Probability problem: "A box contains 20 bottles of Fanta and 4 bottles of Sprite..."

[chijioke]

Here is my work:

Total number of bottles = 24

Number of bottles of Fanta = 20

Number of bottles of Sprite = 4

(a) Probability that a bottle chosen at random is Fanta = [math]\frac{\cancel{20^5}}{\cancel{24^6}}=\frac{5}{6}[/math]
(b)Probability that a bottle chosen at random is Sprite= [math]\frac{\cancel{4^1}}{\cancel{24^6}}=\frac{1}{6}[/math]
(c)Probability that a bottle chosen at random is either Fanta or Sprite = [math]\frac{5}{6}+ \frac{1}{6}=1[/math]
But I am not sure if (c) is correct. Can somebody please check my work?

 

[blamocur]

Looks correct to me: since there is nothing but Fanta and Sprite in the box the probability of picking anything else is 0

 

[Dr.Peterson]

(a) Probability that a bottle chosen at random is Fanta = [math]\frac{\cancel{20^5}}{\cancel{24^6}}=\frac{5}{6}[/math]
(b)Probability that a bottle chosen at random is Sprite= [math]\frac{\cancel{4^1}}{\cancel{24^6}}=\frac{1}{6}[/math]
(c)Probability that a bottle chosen at random is either Fanta or Sprite = [math]\frac{5}{6}+ \frac{1}{6}=1[/math]
But I am not sure if (c) is correct. Can somebody please check my work?

Since you like using Latex, here's a better way to cancel: [math]\frac{\overset{5}{\cancel{20}}}{\underset{6}{\cancel{24}}}=\frac{5}{6}[/math]
I used overset and underset, so it doesn't cancel the new numbers.

 

[chijioke]

I will try to use next time.