1. Each of two urns contains green balls and red balls. Urn 1 contains 8 green balls and 12 red balls. Urn 2 contains 5 green balls and 8 red balls. If a ball is drawn from each urn, what is P(red and red)?
a. 79/65
b. 24/65
c. 20/33
d. 2/13
2. 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
Can someone please explain BOTH of these? Thanks!
P (both) found by multiplying them.
1. P(red,red) = 12/20*8/13 = ?
2. P(1,10) = 5/18 * 6/(18-1) = ?
1 is 24/65, just like PsyDAG said you have to do 12/20 * 8/13 and 24/65 which is B would be the answer
Kat seems to be correct
Of course! Let's start by explaining how to solve the first question:
1. Each of two urns contains green balls and red balls. Urn 1 contains 8 green balls and 12 red balls. Urn 2 contains 5 green balls and 8 red balls. If a ball is drawn from each urn, what is P(red and red)?
To find the probability of drawing a red ball from each urn, we need to find the probability of drawing a red ball from Urn 1 and then drawing a red ball from Urn 2. Since the balls are drawn from different urns and not replaced, we use the multiplication rule of probability.
The probability of drawing a red ball from Urn 1 is 12/(12+8) = 12/20 = 3/5.
After drawing a red ball from Urn 1, the probability of drawing a red ball from Urn 2 is 8/(8+5) = 8/13.
To find the probability of both events happening (red and red), we multiply the probabilities together: (3/5) * (8/13) = 24/65.
Therefore, the answer is (b) 24/65.
Now let's explain how to solve the second question:
2. 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)?
To find the probability of selecting a $1 bill, then a $10 bill without replacement, we need to find the probability of each event happening and multiply them together.
The probability of selecting a $1 bill on the first draw is 5/(5+4+6+3) = 5/18.
After selecting a $1 bill, there are now four $1 bills left in the wallet, out of a total of 4+6+3 = 13 remaining bills.
The probability of selecting a $10 bill on the second draw is 6/13.
To find the probability of both events happening ($1, then $10), we multiply the probabilities together: (5/18) * (6/13) = 30/234 = 5/39.
Therefore, the answer is (b) 5/51.