Can you please check over this problem?

What is the probability of a monkey removing the letters G,S,and W (in any order) from a collections of the 6 letters G,S,W,X,Y, and Z if he removes exactly 3 letters? (exact fraction in lowest terms)
* Is this correct?
GSW SGW = 5/6 Did i miss anything?
WSG
WGS
SWG

What is the probability of a monkey removing the letters G,S,and W (in any order) from a collections of the 6 letters G,S,W,X,Y, and Z if he removes exactly 3 letters?

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Always try to ask yourself if your answer makes sense. Do you think the monkey could choose the correct three letters out of six total 5 out of 6 times consistently? (No.)

There are a couple of ways to approach this problem. I don't know if you've studied Combinations and Permutations yet, so I'll skip that approach.

Consider this: Because there are 6 letters and 3 of them are "correct", the chance of drawing a correct first letter is 3 out of 6, or 1/2.

To choose a second letter correctly, there are now only 2 correct letters left out of 5 total letters, so the chances of getting the second letter correct are 2 out of 5, or 2/5.

By similar logic, there is only 1 chance out of 4 of getting the last correct letter, or 1/4.

So, the probability of getting all three of these events to occur is the product of the three individual probabilities. Does this sound familiar? Hope that helps.

Sue20 said:

What is the probability of a monkey removing the letters G,S,and W (in any order) from a collections of the 6 letters G,S,W,X,Y, and Z if he removes exactly 3 letters? (exact fraction in lowest terms)

Click to expand...

There are \(\displaystyle \dbinom{6}{3}=\dfrac{6\cdot 5\cdot 4}{3\cdot 2\cdot 1}=20\) ways for our monkey to select exactly three of those six letters.
Now \(\displaystyle \{G,S,W\}\) is one of those twenty.

Thank you!

Thank you so much!!