tomatos wrote:Q1 Jack has a jar full of quarters, nickels, and dimes. He picks n coins, totaling 45cents. What is the value of n?

1) the probability of randomly pickin g a nickel from the n coins is 50%.
2) the probability of randomly picking a dime from the n coins is 50%

The answer is B. The explanation from the book is asking students to list all of the possible combination of coins that could add up to 45 cents. On the exam, I am afraid that I won't have the time (or even be that calm) to list out all of the possible combinations. Can anyone please tell me if you have a better way than the book?

Solution 1:
If you don't want to list down all the ways right away, then let me suggest an alternate method.

Statement(1)
essentially says that the number of nickels(n) is equal to the number of other types of coins (d and q). List down the cases when this holds true and the currencies also add up to 45 cents.
n = 2, d = 1, q = 1 ( 5*2 + 10 + 25 ) and
n = 3, d = 3 (5*3 + 10*3).
Two possibilities.
INSUFFICIENT.

Statement(2):
Number of dimes (d) is equal to the number of rest of the coins (c + n).
d = 3, n = 3 (10*3 +5*3) is the only possibility.
SUFFICIENT.

[spoiler](B)[/spoiler] is the correct answer.

While we're at it, I think you must sometimes have the patience to list down the possibilities. It does not take as much time as it looks it will.

tomatos wrote: Q2 Diana is going on a school trip along with her two brothers. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that Diana leaves at the same as at least one of her brothers?

a) 1/27
b) 4/27
c) 5/27
d) 4/9
e) 5/9
Thank you very much!

Solution2:

Required Probability
= 1 - Probability of the two brothers not being in Diana's group
= 1 - First brother not being with Diana AND Second brother not being with Diana
= 1 - (2/3)*(2/3)
= 5/9

[spoiler](E)[/spoiler] is correct.

Alternate Method:

Probability that Diana and the two brothers leave in the same group
=Two brothers are in the same group AND Diana is in that group
= (1/3)*(1/3)
Probability that Diana leaves with the first brother and the second brother is not with them
=Diana and the first brother are in the same group AND the second brother is not
= (1/3)*(2/3)
Probability that Diana leaves with the second brother and the first brother is not with them
=Diana and the second brother are in the same group AND the first brother is not
= (1/3)*(2/3)

Required probability = Sum of all Probabilities above = 1/9 + 2/9 + 2/9 = 5/9
[spoiler](E)[/spoiler] is correct.

P.S.: I know this doubt will arise, so let try to pre-empt it.
Why did I not take the probability of Diana to be in a group to be (1/3)?
Because it does not matter in which group Diana is. If you assume Diana to be in any one of the groups, this probability will be (1/3) but then for three groups you will have to multiply the entire probability by 3. This will nullify the (1/3) because (1/3)*(3) = 1.