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:
107 comments Page 3 of 11.
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
Tulasi 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.
Asiya said:
7 years ago
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.
Anyone please help me to get it.
Sundar said:
2 decades ago
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.
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.
(1)
Gad said:
9 years ago
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.
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.
Shivu said:
7 years ago
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.
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.
(6)
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).
Goutam said:
1 decade ago
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.
Ravi said:
1 decade ago
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.
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.
Sri said:
7 years ago
@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.
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.
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