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the tree diagram below shows all of the possible outcomes for flipping three coins.

a tree diagram has outcomes (h, h, h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t).

what is the probability of one of the coins landing on tails and two of them landing on heads?
1/4
3/8
1/2
3/4

Probability of one of the coin landing on tails and two of them landing on heads is

Step-by-step explanation:

Given:-

Outcomes = (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)

To find:- Probability of 1 coin landing on tails and 2 heads=?

Solution:-

Outcomes = (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)

    ----------------(1)

Favourable outcomes = (H,H,T),(H,T,H),(T,H,H)

   ------------------(2)

Now using probability formula,

Probability of outcomes = No. of favourable outcomes / Total no. of possible outcomes

Probability of 1 tail and 2 heads =

Probability of 1 tail and 2 heads =

Probability of 1 tail and 2 heads =    ------(from 1 and 2)

Therefore probability of one of the coin landing on tails and two of them landing on heads is

(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

nswer: a   y-axis

step-by-step explanation: