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

Question

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 .

Gnpth · Accepted Answer

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.

KarishmaB · Answer

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)

aviram · Answer

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.

EMPOWERgmatRichC · Answer

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

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

Bunuel · Answer

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

mvictor · Answer

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.

Poojapandey0611 · Answer

Can some one answer What is the probability of picking a C or an M?

EMPOWERgmatRichC · Answer

Hi Poojapandey0611,

Based on the question that you are asking, I assume that you want "just C" or "just M" but not 'both C AND M.' That question doesn't actually require much work, since the number of possible outcomes is relatively small, you can quickly list them out....

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

To get C from the Set 1, but NOT M from Set 2, there are 5 possible options: CK, CL, CN, CO and CP

To get M from Set 2, but NOT C from Set 1, there are 4 possible options: AM, BM, DM, and EM

Thus, the probability of getting just C or just M (but not both) is 9/30 = 3/10.

GMAT assassins aren't born, they're made,
Rich

Divyabansal0203 · Answer

Can some one answer What is the probability of picking a C or an M?

Hi Rich,

I think your answer is wrong.

Probability of getting a C or an M, should be like this:

(first no. picked from set1 is c)*(second no picked from set 2 is not M)+(first no picked from set 1 is not c)*(second no picked from set 2 is M)

((1/5)*(5/6))+((4/5)*(1/6))=1/3

Hence, 1/3 is the answer not 3/10.

Let me know if you have some other view.

EMPOWERgmatRichC · Answer

Hi Divyabansal0203,

The calculation that you set up IS correct, but I think that you made a small math mistake. You should have ended up with...

((1/5)*(5/6))+((4/5)*(1/6))
5/30 + 4/30 = 
9/30

That's 3/10 (not 1/3). 
 
GMAT assassins aren't born, they're made,
Rich

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

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