Probability Quant Concept Series Doubts

GRE Quant Problem Solving
1

I was going through the probability quant concept series and I had a couple of questions:

  1. In the below question, if the question was to find the range of possible values of P(A and B and C), what would the answer be?

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    I think the answer should be 0<= x <= 0.1 but I am not sure.

  2. Could someone explain how to solve this question? Ideally, using the “what makes me happy” approach that greg mentions in his probability video? I haven’t studied Combinatorics yet so that’s probably why the method mentioned in the video went over my head.

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    I approached it as :

    P(A * B * C * D * E * not F) + P(A * B * C * D * not F * E) = 2/(6^5 * 4) which is obviously the wrong answer.

    I took P(A) = P(B)= P( C)= P(D)= P(E)=1/6. For P(not F), I did 1/4 because (not F) would mean A,B,C,D,E. But E cannot sit next to E, so we just have 4 options - A,B,C,D and hence 1/4.

    Could someone explain what I did wrong?

2

Hi,

  1. For this question, I simply calculated the probability that Egg sits next to Fur and subtracted it from one. So probability of Egg and fur sitting together would be 5!x2/6! which turns out to be 1/3. So probability of them not sitting together, so 1-1/3 = 2/3.
3

I think your answer to your 1st question is correct as if A, B, C were mutually exclusive than the probability would be 0 and if they completely overlapped than the probability would be 0.1

for the 2nd question i’m not sure how to solve it from a probability perspective but from combinatorics perspective it becomes quite easy, let me show you

the total permutations of A, B, C, D, E, F = \underline{6} \times \underline{5} \times \underline{4} \times \underline{3} \times \underline{2} \times \underline{1} = 6! = 720(choice method)

assume E and F sit together, now calculate the total permutations of A, B, C, D, \fbox{E, F} = \underline{5} \times \underline{4} \times \underline{3} \times \underline{2} \times \underline{1} = 5! = 120(choice method) but you have to also calculate for the case when F and E sit together that will again be 120 hence total permutations where E and F are together = 120 + 120 = 240

probability that E and F sit together is \cfrac{240}{720} or \cfrac{1}{3}

probability that E and F don’t sit together is 1 - \cfrac{1}{3} = \cfrac{2}{3}