Answer:
C. [tex]\frac{1}{18}[/tex]
Step-by-step explanation:
Given: Six cards numbered from [tex]1[/tex] to [tex]6[/tex] are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.
To Find: If the cards are drawn at random and if the sum of the numbers on the cards is [tex]8[/tex], what is the probability that one of the two cards drawn is numbered [tex]5[/tex].
Solution:
Sample space for sum of cards when two cards are drawn at random is [tex]\{(1,1),(1,2),(1,3)......(6,6)\}[/tex]
total number of possible cases [tex]=36[/tex]
Sample space when sum of cards is [tex]8[/tex] is [tex]\{(3,5),(5,3),(6,2),(2,6),(4,4)\}[/tex]
Total number of possible cases [tex]=5[/tex]
Sample space when one of the cards is [tex]5[/tex] is [tex]\{(5,3),(3,5)\}[/tex]
Total number of possible cases [tex]=2[/tex]
Let A be the event that sum of cards is [tex]8[/tex]
[tex]p(\text{A})[/tex] [tex]=\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}[/tex]
[tex]p(\text{A})=\frac{5}{36}[/tex]
Let B be the event when one of the two cards is [tex]5[/tex]
probability than one of two cards is [tex]5[/tex] when sum of cards is [tex]8[/tex]
[tex]p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}[/tex]
[tex]p(\frac{\text{B}}{\text{A}})=\frac{2}{5}[/tex]
Now,
probability that sum of cards [tex]8[/tex] is and one of cards is [tex]5[/tex]
[tex]p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})[/tex]
[tex]p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}[/tex]
[tex]p(\text{A and B})=\frac{1}{18}[/tex]
if sum of cards is [tex]8[/tex] then probability that one of the cards is [tex]8[/tex] is [tex]\frac{1}{18}[/tex], option C is correct.
