You have five 1$ bills four 5$ bills six 10$ bills and three 20$ bills in your wallet. You select a bill at random. Without replacing the bill. You choose a second bill. What is p($1, then 10$)
The probability of selecting a $1 bill first is 5/18 (since there are 5 $1 bills out of a total of 18 bills). If this bill is not replaced, there will now be 17 bills remaining, with 6 of them being $10 bills. Therefore, the probability of selecting a $10 bill second is 6/17.
Using the multiplication rule of probability, we can multiply the probabilities of each event happening to find the probability of both events happening in sequence:
P($1, then 10$) = P($1$) x P($10$ given $1$ was chosen first)
= (5/18) x (6/17)
= 0.098 or approximately 9.8%.
To find the probability of selecting a $1 bill first and a $10 bill second, we need to calculate the probability of each event happening and then multiply them together.
Step 1: Calculate the probability of selecting a $1 bill first.
We have a total of 5 + 4 + 6 + 3 = 18 bills in our wallet. Out of these, there are 5 $1 bills.
So, the probability of selecting a $1 bill first is 5/18.
Step 2: Calculate the probability of selecting a $10 bill second, without replacing the first bill.
After selecting the first bill, we are left with a total of 18 - 1 = 17 bills in our wallet. Out of these, there are 6 $10 bills.
So, the probability of selecting a $10 bill second is 6/17.
Step 3: Multiply the probabilities calculated in Step 1 and Step 2.
p($1, then $10) = (5/18) * (6/17) = 30/306 = 5/51
Therefore, the probability of selecting a $1 bill first and a $10 bill second is 5/51.