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{3}{32}
\frac{1}{64}
\frac{3}{28}
\frac{1}{56}

Explanation:

Step1: Calculate the probability of pulling out a 3 first

The total number of cards is \(1 + 1+2 + 2+3+3 + 3+4=8\) cards. The number of 3 - cards is 3.
The probability of pulling out a 3 first is \(P(3)=\frac{3}{8}\)

Step2: Calculate the probability of pulling out a 2 after pulling out a 3

After pulling out a 3 (without replacement), the number of remaining cards is \(n = 8-1 = 7\) cards. The number of 2 - cards is 2.
The probability of pulling out a 2 after pulling out a 3 is \(P(2|3)=\frac{2}{7}\)

Step3: Calculate the combined probability

By the multiplication rule for dependent events \(P(A\cap B)=P(A)\times P(B|A)\)
Here \(A\) is the event of pulling out a 3 and \(B\) is the event of pulling out a 2.
\(P = \frac{3}{8}\times\frac{2}{7}=\frac{3\times2}{8\times7}=\frac{6}{56}=\frac{3}{28}\)

Answer:

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