Afair die is cast four times. calculate the probability of obtaining exactly two 6's

25/216 or 0.1157

Step-by-step explanation:

The probability of getting a 6 is 1/6.  

The probability of not getting a 6 is 5/6

We roll a die 4 times and want 2 sixes, and 2 not sixes.  

We want a combination of 2 sixes from 4 rolls, so we can use the binomial formula to solve this...

(4C2)(1/6)²(5/6)² = 6(1/36)(25/36) = (1/6)(25/36) = 25/216 = 0.1157

The long hand way of solving this is to list the 6 ways we can roll our desired result along with their probabilities...

We could get

6, 6, not 6, not 6      (1/6)(1/6)(5/6)(5/6) = 25/1296

6, not 6, 6, not 6       (1/6)(5/6)(1/6)(5/6) = 25/1296

6, not 6, not 6, 6       (1/6)(5/6)(5/6)(1/6) = 25/1296

not 6, not 6, 6, 6       (5/6)(5/6)(1/6)(1/6) = 25/1296

not 6, 6, not 6, 6       (5/6)(1/6)(5/6)(1/6) = 25/1296

not 6, 6, 6 not 6        (5/6)(1/6)(1/6)(5/6) = 25/1296

There are 6 different way this can happen, each having a probability of 25/196.  We add up the probabilities, getting 150/1296, which reduces to

25/216

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x