Question

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?

Answer 1

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

$\Rightarrow ^6C_2\times ^7C_1$

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

$\Rightarrow ^{13}C_3$

Probability is

$\Rightarrow P=\dfrac{^6C_2\times ^7C_1}{^{13}C_3}=\dfrac{105}{286}\\\Rightarrow P=0.367$