Let A be the vent that a randomly selected two digit number : Problem Solving (PS)

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Let A be the vent that a randomly selected two digit number

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Intern

Joined: 06 Aug 2015

Posts: 2

Let A be the vent that a randomly selected two digit number [#permalink]

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Updated on: 16 Aug 2015, 11:34

00:00

Difficulty:

25% (medium)

Question Stats:

80% (02:10) correct

20% (02:56) wrong

based on 114 sessions

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Let A be the event that a randomly selected two digit number is divisible by 3 and let B be the event that a randomly selected two digit number is divisible by 5. What is P(A and B)?

A. 1/18
B. 1/15
C. 1/5
D. 1/3
E. 1/2

Spoiler: OA

Originally posted by tonytran14896 on 06 Aug 2015, 11:58.
Last edited by Bunuel on 16 Aug 2015, 11:34, edited 2 times in total.

Edited the question and moved to PS forum.

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Re: Let A be the vent that a randomly selected two digit number [#permalink]

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06 Aug 2015, 12:22

tonytran14896 wrote:

Let A be the event that a randomly selected two digit number is divisible by 3 and let B be the event that a randomly selected two digit number is divisible by 5. What is P(A and B)?
a. 1/18
b.1/15
c.1/5
d.1/3
e.1/2

I know the answer is b:1/15 but i don't know how to explain it. Please clarify for me. Thanks.

Please format the question properly and mention anything related to the solution either in another post or put it under "spoiler" as shown above.

For your question,

For 2 independent events ,

P(A and B) = P(A)*P(B)

where
P(A) = selecting a multiple of 3 in all the 2 digit numbers = Favorable cases/ Total cases = 30/90 = 1/3
P(B) = selecting a multiple of 5 in all the 2 digit numbers = Favorable cases/ Total cases = 18/90 = 1/5

Thus, P(A and B) = P(A)*P(B) = (1/3)(1/5) = 1/15 . B is the correct answer.

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Re: Let A be the vent that a randomly selected two digit number [#permalink]

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10 Aug 2015, 11:09

tonytran14896 wrote:

Spoiler: ::

b:1/15

but i don't know how to explain it. Please clarify for me. Thanks.

We know that

P(A and B) = P(A) * P(B)

Event where a two digit no is divisible by 3 => 30 (12 to 99)
Total two digit no.s = 90
Hence P(A) = 30/90 => 1/3

Similarly, Event where a two digit no is divisible by 5 => 18 (10 to 95)
P(B) = 18/90 => 1/5

P(A and B) = 1/3 * 1/5 => 1/15

Option B

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Re: Let A be the vent that a randomly selected two digit number [#permalink]

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30 Nov 2015, 18:37

Given that P(A) be the probability that a number divisible by 3 and P(B) be the probability that a number is divisible by 5

=> P ( A and B ) is the number divisible by both 3 and 5

the Numbers are 15, 30, 45, 60, 75, 90 => 6

Total No of Events = Count (10 to 99) = 90

Therefore $P(A and B) = \frac{Favorable Events}{Total Events}$$= \frac{6}{90} = \frac{1}{15}$

Ans: B

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Re: Let A be the vent that a randomly selected two digit number [#permalink]

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30 Nov 2015, 23:36

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

Let A be the event that a randomly selected two digit number is divisible by 3 and let B be the event that a randomly selected two digit number is divisible by 5. What is P(A and B)?

A. 1/18
B. 1/15
C. 1/5
D. 1/3
E. 1/2

The event A and B means that a randomly selected two digit number is divisible by both 3 and 5. Since the least common multiple of 3 and 5 is 15, that event is same as a randomly selected two digit number is divisible by 15.

The total number of two digit number is 90(from 10 to 99 inclusively). The number of the multiples of 15 from 10 to 99 is [100/15]=6.

So P(A and B) = 6/90 = 3/45 = 1/15. So the answer is B.
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Re: Let A be the vent that a randomly selected two digit number [#permalink]

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27 Apr 2016, 19:28

can't believe I missed this one because of a silly mistake...
we have 90 numbers.
we need positive outcomes when the numbers are divisible by 3 and 5.
we have: 15, 30, 45, 60, 75, 90 - 6 numbers.
6/90 = 1/15

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Re: Let A be the vent that a randomly selected two digit number [#permalink]

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17 Jul 2018, 02:50

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