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There are these two sets of letters, and you are going to pick exactly

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There are these two sets of letters, and you are going to pick exactly  [#permalink]

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New post 04 Mar 2015, 04:25
1
6
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

75% (01:29) correct 25% (01:37) wrong based on 183 sessions

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Set #1 = {A, B, C, D, E}
Set #2 = {K, L, M, N, O, P}

There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?

A. 1/6
B. 1/3
C. 1/2
D. 2/3
E. 5/6


Kudos for a correct solution.

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Re: There are these two sets of letters, and you are going to pick exactly  [#permalink]

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New post 04 Mar 2015, 04:49
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1
Bunuel wrote:
Set #1 = {A, B, C, D, E}
Set #2 = {K, L, M, N, O, P}

There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?

A. 1/6
B. 1/3
C. 1/2
D. 2/3
E. 5/6


Kudos for a correct solution.



I believe it will be easy to calculate by using the trick for at least scenario: Opposite of at least one is none.


So Not a vowel in Set-1 : \(\frac{3}{5}\)

And not a vowel in Ser-2:\(\frac{5}{6}\)

Now,

\(\frac{3}{5} * \frac{5}{6}= \frac{1}{2}\)


This is for not a vowel.

Then for atleast one vowel will be = \(1- \frac{1}{2}=\frac{1}{2}\)


Answer will be C.
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Re: There are these two sets of letters, and you are going to pick exactly  [#permalink]

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New post 04 Mar 2015, 18:56
1
2
Bunuel wrote:
Set #1 = {A, B, C, D, E}
Set #2 = {K, L, M, N, O, P}

There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?

A. 1/6
B. 1/3
C. 1/2
D. 2/3
E. 5/6


Kudos for a correct solution.


At least questions are best solved by taking the opposite scenario and subtracting it from 1. Probability of choosing no vowel from set 1 is 3/5 and set 2 is 5/6. Multiply these to get 1/2. Therefore, probability of picking at least one vowel = 1-1/2=1/2. Answer C.
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Re: There are these two sets of letters, and you are going to pick exactly  [#permalink]

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New post 04 Mar 2015, 21:42
1
2
Bunuel wrote:
Set #1 = {A, B, C, D, E}
Set #2 = {K, L, M, N, O, P}

There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?

A. 1/6
B. 1/3
C. 1/2
D. 2/3
E. 5/6


Kudos for a correct solution.



We can also use the concept of sets:

Probability of picking a vowel = 2/5 (Picking a vowel from Set 1 and anything from set 2) + 1/6 (Picking a vowel from Set 2 and anything from set 1) - (2/5)*(1/6) = 15/30 = 1/2

The reason we subtract 2/5*1/6 is that it is counted twice - both in 2/5 and in 1/6 - exactly the way it is in sets

Answer (C)
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Re: There are these two sets of letters, and you are going to pick exactly  [#permalink]

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New post 05 Mar 2015, 00:05
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1
Hi All,

In this question, since the number of possible outcomes is relatively small, you can quickly list them out....

We're going to pick one letter from each set; we're asked for the probability that we end up with AT LEAST ONE vowel.

Since Set 1 has 5 letters and Set 2 has 6 letters, there are (5)(6) = 30 possible outcomes.

To get AT LEAST ONE vowel, we could have.....

The letter 'A' from Set 1 AND any of the 6 letters from Set 2 = 6 options
The letter 'E' from Set 1 AND any of the 6 letters from Set 2 = 6 options

The letter 'O' from Set 2 has ALREADY been counted TWICE: once with (A and O) and once with (E and O), so there are only 3 additional options with O....(B and O), (C and O) and (D and O) = 3 additional options.

6 + 6 + 3 = 15 options

15 options/30 total options = 1/2

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Re: There are these two sets of letters, and you are going to pick exactly  [#permalink]

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New post 08 Mar 2015, 14:39
Bunuel wrote:
Set #1 = {A, B, C, D, E}
Set #2 = {K, L, M, N, O, P}

There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?

A. 1/6
B. 1/3
C. 1/2
D. 2/3
E. 5/6


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

P(at least one vowel) = 1 – P(no vowels)

The probability of picking no vowel from the first set is 3/5. The probability of picking no vowel from the second set is 5/6. In order to get no vowels at all, we need no vowels from the first set AND no vowels from the second set. According to the AND rule, we multiply those probabilities.

P(no vowels) = (3/5)*(5/6) = 1/2

P(at least one vowel) = 1 – P(no vowels) = 1 – 1/2 = 1/2

Answer = C
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Re: There are these two sets of letters, and you are going to pick exactly  [#permalink]

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New post 07 Feb 2016, 10:09
I never liked questions with probability and counting...
but this one seems easy to crack, because you can list all the possible outcomes.

ok, so we have 5 letters in first set, and 6 letters in second set. together, we have 5*6 = 30 possible outcomes.
we need to select the number of outcomes, in which at least one vowel is present.
This will be easier to check the possibilities in which no vowels are selected.
we have: B, C, D in first set, and K, L, M, N, P in second set.
we thus have 3*5=15 ways in which no vowel is selected.
that means that 15 ways will be with at least a vowel.
15/30=1/2.

alternatively, let's list of the possible ways, in which a vowel is not selected
BK, BL, BM, BN, BP, CK, CL, CM, CN, CP, DK, DL, DM, DN, DP - 15 possibilities.
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Re: There are these two sets of letters, and you are going to pick exactly  [#permalink]

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Re: There are these two sets of letters, and you are going to pick exactly &nbs [#permalink] 03 Aug 2018, 08:19
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