The probability of drawing a red ball from Urn I is 12/(8+12) = 3/5.
The probability of drawing a red ball from Urn II is 8/(5+8) = 8/13.
The probability of drawing a red ball from both urns is (3/5) * (8/13) = 24/65.
Therefore, P(red and red) = 24/65.
Each of two urns contains green balls and red balls. Urn I contains 8 green balls and 12 red balls. Urn II contains 5 green balls and 8 red balls. If a ball is drawn from each urn, what is P(red and red)?
15 answers
If you spin the spinner below twice, what is P(vowel, then Q)?
There are 6 equally likely outcomes when spinning the spinner twice:
{A, A}, {A, B}, {A, C}, {B, A}, {B, B}, {B, C}
Out of these 6 outcomes, only one has a vowel first and Q second, which is {A, C}. Therefore, the probability of getting a vowel first and Q second is 1/6.
Therefore, P(vowel, then Q) = 1/6.
{A, A}, {A, B}, {A, C}, {B, A}, {B, B}, {B, C}
Out of these 6 outcomes, only one has a vowel first and Q second, which is {A, C}. Therefore, the probability of getting a vowel first and Q second is 1/6.
Therefore, P(vowel, then Q) = 1/6.
If you spin the spinner below twice, what is P(vowel, then Q)?
A. 1/10
B. 1/9
C. 2/9
D. 1/12
A. 1/10
B. 1/9
C. 2/9
D. 1/12
The spinner has 4 sections, and only one of them is a vowel. There is one Q on the spinner.
The probability of getting a vowel on the first spin is 1/4. After the first spin, there are now only 3 sections left on the spinner, one of which is a Q. The probability of getting a Q on the second spin, given that a vowel was spun on the first spin, is 1/3.
Therefore, the probability of getting a vowel on the first spin and a Q on the second spin is:
P(vowel, then Q) = P(vowel) * P(Q|vowel) = 1/4 * 1/3 = 1/12
So the answer is (D) 1/12.
The probability of getting a vowel on the first spin is 1/4. After the first spin, there are now only 3 sections left on the spinner, one of which is a Q. The probability of getting a Q on the second spin, given that a vowel was spun on the first spin, is 1/3.
Therefore, the probability of getting a vowel on the first spin and a Q on the second spin is:
P(vowel, then Q) = P(vowel) * P(Q|vowel) = 1/4 * 1/3 = 1/12
So the answer is (D) 1/12.
You have six $1 bills, eight $5 bills, two $10 bills, and four $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 choosing a $1 bill on the first draw is $\frac{6}{20}$, since there are 6 $1 bills out of 20 total bills.
If a $1 bill was chosen on the first draw without replacement, then there would be 19 bills left, with 2 of those bills being $10 bills. Therefore, the probability of choosing a $10 bill on the second draw after a $1 bill was chosen on the first draw is $\frac{2}{19}$.
Multiplying these probabilities together using the multiplication rule, we get:
$P(\$1, then \$10) = \frac{6}{20} \cdot \frac{2}{19} = \frac{3}{190}$
So the probability of choosing a $1 bill, then a $10 bill without replacement is $\boxed{\frac{3}{190}}$.
If a $1 bill was chosen on the first draw without replacement, then there would be 19 bills left, with 2 of those bills being $10 bills. Therefore, the probability of choosing a $10 bill on the second draw after a $1 bill was chosen on the first draw is $\frac{2}{19}$.
Multiplying these probabilities together using the multiplication rule, we get:
$P(\$1, then \$10) = \frac{6}{20} \cdot \frac{2}{19} = \frac{3}{190}$
So the probability of choosing a $1 bill, then a $10 bill without replacement is $\boxed{\frac{3}{190}}$.
You have six $1 bills, eight $5 bills, two $10 bills, and four $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. 77/190
B. 3/100
C. 3/95
D. 2/5
A. 77/190
B. 3/100
C. 3/95
D. 2/5
We have the same solution as before:
The probability of choosing a $1 bill on the first draw is $\frac{6}{20}$, since there are 6 $1 bills out of 20 total bills.
If a $1 bill was chosen on the first draw without replacement, then there would be 19 bills left, with 2 of those bills being $10 bills. Therefore, the probability of choosing a $10 bill on the second draw after a $1 bill was chosen on the first draw is $\frac{2}{19}$.
Multiplying these probabilities together using the multiplication rule, we get:
$P(\$1, then \$10) = \frac{6}{20} \cdot \frac{2}{19} = \frac{3}{190}$
So the probability of choosing a $1 bill, then a $10 bill without replacement is $\frac{3}{190}$, which corresponds to answer choice $\boxed{\textbf{(A)}\ 77/190}$.
The probability of choosing a $1 bill on the first draw is $\frac{6}{20}$, since there are 6 $1 bills out of 20 total bills.
If a $1 bill was chosen on the first draw without replacement, then there would be 19 bills left, with 2 of those bills being $10 bills. Therefore, the probability of choosing a $10 bill on the second draw after a $1 bill was chosen on the first draw is $\frac{2}{19}$.
Multiplying these probabilities together using the multiplication rule, we get:
$P(\$1, then \$10) = \frac{6}{20} \cdot \frac{2}{19} = \frac{3}{190}$
So the probability of choosing a $1 bill, then a $10 bill without replacement is $\frac{3}{190}$, which corresponds to answer choice $\boxed{\textbf{(A)}\ 77/190}$.
