# Aptitude - Probability - Discussion

Discussion Forum : 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?

Answer: Option

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 |

Discussion:

106 comments Page 1 of 11.
Chandan said:
1 decade ago

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.

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.

(1)

Arjun patidar said:
8 years ago

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.

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.

Bangaru babu said:
1 decade ago

@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.

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.

Saurav said:
1 decade ago

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.

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.

Rajini said:
1 decade ago

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.

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.

Aarti Gosavi said:
1 decade ago

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?

Rajesh Bhadra said:
3 months ago

@Ganapathithanuja.

From 1 to 20 there divisible by 3 is -> 3, 6, 9, 12, 15, 18 and divisible by 5 is -> 5, 10, 15, 20.

Now if you see total numbers which are divisible by 3 or 5 are 10 but if you look in both then you will see 15 is common in both cases so total numbers which are divisible 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20 mean 9.

Hence the answer is 9/20.

From 1 to 20 there divisible by 3 is -> 3, 6, 9, 12, 15, 18 and divisible by 5 is -> 5, 10, 15, 20.

Now if you see total numbers which are divisible by 3 or 5 are 10 but if you look in both then you will see 15 is common in both cases so total numbers which are divisible 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20 mean 9.

Hence the answer is 9/20.

(7)

Azoya said:
6 years ago

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.

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.

Ramesh said:
6 years ago

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.

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.

Saurav said:
8 years ago

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.

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.

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