A box contains four tiles, numbered 1,4,5, and 8. Kelly randomly chooses one tile, places it back in the box, then chooses a second tile. What is the probability that the sum of the two chosen tiles is greater than 7?

Given:

A box contains four tiles, numbered 1,4,5, and 8. Kelly randomly chooses one tile, places it back in the box, then chooses a second tile.

To find:

The probability that the sum of the two chosen tiles is greater than 7.

Solution:

A box contains four tiles, numbered 1,4,5, and 8. So, the total possible outcomes are:

S = {(1,1),(1,4),(1,5),(1,8),(4,1),(4,4),(4,5),(4,8),(5,1),(5,4),(5,5),(5,8),(8,1),(8,4),(8,5),(8,8)}

n(S) = 16

A : Sum of the two chosen tiles is greater than 7.

A = {(1,8),(4,4),(4,5),(4,8),(5,4),(5,5),(5,8),(8,1),(8,4),(8,5),(8,8)}

n(A) = 11

So, probability that the sum of the two chosen tiles is greater than 7 is

Therefore, the required probability is .

The probability that the sum of the two chosen tiles is greater than 7 is;

P(sum of the two chosen tiles is greater than 7) = 11/16

We are told that the four tiles are numbered as;

1, 4, 5 and 8.

Now, let us look at all the possible combinations that are greater than 7 and they are;

(1, 8), (8, 1), (4, 5), (4, 4), (5, 4), (5, 5), (4, 8), (8, 4), (5, 8), (8, 5), (8, 8)

Since there are 4 number, then the total number of possible two digit number that can be formed if the first tile is replaced after choosing before a second one is chosen is; 4 × 4 = 16 possible ways

There are 11 possible combinations that are greater than 7.

Thus;

P(sum of the two chosen tiles is greater than 7) = 11/16

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answer is:

1.75 pounds

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2