Respuesta :
Total cards = 8
Probability of pulling number 3 card first = 3/8
Probability of pulling number 2 card second= 2/7
P(3, then 2) = (3/8)(2/7) = 3/28
Answer: 3/28
Answer: The correct option is (D) [tex]\dfrac{3}{28}.[/tex]
Step-by-step explanation: Given that Hiro has a stack of cards with one number from the set 1, 1, 2, 2, 3, 3, 3, 4 written on each card.
We are to find the probability that Hiro pulls out a 3 first and then pulls out a 2 without replacing them.
Let, 'S' be the sample space for the experiment.
Then, S = {1, 1, 2, 2, 3, 3, 3, 4}
⇒ n(S) = 8.
Let, 'E' and 'F' be the events of pulling a 3 and a 2 respectively.
Then, E = {3, 3, 3} ⇒ n(E) = 3
and
F = {2, 2} ⇒ n(F) = 2.
So, the probability that he pulls out a 3 is given by
[tex]P(E)=\dfrac{n(E)}{n(S)}=\dfrac{3}{8},[/tex]
and the probability that he pulls out a 2 without replacing the 3 is given by
[tex]P(F)=\dfrac{n(F)}{n(S)}=\dfrac{2}{7}.[/tex]
Therefore, the probability that he pulls out a 3 first and then pulls out a 2 without replacing them is
[tex]P=P(E)\times P(F)=\dfrac{3}{8}\times \dfrac{2}{7}=\dfrac{3}{28}.[/tex]
Hence, the required probability is [tex]\dfrac{3}{28}.[/tex]
Thus, (D) is the correct option.
