En una caja de muchos bombones hay hasta tres sabores de ellos. ¿Cuántos se deben extraer como mínimo al azar para obtener con seguridad 5 bombones de un mismo sabor?

At least 13 chocolates must be removed

Step-by-step explanation:

If there are three flavors, the probability of drawing 1 would be: 1

1/3 = 0.333

Which means, that every 3 attempts, theoretically you should do 1 of each, but how they ask for 5 chocolates of each would be:

3 * 5 = 15

At least 15 chocolates must be extracted to theoretically guarantee 5 chocolates each, but how we are interested in knowing is that a single flavor has 5 chocolates, so we discard the last two chocolates that represent the other two flavors

Therefore, for there to be safely 5 chocolates of the same flavor, at least 13 chocolates must be removed.

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).

ok then what happened.