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)?

A. 11/35
B. 5/51
C. 5/54
D. 193/306

Question

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)?

A. 11/35
B. 5/51
C. 5/54
D. 193/306

kendricklamar2014 · Answer

@kropot72

kendricklamar2014 · Answer

@Maria195

anonymous · Answer

Don't worry mah friend. Let me check in my previous lessons.

anonymous · Answer

Aww my numbers are different than yours X(

kendricklamar2014 · Answer

:(

kropot72 · Answer

The probability of choosing a $1 bill first is 5/18. the probability of choosing a $10 bill next is 6/17.
The required probability is:
$$P($1\ then\ $10)=\frac{5\times6}{18\times17}=you\ can\ calculate$$

kendricklamar2014 · Answer

5*6=30
18*17=306

anonymous · Answer

I think it's B.  The probabilities change every time a bill is pulled out. The probability of pulling out a 1 first is 5 out of 18 which is .227 then the probability of pulling out a 10 is 6 out of 17 because you don't replace the bill. since the events are independent of each other, you get .35, then you multiply them to get .098 which in fraction form is answer B.

kropot72 · Answer

$$\frac{30}{306}=\frac{5}{51}$$

kendricklamar2014 · Answer

Thanks @kropot72

kropot72 · Answer

Dividing the numerator an the denominator of the left hand fraction by 6.

kropot72 · Answer

You're welcome :)