 multiplies are 5,10,15,20. 3 multiplies are 3,9,12,15,18. Answer = 9/20.

 Anonymous said: (Mar 20, 2018) Why didnt we apply the (not A) formula? i.e. (Not blue ball) = 1 - (blue ball). = 1 - 2/7. = 5/7.

 Azoya said: (Jun 4, 2018) Multiple of 3 = {3,6,9,12,15,18}, Multiple of 5 ={5,10,15}, Multiple of both 3 and 5 = {15}, So p(3) = 6/20, p(5) = 3/20, p(3 and 5) = 1/20. So, answer is = 6/20 + 3/20 -1/20. i,e. 8/20 which is 2/5 according to options. I am getting this answer because the question is multiple of 3 or 5 . So we need to exclude multiple of 3 and 5.

 Deepika Balaji said: (Jun 8, 2018) Can anyone help me in solving this problem. What is the probability of getting at least one six in a single throw of three unbiased dice?

 Asiya said: (Jul 28, 2018) My answer is 8/20 because in question it is stated that we take multiple of 3 or 5 not multiple of both so we should neglect 15 because it is the multiple of 3 and 5. Anyone please help me to get it.

 Siva said: (Aug 8, 2018) How come 4/20 & 6/20?

 V S Abhirame said: (Aug 17, 2018) @Souravbaidya. Circular board of i feet has area Pi. The area of the region containing all points which are closed to center than to periphery will be Pi/4. Hence the proportion to hit near the center is 1/4=.25. So, D is the correct answer.

 Sagaana said: (Aug 18, 2018) How to solve the same problem when the 1000 tickets mixed in the pack what is the probability of getting multiples of 3 and 5?

 Sri said: (Aug 21, 2018) @All. Hi, Why don't we get 15? It is a multiple of 5. So P(E)={3,6,9,12,15,18,5,10,15,20}. So these are the multiples of 3 and 5. Total is 10. So P(E)/P(S) = 10/20 .That is 1/2.

 Tushar said: (Sep 8, 2018) Among the 20 numbers multiples of 3=20/3 which is 6. And similarly, multiples of 5 are=20/5 which are 4. Now total multiples are 10. Now among these 10 multiples, we have one common multiple that is 15. Now remaining multiples=total multiples -common multiple. =10-1. =9. Now required prob =9/20.

 Ramesh said: (Oct 31, 2018) Multiples of 3 = 3,6,9,12,15,18 ----> event A. Multiples of 5 = 5,10,15,20 ------> event B. Both are having 15 as common so these are dependent events. P(A)= 6/20. P(B)= 4/20. P(AUB)= P(A)+P(B)-P(A intersection B). = P(A)+P(B)-P(A)*P(B/A), = 6/20 + 4/20 - {(6/20)*(1/6)}. = 9/20.

 Lalitha said: (Nov 14, 2018) If the ticket numbers are taken from 1 2 3 ..... 100 what's the probability of getting the probability of card which multiple by 3 or 5. Can anyone solve this?

 Shraddha said: (Dec 8, 2018) But why is the number 5 not considered? 5 is a multiple of 5!

 Anitha said: (Jan 3, 2019) @Shraddha. 5 is also considered so that we get probability 9/20. If 5 is not considered we can't get 9/20.

 Shivu said: (Feb 24, 2019) Ans is 9/20 because; Multiple of 3 (3 6 9 12 15 18)=possiblity 6 so 6/20, Multiple of 5 (5 10 15 20)=possibility 4 so 4/20, 15 is both sides so eliminate, so -1/20. 6/20+4/20-1/20 = 9/20.

 Uday said: (Sep 18, 2019) (3, 6,9,12,15,18,5,10,15,20). = 10/20, = 1/2.

 Kousalya said: (Jan 23, 2020) Yes, I too agree 1/2 is the right answer.

 Kavya said: (Feb 21, 2020) @Uday, 15 is repeated. And 5 is missing. Anyway, the answer is correct.

 Avaneet Agrahari said: (Jun 27, 2020) Multiple of 3=[,3,6,912,15,18]=6 M....o. 5=[5,10,15,20]=4 but 15 is already in 3 so it count 3 Ans= 6+3/20=9/20