+12 Four letters are selected at random with replacement from the word MATHCOUNTS. What is the probability that the letters in the order they are selected will form the word MATH?
The probability that the first letter is M is 1/9. The probability that hte second letter is A is 1/8. The probability that the third letter is T is 2/7. The probability that the fourth letter is H is 1/6. Multiply these products, and you get 1/1512.
+289 The first letter we have to choose is M which is our only option. This gives us a probabililty of 1/10.
Next, the only letter where we can get our A is from the second letter. Again, this gives a probability of 1/10.
When choosing the letter T, however, we have to consider the fact that there are two T's in MATHCOUNTS. This gives a probability of 1/5 of choosing it next.
Again, there is only one H so there is 1/10 probability
Multiplying all of these gives us: 1/5000.
Which I think is somehow wrong, someone please correct me.
+2668 No, I agree with your answer...
There are \(10^4 = 10000\) ways to choose the 4 letters, and there are only 2 ways to choose the word "MATH" (there are 2 T's), so the probability is \(\color{brown}\boxed{1 \over 5000}\)
+12 There are \(10^4=10,000\) ways to choose M, A, T, and H. There is only one way to choose M, A, and H and two ways to choose T in MATHCOUNTS.
That means that there are \(1 \cdot 1 \cdot 1 \cdot 2 = 2\) ways to spell MATH. Then, the probability of getting MATH is
\(\dfrac{2}{10,000} = \boxed{\dfrac{1}{5,000}}.\)
And yes, HeWhoShallNotBeNamed (@Voldemort)s' answer was correct.
