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

idk

step-by-step explanation:

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2