Question

5 Garry has these cans of soup in his kitchen cabinet.. 3 cans of tomato soup• 4 cans of chicken soup• 6 cans of cheese soup• 7 can of beef soupIGarry will randomly choose one can of soup. Then he will put it back and randomly choose anothercan of soup. What is the probability that he will choose a can of chicken soup and then a can of beefsoup?

Answer 1

hello, this question is about probability.

First of all, we must to sum all the cans of soup that Garry has in his kitchen cabinet.

`\text{3 tomato + 4 chicken + 6 cheese + 7 beef = }20\text{ cans of soup}`

Now, we know that he will choose a random can of soup.

The probability that he will choose a can of chicken soup:

If he has 4 cans of chicken soup, he can take any of them out of a total of 20, right? So we can write this as:

`\frac{4\text{ chicken soup}}{20\text{ total of cans}}=\text{ }(4)/(20)=(2)/(10)=(1)/(5)=0.2\text{ = 20\% probability.}`

For the beef soup, we will do the same process:

`\frac{7\text{ beef soup}}{20\text{ total of cans}}=\text{ }(7)/(20)\text{ }=\text{ 35\% of probability}`

But we have a problem, the question asks how likely he is to choose these cans in a row, we can see this by the connective "e".

It's simple, we just need to multiplicate the probabilities of the first case happen by the probability of the second case:

`(4)/(20)*(7)/(20)\text{ = }(28)/(400)=(14)/(200)=\text{ }(7)/(100)=\text{ 7\%}`