Question

Arnold is playing a game where he is trying to roll a two with a standard die. If he gets a two in any of his first 6 rolls, he wins; otherwise he loses. What is the probability that Arnold wins the game? Round your answer to the nearest tenth of a percent.
A. 42.1%
B. 51.8%
C. 66.5%
D. 59.8%

Answer 1

i have $20 saying arnolds gonna lose........but your answer is C....66.5 because P(lose) = 5^6 / 6^6 = (5/6)^6

P(win) = 1 - (5/6)^6 = 66.5%

Answer 2

Answer:  C. 66.5%

Step-by-step explanation:

Given: Arnold is playing a game where he is trying to roll a two with a standard die.

We know that for any event P(A) = 1 - P(A')  

Therefore, P(win)=1-P(lose)

Thus, he loses if he never rolls a 2 in the 6 rolls.  

The total number of outcomes = 6

There number of losing outcomes (1, 3, 4, 5, 6)=5

Therefore, the probability that Arnold loses the game is given by :-

$(\dfrac{5}{6})^6$

Now, the probability that Arnold wins the game is given by :-

$\text{P(win)}=1-(\dfrac{5}{6})^6=0.66510$

In percent, the probability that Arnold wins the game is given by :-

$\text{P(win)}=60.66510\times100=66.5\%$