A carnival card game gives the contestant a one in three probability

For which value of the answer choices do we have the lowest value of "n" such that \((1/3)^n\) yields \((1/3)^n<1/100\)? The answer is 5.

Probability of win is \frac{1}{3}
Let n time he must play so that the probability that he never loses (Win) is less than 1%

\((\frac{1}{3})^n\) < 1/100

100 < \(3^n\)

so the minimum value n which satisfy the above equation is 5.

Option

MANHATTAN GMAT OFFICIAL SOLUTION:

This problem tests primarily on Probability theory. In addition, Exponents are needed to represent the impact of playing multiple games on the probability of the outcomes. The probabilities are given in fractions, yet the question is asked in terms of percents, so FDP Connections are relevant.

Finally, the question is phrased in terms of an inequality, so Inequalities come into play. However, all the secondary issues can be resolved relatively easily. Therefore, we can consider this question a minor hybrid problem, in which one of the content areas (probability) is much more important than the others.

“The probability that he never loses” can be rephrased as “the probability that he always wins.” This probability can be expressed as (1/3)^n, where 1/3 is the chance of winning on a single play and n is the number of times the contestant plays.

Let's track it in a chart:


We might also have noticed that only the denominator in the probabilities mattered, as the numerator was always 1. To have a probability of less than 1%, the fractional probability must be “1 over something greater than 100.” In order for 3^n >100, n must be at least 5.

The correct answer is C.

oh man...the wording at first might be so confusing..but it is so easy to solve...
probability of not losing = probability of winning.
1/3 * 1/3 = 1/9 > 1%
1/9 * 1/3 = 1/81 > 1%
1/81 *1 = 1/243 <1%
1/243 = 1/(3^5)

so 5 games are minimum necessary.

The probability that he never loses is equivalent to the probability that he always wins. The probability that he will win a game is 1/3; thus, the probability that he will win the game n consecutive times is (1/3)^n and we want to determine the minimum value of n such that (1/3)^n < 0.01 = 1/100. Let’s analyze the answer choices:

If n = 3, then (1/3)^3 = 1/27 > 1/100.

If n = 4, then (1/3)^4 = 1/81 > 1/100.

If n = 5, then (1/3)^5 = 1/243 < 1/100.

We see that (1/3)^5 < 1/100, thus the minimum value of n is 5.

Answer: C

hi guys. I have easy question.
I do not know why if we increase the chance of success but the probability of success go down !
if we play 5 times we must have higher chance to success but WE got 1% ( why chance goes down instead of higher ?!)

P(win) = 1/3
The question is asking how many times can he play until P(win) < 1/100
So, (1/3)^n < 1/100, n = ?

A) 1/27 < 1/100, no
B) 1/81 < 1/100, no
C) 1/243 < 1/100, yes

09173140521
you misread the question, it's talking about the probability of winning every single time (i.e. never losing) NOT the probability of winning at least once (which would indeed be higher the more plays he made)