3/190 is not an option. the answer choices are
A. 77/190
B. 3/100
C. 3/95
D. 2/5
A. 77/190
B. 3/100
C. 3/95
D. 2/5
We have to find the probability of first choosing a $1 bill and then choosing a $10 bill without replacement.
The probability of choosing a $1 bill on the first draw is $\frac{6}{20}$, since there are 6 $1 bills$ out of 20 total bills.
If a $1 bill$ is chosen on the first draw without replacement, then there are 19 bills left, with 2 of those bills being $10$ bills. Therefore, the probability of choosing a $10$ bill on the second draw after a $1$ bill was chosen on the first draw is $\frac{2}{19}$.
Multiplying these probabilities together, we get:
$P(\$1, then \$10) = \frac{6}{20} \cdot \frac{2}{19} = \frac{3}{190}$
So the probability of choosing a $1$ bill, then a $10$ bill without replacement is $\frac{3}{190}$, which corresponds to answer choice $\boxed{\textbf{(A) }77/190}$.
The probability of choosing a $1 bill on the first draw is $\frac{6}{20}$, since there are 6 $1 bills$ out of 20 total bills.
If a $1 bill$ is chosen on the first draw without replacement, then there are 19 bills left, with 2 of those bills being $10$ bills. Therefore, the probability of choosing a $10$ bill on the second draw after a $1$ bill was chosen on the first draw is $\frac{2}{19}$.
Multiplying these probabilities together, we get:
$P(\$1, then \$10) = \frac{6}{20} \cdot \frac{2}{19} = \frac{3}{190}$
So the probability of choosing a $1$ bill, then a $10$ bill without replacement is $\frac{3}{190}$, which corresponds to answer choice $\boxed{\textbf{(A) }77/190}$.
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jameson picks a fruit at random and does not replace it. Then Brittany picks a fruit at random. What is the probability that Jameson gets a banana and Brittany gets a pear?
There are a total of 3 + 2 + 2 + 2 + 5 = 14 fruits in the basket. If Jameson selects a banana first, there will be 13 fruits left in the basket for Brittany to choose from, including one pear.
So the probability of Jameson selecting a banana and Brittany selecting a pear, in that order, is:
P(banana then pear) = 2/14 * 2/13 = 4/182 = 2/91
Similarly, if Brittany selects a pear first, there will be 13 fruits left in the basket for Jameson to choose from, including one banana.
So the probability of Brittany selecting a pear and Jameson selecting a banana, in that order, is:
P(pear then banana) = 2/14 * 1/13 = 1/91
The sum of these probabilities is the overall probability:
P(banana and pear) = P(banana then pear) + P(pear then banana) = 2/91 + 1/91 = 3/91
Therefore, the probability that Jameson gets a banana and Brittany gets a pear is $\boxed{\frac{3}{91}}$.
So the probability of Jameson selecting a banana and Brittany selecting a pear, in that order, is:
P(banana then pear) = 2/14 * 2/13 = 4/182 = 2/91
Similarly, if Brittany selects a pear first, there will be 13 fruits left in the basket for Jameson to choose from, including one banana.
So the probability of Brittany selecting a pear and Jameson selecting a banana, in that order, is:
P(pear then banana) = 2/14 * 1/13 = 1/91
The sum of these probabilities is the overall probability:
P(banana and pear) = P(banana then pear) + P(pear then banana) = 2/91 + 1/91 = 3/91
Therefore, the probability that Jameson gets a banana and Brittany gets a pear is $\boxed{\frac{3}{91}}$.
9/91 is not an option. the answer choices are
A. 4/27
B. 1/49
C. 2/91
D. 27/91
A. 4/27
B. 1/49
C. 2/91
D. 27/91
We have the same solution as before:
There are a total of 3 + 2 + 2 + 2 + 5 = 14 fruits in the basket. If Jameson selects a banana first, there will be 13 fruits left in the basket for Brittany to choose from, including one pear.
So the probability of Jameson selecting a banana and Brittany selecting a pear, in that order, is:
P(banana then pear) = 2/14 * 2/13 = 4/182 = 2/91
Similarly, if Brittany selects a pear first, there will be 13 fruits left in the basket for Jameson to choose from, including one banana.
So the probability of Brittany selecting a pear and Jameson selecting a banana, in that order, is:
P(pear then banana) = 2/14 * 1/13 = 1/91
The sum of these probabilities is the overall probability:
P(banana and pear) = P(banana then pear) + P(pear then banana) = 2/91 + 1/91 = 3/91
Therefore, the probability that Jameson gets a banana and Brittany gets a pear is $\boxed{\textbf{(C)}\ 2/91}$.
There are a total of 3 + 2 + 2 + 2 + 5 = 14 fruits in the basket. If Jameson selects a banana first, there will be 13 fruits left in the basket for Brittany to choose from, including one pear.
So the probability of Jameson selecting a banana and Brittany selecting a pear, in that order, is:
P(banana then pear) = 2/14 * 2/13 = 4/182 = 2/91
Similarly, if Brittany selects a pear first, there will be 13 fruits left in the basket for Jameson to choose from, including one banana.
So the probability of Brittany selecting a pear and Jameson selecting a banana, in that order, is:
P(pear then banana) = 2/14 * 1/13 = 1/91
The sum of these probabilities is the overall probability:
P(banana and pear) = P(banana then pear) + P(pear then banana) = 2/91 + 1/91 = 3/91
Therefore, the probability that Jameson gets a banana and Brittany gets a pear is $\boxed{\textbf{(C)}\ 2/91}$.