A spinner labeled with the letters of the word HONEST is to be spun once. Describe the probability of landing on an unshaded section in the sample space.
the spinner has 3 shades and 3 unshaded
Therefore, the probability of landing on an unshaded section is 3/6 or 1/2.
a basket of fruit contains 5 bananas, 8 mangoes, and 7 apples. Two fruits are randomly selected without replacement. Calculate the probability of selecting an apple and a banana
First, let's find the probability of selecting an apple on the first try:
P(apple on first try) = 7/20
Since we are not replacing the fruit, there are now 19 fruits left, including 5 bananas.
So the probability of selecting a banana on the second try, given that we already selected an apple on the first try, is:
P(banana on second try) = 5/19
To find the probability of both events happening (selecting an apple and a banana), we multiply these probabilities:
P(apple and banana) = P(apple on first try) x P(banana on second try)
P(apple and banana) = (7/20) x (5/19)
P(apple and banana) = 35/380
Simplifying the fraction, we get:
P(apple and banana) = 7/76
Therefore, the probability of selecting an apple and a banana is 7/76.
A basket of fruits contains 10 bananas, 5 mangoes, and 5 apples. A fruit is selected at random. Calculate the probability of selecting an apple or a banana. Percentage form
The probability of selecting an apple or a banana can be found by adding the probability of selecting an apple to the probability of selecting a banana, since these events are mutually exclusive (i.e. we cannot select both at the same time).
P(apple or banana) = P(apple) + P(banana)
= 5/20 + 10/20
= 15/20
= 0.75
This can be simplified to 75% in percentage form.
Therefore, the probability of selecting an apple or a banana is 75%, or 0.75 as a decimal.
a basket of fruits contains 5 apples and 3 pears. Sharon took two fruit at random. What is the probability both fruit are apple. Simple fraction form
The total number of fruits in the basket is 5 + 3 = 8.
Since Sharon takes two fruits at random, the total number of ways she can do so is:
8C2 = (8 x 7) / (2 x 1) = 28
This is the total number of possible outcomes in the sample space.
The number of ways Sharon can select two apples out of the five is:
5C2 = (5 x 4) / (2 x 1) = 10
Therefore, the probability of selecting two apples out of the basket is:
P(2 apples) = 10/28
Simplifying the fraction, we get:
P(2 apples) = 5/14
Therefore, the probability of Sharon selecting both fruit as apples is 5/14.
A coin is flipped and a spinner is spun simultaneously. What is the probability of flipping a tail and landing on 5
6 sections spinner
The probability of flipping a tail is 1/2, since there are two possible outcomes (heads or tails) and both are equally likely.
The probability of landing on 5 in a 6-section spinner is 1/6, since there is one section labeled 5 out of a total of six sections.
Since the coin flip and spinner spin are independent events, we can multiply their probabilities to find the probability of both events happening simultaneously:
P(tail and 5) = P(tail) x P(landing on 5)
P(tail and 5) = (1/2) x (1/6)
P(tail and 5) = 1/12
Therefore, the probability of flipping a tail and landing on 5 on the spinner simultaneously is 1/12.
To find the probability of landing on an unshaded section of the spinner, we first need to determine the total number of possible outcomes. In this case, the spinner has 6 sections, each labeled with a letter from the word "HONEST."
Out of the 6 sections, there are 3 sections that are shaded and 3 sections that are unshaded.
Therefore, the probability of landing on an unshaded section is:
Number of favorable outcomes (unshaded sections) / Total number of possible outcomes
= 3 (unshaded sections) / 6 (total sections)
Simplifying the fraction, we get:
1/2
Therefore, the probability of landing on an unshaded section is 1/2 or 50%.