25.
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black? 1/4
1/2
5/8
2/3
3/4
Solution:
SHORTCUT: The initial probability of drawing a certain ball is the same as the probability of drawing that certain ball at any future point. Therefore answer: 5/8
26.
On a business trip, 30 percent of 60 sales representatives will be given accommodations at Hotel XYZ and the remaining 70 percent will be given accommodations at Hotel ABC. However, 55 percent of the sales representatives prefer to stay at Hotel XYZ and 45 percent prefer to stay at Hotel ABC. What is the highest possible number of sales representatives NOT given accommodations at the hotel they prefer?
A) 11 B) 18 C) 36 D) 45 E) 51
Solution:
Let's start by converting percents into numbers, since the final answer needs to be in number form.
18 will end up in XYZ
42 will end up in ABC
33 WANT to be in XYZ
27 WANT to be in ABC
Now, let's see what we can do to maximize customer dissatisfaction!
All 33 who want to be in XYZ can fit into ABC, so let's throw them over there - that's 33 unhappy customers.
There's only room for 18 people in XYZ, so sadly we can't make all the pro-ABC people unhappy - but we can certainly upset 18 of the 27 who want to be in ABC by putting them in XYZ instead - another 18 unhappy people.
33 + 18 = 51 people who won't get a good night's sleep - choose (E).
27.
A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?
A. 3 B. 4 C. 5 D. 6 E. 8
Solution:
While we could use formulas, a lot of combination questions are easier to answer through a bit of common sense and brute force. On this question, it's much quicker to just list the possible committees than it is to use a formula approach.
No Paul means no Jane, which leaves us with only Joan/Stuart/Jessica. Sadly, Stuart cannot appear in any other committees.
Paul then needs 2 co-members and has 3 possible buddies for those 2 spots. We could use 3C2=3 to calculate or we could just brute force: Paul/Joan/Jessica Paul/Joan/Jane Paul/Jane/Jessica So, as others have noted, there are 4 possible committees - choose (B).
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black? 1/4
1/2
5/8
2/3
3/4
Solution:
SHORTCUT: The initial probability of drawing a certain ball is the same as the probability of drawing that certain ball at any future point. Therefore answer: 5/8
26.
On a business trip, 30 percent of 60 sales representatives will be given accommodations at Hotel XYZ and the remaining 70 percent will be given accommodations at Hotel ABC. However, 55 percent of the sales representatives prefer to stay at Hotel XYZ and 45 percent prefer to stay at Hotel ABC. What is the highest possible number of sales representatives NOT given accommodations at the hotel they prefer?
A) 11 B) 18 C) 36 D) 45 E) 51
Solution:
Let's start by converting percents into numbers, since the final answer needs to be in number form.
18 will end up in XYZ
42 will end up in ABC
33 WANT to be in XYZ
27 WANT to be in ABC
Now, let's see what we can do to maximize customer dissatisfaction!
All 33 who want to be in XYZ can fit into ABC, so let's throw them over there - that's 33 unhappy customers.
There's only room for 18 people in XYZ, so sadly we can't make all the pro-ABC people unhappy - but we can certainly upset 18 of the 27 who want to be in ABC by putting them in XYZ instead - another 18 unhappy people.
33 + 18 = 51 people who won't get a good night's sleep - choose (E).
27.
A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?
A. 3 B. 4 C. 5 D. 6 E. 8
Solution:
While we could use formulas, a lot of combination questions are easier to answer through a bit of common sense and brute force. On this question, it's much quicker to just list the possible committees than it is to use a formula approach.
No Paul means no Jane, which leaves us with only Joan/Stuart/Jessica. Sadly, Stuart cannot appear in any other committees.
Paul then needs 2 co-members and has 3 possible buddies for those 2 spots. We could use 3C2=3 to calculate or we could just brute force: Paul/Joan/Jessica Paul/Joan/Jane Paul/Jane/Jessica So, as others have noted, there are 4 possible committees - choose (B).
Hello, my name is Jack Sparrow. I'm a 50 year old self-employed Pirate from the Caribbean.
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