Will wanted to track the growth of various fruits in his garden, so he decided to label them. His garden had APPLES labeled 1,2,3,4,5,6, LEMONS labeled 1,2, and MELONS labeled 1,2,3. If a single fruit is picked at random, what is the probability that the fruit is an APPLE or has an EVEN number?

Between 6 apples. 2 lemons and 3 melons, Will has a total of 11 fruits in his garden. There are three apples labeled with even numbers (2,4 and 6), one melon (2) and one lemon (2), for a total of five fruits.

The probability that a randomly selected fruit is an APPLE or has an EVEN number is given by the probability that it is an apple, P(A), added to the probability that it is even, P(E), minus the probability that it is an even apple, P(A and E):

[tex]P(A\ or\ E) = P(A) +P(E) - P(A\ and\ E)\\P(A\ or\ E) = \frac{6}{11}+\frac{5}{11}-\frac{3}{11} \\P(A\ or\ E) = 0.7272[/tex]

There is a 8 in 11, or a 0.7272 chance that the fruit is an APPLE or has an EVEN number.

joaobezerra joaobezerra · Answer 2 · 2021-11-23T12:04:00+01:00

Using the probability concept, it is found that there is a 0.7273 = 72.73% probability that the fruit is an APPLE or has an EVEN number.

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem:

There are 11 options, hence [tex]T = 11[/tex].
Of those, there are 6 options which land on Apple, plus 2 options which does not land on apple but have an even number, hence [tex]D = 6 + 2 = 8[/tex]

Then, the probability is:

[tex]p = \frac{D}{T} = \frac{8}{11} = 0.7273[/tex]

0.7273 = 72.73% probability that the fruit is an APPLE or has an EVEN number.

A similar problem involving the probability concept is given at https://brainly.com/question/15536019