Question

Can someone help me with this question? Thanks!

Answer 1

Answer:  50%

You won't type in the percent sign since it's already taken care of by your teacher.

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Explanation:

We know 100% that a fruit was chosen first. That means the three original fruits drops down to 3-1 = 2 fruits.

There are 2 fruits left and 2 vegetables. That gives 2+2 = 4 items to pick from for the second slot.

We have 2 things we want (those vegetables) out of 4 items to pick from. The chances of getting a vegetable here is 2/4 = 1/2 = 0.50 = 50%

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This is an optional section, but here's another way to look at it. This method is slightly longer.

Let A,B,C be the three fruits and D,E be the two vegetables.

Here are all of the ways to have a fruit chosen first followed by a vegetable

• AD and AE
• BD and BE
• CD and CE

There are m = 3*2 = 6 ways to do this.

Now consider all of the ways to have a fruit first (either A,B, or C) and the second slot can be a fruit or a vegetable.

Here are all of the ways to do that

• AB, AC, AD, AE
• BA, BC, BD, BE
• CA, CB, CD, CE

Each row has 4 items, so we have n = 3*4 = 12 different ways to have a fruit go first and the second item is a fruit or a vegetable.

Therefore, m/n = 6/12 = 1/2 = 0.5 = 50% is the probability of getting a vegetable second given that a fruit was chosen first.