Question

hiro has a stack of cards with one number from the set 1, 1, 2, 2, 3, 3, 3, 4 written on each card. what is the probability that he pulls out a 3 first and then pulls out a 2 without replacing them? \\(\frac{1}{64}\\) \\(\frac{1}{56}\\) \\(\frac{3}{32}\\) \\(\frac{3}{28}\\)

Explanation:

Step1: Find total number of cards

Count the number of cards: \(1,1,2,2,3,3,3,4\) has \(2 + 2+ 3 + 1=8\) cards? Wait, no: 1 appears 2 times, 2 appears 2 times, 3 appears 3 times, 4 appears 1 time. So total cards \(n = 2+2 + 3+1=8\)? Wait, 2+2=4, 4+3=7, 7+1=8? Wait, no: 1,1 (2), 2,2 (2), 3,3,3 (3), 4 (1). So 2+2+3+1=8? Wait, 2+2 is 4, 4+3 is 7, 7+1 is 8. Yes, total 8 cards? Wait, no, wait the set is 1,1,2,2,3,3,3,4 – that's 8 elements? Wait, 1,1 (2), 2,2 (2), 3,3,3 (3), 4 (1). 2+2=4, 4+3=7, 7+1=8. So total 8 cards.

Step2: Probability of first card (3)

Number of 3s: 3. So probability \(P(\text{first 3})=\frac{\text{number of 3s}}{\text{total cards}}=\frac{3}{8}\).

Step3: Probability of second card (2) without replacement

After removing one 3, total cards left: \(8 - 1=7\). Number of 2s: 2. So probability \(P(\text{second 2})=\frac{\text{number of 2s}}{\text{remaining cards}}=\frac{2}{7}\).

Step4: Multiply the probabilities

Since these are dependent events, multiply the two probabilities: \(P=\frac{3}{8}\times\frac{2}{7}=\frac{6}{56}=\frac{3}{28}\).

Answer:

\(\frac{3}{28}\)