The total number of fruits in the basket is 5 apples + 3 pears = 8 fruits.
The probability of choosing an apple on the first pick is 5 apples / 8 total fruits = 5/8.
After taking one fruit, there are 7 fruits remaining in the basket (4 apples and 3 pears).
The probability of choosing an apple on the second pick is 4 apples / 7 total fruits = 4/7.
Therefore, the probability of choosing both apples is (5/8) * (4/7) = 20/56 = 5/14.
So, the probability of both fruits being apples is 5/14.
A basket of fruits contains 5 apples and 3 pears. Sharon took two fruits at random. What is the probability that both fruits are apples? Write you answer in the simplest form of a fraction.
11 answers
A coin flipped and a spinner is spun simultaneously. The spinner is divided into six equally sized sections labeled with 1, 2, 3, 4, 5, 6,. What is the probability of flipping a tail and landing on 5?
The probability of flipping a tail is 1/2 because there are two equally likely outcomes (heads or tails) when flipping a coin.
The spinner has 6 equally likely outcomes: 1, 2, 3, 4, 5, 6.
The probability of landing on 5 when spinning the spinner is 1/6 because there is only one section labeled 5 out of 6 sections.
To find the probability of flipping a tail and landing on 5 simultaneously, we multiply the individual probabilities:
P(Tail and 5) = P(Tail) * P(5) = (1/2) * (1/6) = 1/12.
Therefore, the probability of flipping a tail and landing on 5 is 1/12.
The spinner has 6 equally likely outcomes: 1, 2, 3, 4, 5, 6.
The probability of landing on 5 when spinning the spinner is 1/6 because there is only one section labeled 5 out of 6 sections.
To find the probability of flipping a tail and landing on 5 simultaneously, we multiply the individual probabilities:
P(Tail and 5) = P(Tail) * P(5) = (1/2) * (1/6) = 1/12.
Therefore, the probability of flipping a tail and landing on 5 is 1/12.
If two events are independent, the probability of both events occurring is the product of their Individual probability. Events A and B are independent with P(A) = 0.3 and P(B)=0.25. what is the probability of events A and B happening at the same time? Write your answer in decimal form.
Since events A and B are independent, the probability of both events occurring at the same time is the product of their individual probabilities.
P(A and B) = P(A) * P(B) = 0.3 * 0.25 = 0.075.
Therefore, the probability of events A and B happening at the same time is 0.075.
P(A and B) = P(A) * P(B) = 0.3 * 0.25 = 0.075.
Therefore, the probability of events A and B happening at the same time is 0.075.
Girls Boys Row total
Want the option to keep studying online 10 3 13
Don't want the option to keep studying 5 7 12
Column total 15 10 25
What is the probability of selecting a student who want the option to keep studying online? Write your answer in percent form.
Want the option to keep studying online 10 3 13
Don't want the option to keep studying 5 7 12
Column total 15 10 25
What is the probability of selecting a student who want the option to keep studying online? Write your answer in percent form.
The total number of students who want the option to keep studying online is 13.
The total number of students in the school is 25.
The probability of selecting a student who wants the option to keep studying online is the number of students who want the option divided by the total number of students in the school.
So, P(selecting a student who wants the option to keep studying online) = 13/25
Converting this fraction to a percentage:
(13/25) * 100% = 52%
Therefore, the probability of selecting a student who wants the option to keep studying online is 52%.
The total number of students in the school is 25.
The probability of selecting a student who wants the option to keep studying online is the number of students who want the option divided by the total number of students in the school.
So, P(selecting a student who wants the option to keep studying online) = 13/25
Converting this fraction to a percentage:
(13/25) * 100% = 52%
Therefore, the probability of selecting a student who wants the option to keep studying online is 52%.
A money envelope contains five $10 bills, ten $20 bills, and fifteen $50 bills. Sandra randomly selects two bills without returning the first bill. what is the probability of getting $10 bill then a $50 bill? Write your answer in the simplest form of a fraction.
First, let's calculate the total number of bills in the envelope:
Total number of bills = 5 (10-dollar bills) + 10 (20-dollar bills) + 15 (50-dollar bills) = 30 bills
The probability of Sandra selecting a $10 bill on her first pick is 5/30 = 1/6.
After selecting one bill without returning it, there are now 29 bills left, with 4 $10 bills and 15 $50 bills.
The probability of selecting a $50 bill on the second pick is 15/29.
Therefore, the overall probability of getting a $10 bill first and then a $50 bill is:
P($10 then $50) = P($10) * P($50) = (1/6) * (15/29) = 15/174
So, the probability of getting a $10 bill first and then a $50 bill is 15/174.
Total number of bills = 5 (10-dollar bills) + 10 (20-dollar bills) + 15 (50-dollar bills) = 30 bills
The probability of Sandra selecting a $10 bill on her first pick is 5/30 = 1/6.
After selecting one bill without returning it, there are now 29 bills left, with 4 $10 bills and 15 $50 bills.
The probability of selecting a $50 bill on the second pick is 15/29.
Therefore, the overall probability of getting a $10 bill first and then a $50 bill is:
P($10 then $50) = P($10) * P($50) = (1/6) * (15/29) = 15/174
So, the probability of getting a $10 bill first and then a $50 bill is 15/174.
the real answer is 5/58 ._.
I apologize for the error in the previous calculation. Let's recalculate the probability.
The probability of selecting a $10 bill on the first pick is 5/30 = 1/6.
After selecting one bill without returning it, there are now 29 bills left, 15 of which are $50 bills.
The probability of selecting a $50 bill on the second pick is 15/29.
Therefore, the overall probability of getting a $10 bill first and then a $50 bill is:
P($10 then $50) = P($10) * P($50) = (1/6) * (15/29) = 15/174.
Converting 15/174 to the simplest form of a fraction, we have:
15/174 = 5/58.
So, the correct probability of getting a $10 bill first and then a $50 bill is 5/58. Thank you for pointing out the error.
The probability of selecting a $10 bill on the first pick is 5/30 = 1/6.
After selecting one bill without returning it, there are now 29 bills left, 15 of which are $50 bills.
The probability of selecting a $50 bill on the second pick is 15/29.
Therefore, the overall probability of getting a $10 bill first and then a $50 bill is:
P($10 then $50) = P($10) * P($50) = (1/6) * (15/29) = 15/174.
Converting 15/174 to the simplest form of a fraction, we have:
15/174 = 5/58.
So, the correct probability of getting a $10 bill first and then a $50 bill is 5/58. Thank you for pointing out the error.