A carnival card game gives the contestant a one in three probability of successfully choosing the right card and thereby winning the game. If a contestant plays the game repeatedly, what is the minimum number of times that he must play the game so that the probability that he never loses is less than 1%?	Choose the correct choice from the following choices.
A) 3
B) 4
C) 5
D) 6
E) 7
Probability of Choosing Right card hence winning = 1/3   i.e. Probability of NOT Choosing Right card hence Losing = 2/3   We want that the probability that he never loses should be &lt; 1%   i.e. We want that the probability that he never loses should be &lt; 1/100   The probability of him losing can be calculated by taking the case that the right card is never picked   i.e. Probability of first card being the Right card = (1/3)   i.e. Probability of first as well as second card NOT being the Right card = (1/3)*(1/3)   i.e. i.e. Probability of all n attempts being the Right card (i.e. Never Losing) = (1/3)^n   and (1/3)^n&lt;1/100   But(1/3)^4=1/81   And (1/3)^5=1/243   i.e. (1/3)^5&lt;1/100   i.e. Minimum value of 'n' can be = 5   Answer:5
<result>C</result>
A carnival card game gives the contestant a one in three probability of successfully choosing the right card and thereby winning the game. If a contestant plays the game repeatedly, what is the minimum number of times that he must play the game so that the probability that he never loses is less than 1%?