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

Answer: Option D

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.

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