Elizabeth brought a box of donuts to share. There are two-dozen (24) donuts in the box, all identical in size, shape, and color. 6 are jelly-filled, 10 are lemon-filled, and 8 are custard-filled. You randomly select one donut, eat it, and select another donut. Find the probability of selecting two custard-filled donuts in a row. Write your answer as a fraction in simplest form.

The probability of the first selection being custard-filled is 8/24. If a custard-filled donut was selected first, there are 7 custard-filled donuts remaining among a total of 23 donuts. Therefore the probability of the second selection being a custard- filled donut is 7/23. The required probability is therefore:
(8/24) x (7/23) = 7/69

The answer is 7/69

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MrRoyal MrRoyal · Answer 2 · 2021-12-07T04:40:13-05:00

The probability of selecting two custard-filled donuts is: 7/69

The distribution of donuts is given as:

\(\mathbf{Jelly = 6}\)

\(\mathbf{Lemon = 10}\)

\(\mathbf{Custard= 8}\)

\(\mathbf{Total= 24}\)

Eating a selected donut is an illustration of a selection without replacement.

So, the probability of selecting two custard-filled donuts is:

\(\mathbf{P = \frac{Custard}{Total} \times \frac{Custard -1}{Total - 1}}\)

Substitute known values

\(\mathbf{P = \frac{8}{24} \times \frac{8 -1}{24- 1}}\)

Simplify

\(\mathbf{P = \frac{8}{24} \times \frac7{23}}\)

Further, simplify

\(\mathbf{P = \frac{1}{3} \times \frac7{23}}\)

Calculate the product

\(\mathbf{P = \frac7{69}}\)

Hence, the probability is 7/69