If a fair die is rolled twice and two numbers X1=result of the first roll and X2=result of the second roll are obtained. Given that X1+X2=7, what is the probability that X1 =4 or X2 =4

P = 2/3 = 0.67

Step-by-step explanation:

Assuming that X1 + X2 = 7

The total number of outcomes such that X1 + X2 = 7 are:

X1 = 3, X2 = 4

X1 = 4, X2 = 3

X1 = 1, X2 = 6

X1 = 6, X2 = 1

X1 = 2, X2 = 5

X1 = 5, X2 = 2

So there are 6 total outcomes such that X1 + X2 = 7

Now, the probability that X1 = 4 is equal to the quotient between the number of outcomes where X1 = 4 (only one outcome) and the total number of outcomes (6)

Then:

p = 1/6

And the probability that X2 = 4 is calculated in the same way, then the probability is:

q = 1/6

Now, the probability that X1 = 4 or X2 = 4 is the sum of both probabilities we've found above, then:

P = p + q = 1/6 + 1/6 = 2/6 = 2/3 = 0.67

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12