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

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. |

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