A fruit stand at the checkout line has 5 bananas, 3 apples, and 4 oranges. Find the probability that if you randomly chose 3 fruits: 1) Two oranges and one banana are chosen

2) One of each fruit is chosen

3) All three are bananas

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7  -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so   14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as  -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84  then x = - 0.92 or around -0.9best luck!