Answer: [tex]\dfrac{20}{171}[/tex]
Step-by-step explanation:
Given : Number of nickels in the bag = 10
Number of dimes in the bag = 4
The total number of coins in the bag =[tex]10+4+5=19[/tex]
The probability of drawing nickel :-
[tex]\text{P(nickel)}=\dfrac{10}{19}[/tex]
Total number coins left = 18
The conditional probability of drawing dime, given that a nickel is already drawn :-
[tex]\text{P(dime}|\text{nickel)}=\dfrac{4}{18}[/tex]
Now, the probability you withdrew a nickel and then a dime is given by :-
[tex]\text{P(dime and nickel)}=\text{P(dime}|\text{nickel)}\times\text{P(nickel)}\\\\=\dfrac{10}{19}\times\dfrac{4}{18}=\dfrac{20}{171}[/tex]
Now, the probability you withdrew a nickel and then a dime [tex]}=\dfrac{20}{171}[/tex]
