Apack of confectioneries contains candies in four (4) flavors (berry, lemon, orange and cacao), 3 pieces each. candies are distributed randomly in the pack. you draw the first three (3) pieces from the pack. what is the probability that they are all different flavors?

The probability is 0.4909

Step-by-step explanation:

The following equation for nCk give as the number of ways in which we can select k elements from a group of n elements:

Then, there are 220 ways in which we can select 3 candies from the 12 that are in the pack. It is calculated as:

On the other hand, there are 108 different ways to select the 3 candies in which they are all different flavors. It is calculated as:

4C3 * 3C1 * 3C1 * 3C1 = 108

Because, 4C3 give us the number of ways to select 3 flavors from the 4 flavors. From this 3 flavors selected, we are going to select one candie from each one, so we multiply 3 times by 3C1, one for each flavor.

Finally, the probability is the division between the number of ways in which we can select 3 candies with different flavors and the total number of ways in which we can select 3 candies from the 12 in the pack. This is:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

its c or d   take a lucky guess my dude