Veronica wrote:Each of the 25 balls in a certain box is either read, blue, or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?
1. Probability that the ball will both be white and have an even number painted on it is 0.
2. Probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2

What is the 100th digit of decimal z?
1. the 10th digit of 100z is 2
2. the units digit of 1000z is 2

Please help me how to solve these problems, thank you!

Problem 1: P(W or E) = P(W) + P(E) - P(W and E)
and P(W) = number of white balls/25 , P(E) = number of even numbered balls/25
Statement (1) - P(W and E) = 0, but as the number of white balls and number of even numbered balls is unknown. INSUFFICIENT
Statement (2) - P(W) - P(E) = 0, as P(W and E) is unknown INSUFFICIENT
Statement (1) and (2) - INSUFFICIENT
IMO E

Problem 2:
Statement (1) - 1000th digit of z is 2. 100th unknown. INSUFFICIENT
Statement (2) - 1000th digit of z is 2. 100th unknown. INSUFFICIENT
Statement (1) and (2) - INSUFFICIENT
IMO E

What are the OA?

Veronica wrote:Each of the 25 balls in a certain box is either read, blue, or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?
1. Probability that the ball will both be white and have an even number painted on it is 0.
2. Probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2

Answer: E

Focus: P(W or E) = ?

Let us BIFURCATE the statements (1) AND (2) together, what I call (1+2), to guarantee the answer is (E) without having to look at the official answer!!!

Note that (1) is equivalent to "(WHITE --> ODD) and (Even --> Red or Blue)" (*)

> Take (10 white balls, with odd printed in it) AND (5 even printed red balls) AND (10 blue with odd printed).

Then P(W) -P(E) = 2/5 - 1/5 = 1/5 = 0,2 and both statement s are satisfied, with answer (to the question stem): 3/5


> Take (15 white balls, with odd printed in it) AND (10 even printed red balls) AND (no balls are left ).

Then P(W) -P(E) = 3/5 - 2/5 = 1/5 = 0,2 and both statements are satisfied, with answer (to the question stem): 1


From the fact that two viable scenarios (that is, both satisfy statements (1) and (2) and the question stem) turn out to have different answers (to the question asked), we are SURE the answer is "E".

Hope you like it!

Have good studies,
Fábio.

P.S.: hint for understanding my reasoning: first of all, I created the 2/5 - 1/5 = 1/5 and 3/5 - 2/5 = 1/5 situations and, afterwards, I created the scenarios to have this calculations validated and satisfying the (*) restrictions.

Veronica wrote:What is the 100th digit of decimal z?
1. the 10th digit of 100z is 2
2. the units digit of 1000z is 2

Answer: E

"Focus " on the question: 100th digit of decimal z (value) ?

This kind of problem is better "tamed" when we put our hands in specific and representative examples! See below:

> Take 100 z = 1.23 , then 1000z = 12.3 and both statements are satisfied

In this case, z = 0.0123 therefore the answer to the question asked is 1

> Take 100 z = 4.23 , then 1000z = 42.3 and both statements are satisfied

In this case, z = 0.0423 therefore the answer to the question asked is 4

Again this BIFURCATION guarantees that the answer to this problem is "E".

Nice?

Regards,
Fábio.

P.S.: my method (presented at www.GMATH.net) is "bullet-proof" and my experience as a teacher is that this sort of reasoning is possible to be done pretty quick ONLY IF practiced in ALL training situations (to put your brain to work on this "tune" requires some weeks), therefore ALL Data Sufficiency problems presented at my prep course are dealt with this "care"/rigour and I give lots of suggestions to "provoke" this sort of reasoning in the student´s head!