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Data Sufficiency Butler: January 2024 January 29 DS 1 DS 2

Problem Solving Butler: January 2024 January 29 PS 1 PS 2

Hello guys, don’t you think that the free official practice exam is a little bit harder than the real test (to make you buy the other material that aren’t free) ?

Hi guys, please give this one a shot: survey-conducted-at-city-club-gathered-the-following-information-425050.html?fl=notifications#p3343127 I’ve talked to friends and they say the Free Official mocks have the exact same algorithm as the Original exam

Data Sufficiency Butler: January 2024 January 30 DS 1 DS 2

Problem Solving Butler: January 2023 January 30 PS 1 PS 2

From a group of J employees, K will be selected, at random, to sit in a line of K chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Lisa is seated exactly next to employee Phillip? (1) K = 15 (2) K = J In this DS question can anyone explain the calculation. How do I solve for the numerator?

Have you check this topic: https://gmatclub.com/forum/from-a-group ... 93411.html ?

Hello All, I had a small question in the quant section while doing probability I can across this following example, I am not sure how did we reach the conclusion that P(A and C) < P(A) I.e 0.23 Here is the solution in the example: Example: In an experiment with events A, B, and C, suppose P (A) = 0.23, P (B) = 0.40, and P(C) = 0.85. Also suppose events A and B are mutually exclusive, and events B and C are independent. Since A and B are mutually exclusive, P(A or B) = P(A) + P(B) = 0.23 + 0.40 = 0.63. Since B and C are independent, P(B or G) = P(B) + P(C) -P(B)P(C) = 0.40 + 0.85 - (0.40)(0.85) = 0.91. P(4 or G) and P(4 and C) can’t be found from the information given. But we can find that P(A) + P(G) = 1.08 > 1. So P(A) + P(C) can’t equal P(A or C), which like any probability must be less than or equal to 1. This means that A and C can’t be mutually exclusive, and that P(Aand C) ≥ 0.08. • Since An Bis a subset of 4, we can also find that P(A and C.) ≤ P (A) = 0.23.? And Cis a subset of A U C, so P(A or C) ≥ P(C) = 0.85. Thus, we’ve found that 0.85 ≤ P(A or C) ≤ 1 and that 0.08 ≤ P(A and G) ≤ 0.23. 55

I did check out but i couldn’t understand the solutions provided. Especially how to solve for the numerator

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