# 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:

101 comments Page 1 of 11.
Sundar said:
1 decade 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.

Karthik said:
1 decade ago

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.

Because, 15 has already consider once so there is no need to consider it again.

So the answer is 9/20.

Veeru said:
1 decade ago

Sundar & Manju thought wrong because 15 already be consider.

So ans will be 9/20.

So ans will be 9/20.

Sneha said:
1 decade ago

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:
1 decade ago

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:
1 decade ago

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

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:
1 decade ago

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

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:
1 decade ago

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.

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:
1 decade ago

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:
1 decade ago

Answer will be 9/20, because 15 already be considered, so it can not be consider next time.

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