You are randomly selecting 3 cans of regular and diet cola, but assume that there are 6 cans of regular cola and 7 cans of diet cola.

What is the probability that 2 cans of regular cola and 1 can of diet cola are selected?

Respuesta :

Answer:0.367

Step-by-step explanation:

Given

There are 6 cans of regular and 7 cans of diet cola

number of ways of selecting 2 regular colas and 1 can of diet cola is

[tex]\Rightarrow ^6C_2\times ^7C_1[/tex]

The total no of ways of selecting 3 items out of 13 items is

[tex]\Rightarrow ^{13}C_3[/tex]

Probability is

[tex]\Rightarrow P=\dfrac{^6C_2\times ^7C_1}{^{13}C_3}=\dfrac{105}{286}\\\Rightarrow P=0.367[/tex]