Answer:
C. ![\frac{1}{18}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B18%7D)
Step-by-step explanation:
Given: Six cards numbered from
to
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
, what is the probability that one of the two cards drawn is numbered
.
Solution:
Sample space for sum of cards when two cards are drawn at random is ![\{(1,1),(1,2),(1,3)......(6,6)\}](https://tex.z-dn.net/?f=%5C%7B%281%2C1%29%2C%281%2C2%29%2C%281%2C3%29......%286%2C6%29%5C%7D)
total number of possible cases
Sample space when sum of cards is
is ![\{(3,5),(5,3),(6,2),(2,6),(4,4)\}](https://tex.z-dn.net/?f=%5C%7B%283%2C5%29%2C%285%2C3%29%2C%286%2C2%29%2C%282%2C6%29%2C%284%2C4%29%5C%7D)
Total number of possible cases ![=5](https://tex.z-dn.net/?f=%3D5)
Sample space when one of the cards is
is ![\{(5,3),(3,5)\}](https://tex.z-dn.net/?f=%5C%7B%285%2C3%29%2C%283%2C5%29%5C%7D)
Total number of possible cases ![=2](https://tex.z-dn.net/?f=%3D2)
Let A be the event that sum of cards is ![8](https://tex.z-dn.net/?f=8)
![=\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctext%7Btotal%20cases%20when%20sum%20of%20cards%20is%208%7D%7D%7B%5Ctext%7Ball%20possible%20cases%7D%7D)
![p(\text{A})=\frac{5}{36}](https://tex.z-dn.net/?f=p%28%5Ctext%7BA%7D%29%3D%5Cfrac%7B5%7D%7B36%7D)
Let B be the event when one of the two cards is ![5](https://tex.z-dn.net/?f=5)
probability than one of two cards is
when sum of cards is ![8](https://tex.z-dn.net/?f=8)
![p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}](https://tex.z-dn.net/?f=p%28%5Cfrac%7B%5Ctext%7BB%7D%7D%7B%5Ctext%7BA%7D%7D%29%3D%5Cfrac%7B%5Ctext%7Btotal%20case%20when%20one%20of%20the%20number%20is%205%7D%7D%7B%5Ctext%7Btotal%20case%20when%20sum%20is%208%7D%7D)
![p(\frac{\text{B}}{\text{A}})=\frac{2}{5}](https://tex.z-dn.net/?f=p%28%5Cfrac%7B%5Ctext%7BB%7D%7D%7B%5Ctext%7BA%7D%7D%29%3D%5Cfrac%7B2%7D%7B5%7D)
Now,
probability that sum of cards
is and one of cards is
![p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})](https://tex.z-dn.net/?f=p%28%5Ctext%7BA%20and%20B%7D%3Dp%28%5Ctext%7BA%7D%29%5Ctimes%20p%28%5Cfrac%7B%5Ctext%7BB%7D%7D%7B%5Ctext%7BA%7D%7D%29)
![p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}](https://tex.z-dn.net/?f=p%28%5Ctext%7BA%20and%20B%7D%29%3D%5Cfrac%7B5%7D%7B36%7D%5Ctimes%5Cfrac%7B2%7D%7B5%7D)
![p(\text{A and B})=\frac{1}{18}](https://tex.z-dn.net/?f=p%28%5Ctext%7BA%20and%20B%7D%29%3D%5Cfrac%7B1%7D%7B18%7D)
if sum of cards is
then probability that one of the cards is
is
, option C is correct.