what is the probability of selecting two S's if the first card is not replaced before selecting the second card in the word Massachusetts
I don't quite think you are 100% correct. After all I am a teacher. :(
16/169
To determine the probability of selecting two S's in the word "Massachusetts" without replacement, we need to calculate the probability of selecting an S as the first card, and then the probability of selecting another S as the second card, given that the first card was not replaced.
Step 1: Calculate the probability of selecting an S as the first card.
In the word "Massachusetts," there are a total of 4 S's out of 13 letters.
So, the probability of selecting an S as the first card is 4/13.
Step 2: Calculate the probability of selecting another S as the second card, given that the first card was not replaced.
After the first card is selected, there are 12 remaining cards, out of which 3 are S's.
So, the probability of selecting another S as the second card, without replacement, is 3/12.
Step 3: Multiply the probabilities from steps 1 and 2.
To find the probability of both events occurring (selecting an S as the first card and another S as the second card), multiply the probabilities from steps 1 and 2 together:
(4/13) * (3/12) = 12/156 = 1/13 ≈ 0.0769
Therefore, the probability of selecting two S's if the first card is not replaced before selecting the second card in the word "Massachusetts" is approximately 0.0769 or 1/13.
4 esses in 13 letters, so
p = 4/13 * 3/12