Suppose that distinct integer values are written on each of 3 cards. These cards arethen randomly given the designations A, B, and C. The values on cards A and B arethen compared. If the smaller of these values is then compared with the value on cardC, what is the probability that it is also smaller than the value on card C?

P[min(A,B) < C] = (2/3)

Step-by-step explanation:

We will list the possible outcomes of assigning three distinct integers to cards A, B and C.

In our possible outcomes, we will rightly assume that the cards are arranged in the order of increasing value. That is, the first number on the first card is the least, the next one is higher than the first number but still less than the third and the third number on the third card is the highest.

The possible outcomes include

(A, B, C)

(here, A has the least integer, followed by the integer on B, then on C)

(A, C, B)

(here, A has the least integer, followed by the integer on C, then on B)

The other possible outcomes

(B, A, C)

(B, C, A)

(C, A, B)

(C, B, A)

The scope of the possible outcomes is now clear.

We require the probability that the smaller integer of A and B is also the smaller integer when compared to the integer on C.

P[min(A,B) < C]

Considering each of the outcomes one, at a time,

- [In ABC, the smaller of A & B is A, and it is smaller than C]

- [In ACB, the smaller of A & B is A, and it is still smaller than C]

- [In BAC, the smaller of A & B is B, and it is smaller than C]

- [In BCA, the smaller of A & B is A, and it is smaller than C]

- [In CAB, the smaller of A & B is A, but it isn't smaller than C]

- [In CBA, the smaller of A & B is B, but it isn't smaller than C]

Mathematically, P[min(A,B) < C]

The outcomes in which, the smaller integer of A and B are also smaller than C include

(A,B,C), (A,C,B), (B,A,C), (B,C,A)

The required probability,

P[min(A,B) < C] = (4/6) = (2/3)

Hope this Helps!!!

what is all that apply?

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3