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}.
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n(E) | = | 9 | . |
n(S) | 20 |
Discussion:
107 comments Page 9 of 11.
Shekar said:
1 decade ago
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.
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.
Subrata said:
1 decade ago
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
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
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.
Aneel said:
1 decade ago
Thank you murugesh for clarifying the doubt.
Sohail said:
1 decade ago
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
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
Aruna said:
1 decade ago
Hi murugesh what you explained is clear. Thank you.
Arun said:
1 decade ago
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}.
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}.
Rammu said:
1 decade ago
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).
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).
Geeta chauhan said:
1 decade ago
I cant understand the ans please explain me.
Pinki said:
1 decade ago
How we calculate n(E) and n(S). ?
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