Question

At a game booth, a student gets a box of candy as the prize for winning a game. The boxes come in four colors: white, red, green, and blue. There are 10 boxes of each color. All the boxes are equally likely to be given away as prizes. Which expression shows the probability of the first winner receiving a white box and the second winner also receiving a box of the same color?

Answer 1

Answer:

First Winner Getting white box = $\frac{10}{40}=\frac{1}{4}$

Second Winner Getting white box = $\frac{9}{39}=\frac{3}{13}$

Both getting the white box = $\frac{3}{52}$

Step-by-step explanation:

Number of available colors = 4

Number of boxes of each color = 10

So total boxes available = 40

Number of white boxes = 10

Probability is defined as the ratio of favorable outcomes to total possible outcomes. In this case, getting a white box is favorable outcome, so number of favorable outcomes is 10 and total possible boxes are 40. So,

Probability that first winner will receive a white box = $\frac{10}{40}=\frac{1}{4}$

After the first winner has been given his box, there are 39 total boxes remaining with 9 white boxes. So,

Probability that second winner will receive a white box =$\frac{9}{39}=\frac{3}{13}$

The probability that both the winners will receive the white boxes will be the product of their individual probabilities i.e.

Probability that the first winner receiving a white box and the second winner also receiving a box of the same color = $\frac{10}{40} \times \frac{9}{39}=\frac{3}{52}$

Answer 2

Step-by-step explanation:

3/52

Answer:

The answer is 52  because the fundamental theorem of calculus states The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives (also called indefinite integral), say F, of some function f may be obtained as the integral of f with a variable bound of integration. Following that you get the answer 3/52 to the nearest thousandths, remember to follow green's theorem while solving.

∛≅ΔΔΔr⇆=21-3∛21×3÷33±8

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