65% of the time, richard eats breakfast. what is the probability he eats breakfast at least once during the work week, monday through friday? (round your answer to the nearest thousandth)

Short P(B) = 0.995
Remark
Start the problem by assuming he never eats breakfast during the work week. 

Step One.
Figure out the probability that on day one (Monday) he does not eat breakfast

1 - 65/100  is the probability that he will not eat breakfast on Monday.
P(NB) = 100/100 - 65/100 = 35/100  NB stands for no breakfast.

Step Two 
Reduce 35/100
35/100 = 7/20

Step 3
For each day following the previous result  is multiplied by 7/20
P(NBW) = (7/20)^5  NBW = No Breakfast all week.
P(NBW) = 0.005252

Step 4 
Find the probability he will have at least one Breakfast.
P(B) = 1 - 0.005252
P(B) = 0.995 to the nearest 1000th. <<<< answer.

First subtract 10 from both side
y^2+6y-16=0
(y+8)(y-2)=0
-8 and 2