Question

a student picks a random letter from the word $pi$ and a random letter from the word $please$. how many possible outcomes are in the sample space? \\(\circ\\) a) 2 \\(\circ\\) b) 8 \\(\circ\\) c) 12 \\(\circ\\) d) 20

Explanation:

Step1: Determine number of letters in each word

The word "PI" has 2 letters. The word "PLEASE" has 6 letters (P, L, E, A, S, E – note that repeated letters still count as separate when picking randomly, but here we count the number of distinct choices? Wait, no, when picking a random letter, for "PI" there are 2 possible letters (P and I), and for "PLEASE" let's check: P, L, E, A, S, E – but the number of distinct letters? Wait, no, the problem is about possible outcomes when picking one from each. Wait, actually, when picking a letter from "PI", there are 2 possible results (P or I). When picking a letter from "PLEASE", let's see the letters: P, L, E, A, S, E – but how many distinct letters? Wait, no, the word "PLEASE" has 6 letters (including the repeated E? Wait, no, the length of "PLEASE" is 6? Wait, P-L-E-A-S-E: that's 6 letters (P, L, E, A, S, E). But when picking a random letter, each position is a possible outcome? Wait, no, the problem is about the number of possible pairs. So first, number of choices from "PI": 2 (P and I). Number of choices from "PLEASE": let's see, the letters in "PLEASE" are P, L, E, A, S, E – but how many distinct letters? Wait, no, maybe we should count the number of distinct letters? Wait, no, the problem is about possible outcomes when picking one from each word. So for "PI", there are 2 letters (P, I), so 2 possible choices. For "PLEASE", let's list the letters: P, L, E, A, S, E – but the distinct letters are P, L, E, A, S (since E is repeated, but when picking a random letter, does the repeated E count as two separate outcomes? Wait, no, when you pick a letter from a word, each occurrence is a possible outcome? Wait, no, the problem is about the sample space of possible pairs (letter from PI, letter from PLEASE). So first, determine the number of possible letters in each word.

Wait, "PI" has 2 letters (P, I) – so 2 possible choices. "PLEASE" – let's check the letters: P, L, E, A, S, E. But the number of distinct letters? Wait, no, the problem is probably considering that when picking a letter from "PLEASE", how many different letters are there? Wait, P, L, E, A, S – that's 5? Wait, no, E is repeated, but when you pick a letter, if you pick E, is it considered the same outcome regardless of which E? Wait, the problem says "a random letter", so for "PLEASE", the possible letters are P, L, E, A, S, E – but the distinct letters are P, L, E, A, S (since E is repeated, but maybe the problem considers the distinct letters? Wait, no, let's check the answer options. The options are 2, 8, 12, 20. Let's think again.

Wait, maybe I made a mistake. Let's count the number of letters in each word: "PI" has 2 letters (so 2 possible choices: P or I). "PLEASE" – let's spell it: P (1), L (2), E (3), A (4), S (5), E (6). So there are 6 letters, but some are repeated. However, when forming the sample space, each outcome is a pair (letter from PI, letter from PLEASE). So we need to find the number of possible pairs.

But wait, maybe the problem is considering the number of distinct letters in each word. Wait, "PI" has 2 distinct letters (P, I). "PLEASE" – let's see: P, L, E, A, S – that's 5 distinct letters? But 2*5=10, which is not an option. Alternatively, maybe "PLEASE" has 6 letters (including the repeated E), but when picking a letter, the repeated E is considered the same? No, that doesn't make sense. Wait, maybe I miscounted the letters in "PLEASE". Wait, P-L-E-A-S-E: that's 6 letters (P, L, E, A, S, E). So the number of distinct letters is 5 (P, L, E, A, S), but the number of possible letters when picking randomly is 6? No, because when you pick a letter from "PLEASE", each position is a possible outcome, but the problem is about the letter itself, not the position. So if the letter is E, picking the first E or the second E would result in the same letter, so we should consider distinct letters. Wait, this is confusing. Let's check the answer options. The options include 12, which is 2*6. Wait, maybe the problem is considering that "PLEASE" has 6 letters (including the repeated E as separate? No, that can't be. Wait, maybe I made a mistake in the number of letters in "PLEASE". Wait, P-L-E-A-S-E: that's 6 letters (P, L, E, A, S, E). So 6 letters. Then, the number of possible outcomes is the number of ways to pick one from 2 and one from 6, so 2*6=12. Ah, that must be it. So even though E is repeated, when we pick a letter from "PLEASE", each occurrence is a possible outcome? Wait, no, but the letter is the same. Wait, no, the problem is about the sample space of possible pairs (letter from PI, letter from PLEASE). So for each letter in PI (2 options) and each letter in PLEASE (6 options, considering that even though E is repeated, when you pick a letter, the result is the letter, so if you pick E from PLEASE, it's just E, but the number of possible letters in PLEASE is 6? Wait, no, the word "PLEASE" has 6 letters, so when you pick a random letter, there are 6 possible outcomes? No, because two of them are E. Wait, no, the number of distinct letters in "PLEASE" is 5 (P, L, E, A, S), but the problem might be considering the number of letters (including repeats) as the number of choices. Wait, but that would be incorrect, because picking the first E or the second E would result in the same letter, so the outcome (P, E) would be the same regardless of which E you picked. So the correct way is to count the number of distinct letters in each word. Wait, "PI" has 2 distinct letters (P, I). "PLEASE" has 5 distinct letters (P, L, E, A, S). Then the number of possible outcomes is 2*5=10, but 10 is not an option. So I must have made a mistake.

Wait, wait, let's check the word "PLEASE" again. P-L-E-A-S-E: that's 6 letters. Let's list the letters with their counts: P:1, L:1, E:2, A:1, S:1. So the distinct letters are P, L, E, A, S (5 letters). But the answer options include 12, which is 2*6. So maybe the problem is considering that when picking a letter from "PLEASE", each of the 6 positions is a possible outcome, even if the letter is repeated. So for example, picking the first E or the second E are considered different outcomes? But that doesn't make sense because the letter is the same. Wait, no, the problem says "a random letter", so maybe it's about the number of possible letters, not the positions. So maybe the problem has a typo, or I'm misunderstanding. Wait, let's check the answer options. The options are 2, 8, 12, 20. 2*6=12, which is option c. So maybe the problem considers that "PLEASE" has 6 letters (including the repeated E) as 6 possible choices (even though E is repeated, but when forming the sample space, each letter from PI (2 choices) paired with each letter from PLEASE (6 choices) gives 2*6=12 possible outcomes. So that must be it.

Step2: Calculate the number of possible outcomes

To find the number of possible outcomes in the sample space, we multiply the number of possible choices from each word. The number of choices from "PI" is 2, and from "PLEASE" is 6. So the total number of outcomes is \( 2 \times 6 = 12 \).

Answer:

c) 12