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?
10 over 40 multiplied by 9 over 39
10 over 40 multiplied by 10 over 39
10 over 40 plus 9 over 39
10 over 40 plus 10 over 39

Answer 1

Answer: 10 over 40 plus 9 over 39

Step-by-step explanation:

Since, the total number of white box = 10,

Red box = 10,

Green box = 10,

And, Blue box = 10,

Thus, the total number of boxes = 10 + 10 + 10 + 10 = 40

The probability of receiving white box by first winner =  White box / Total box

= $\frac{10}{40}$

Since, the replacing is not allowed ( Because, here boxes are allotted )

Thus, the remaining white box = 10 - 1 = 9,

And, the remaining total boxes = 40 - 1 = 39,

Thus, the probability of receiving white box by second winner = Remaining white box / Remaining total box

= $\frac{9}{39}$

Hence, the probability of the first winner receiving a white box and the second winner also receiving a box of the same color

= $\frac{10}{40}\times \frac{9}{39}$

Which is required expression.

Answer 2

Your answer is:

$P\left( E \right) =\frac { 10 }{ 40 } \times \frac { 9 }{ 39 }$

This is because there are 40 boxes in total. The chance of a white box being handed out is 10/40. If a white box is handed out as a prize, there are going to be 39 boxes left, but only 9 of those 39 boxes would be white.

Ideally, tree diagrams should be used to solve such problems.